The probability that a filled bag will weigh less than 31.7 kilograms is 0.3674 (approx) when rounded to four decimal places.
Given that mean of filled bag is 32 kg and the standard deviation is 0.89 kg.
To find the probability that a filled bag will weigh less than 31.7 kg, we need to standardize the variable and find the area under the standard normal curve.The standard normal distribution has mean 0 and standard deviation 1.
Let X be the weight of the filled bag.Then,
Z = (X - μ) / σ
= (31.7 - 32) / 0.89
= -0.337079
Possible ways of rounding depends on the instructions provided to the students. Here, rounding to 4 decimal places is necessary because it has been explicitly mentioned in the question.
Now, we need to find the probability that a filled bag will weigh less than 31.7 kilograms, which is same as finding the probability that Z is less than -0.337079 from the standard normal distribution table using the standard normal distribution.p(Z < -0.337079) = 0.3674 (approx)
Therefore, the probability that a filled bag will weigh less than 31.7 kilograms is 0.3674 (approx) when rounded to four decimal places.Note: The question is asking to round the answer to four decimal places, which makes it important to give the answer as 0.3674 (approx).
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If two dice are rolled, find the probability of getting at least one 5 or a sum of 10.
The probability of getting at least one 5 or a sum of 10 when two dice are rolled is 13/36.
When two dice are rolled, the possible outcomes are 6*6 = 36.
The sample space, in this case, is 36.
Now, we can calculate the probability of getting at least one 5 or a sum of 10 by using the main answer.
In this case, the number of events that we need to count is:
Getting at least one 5: (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5). The total is 11.Sum of 10: (4, 6), (5, 5), (6, 4).
The total is 3.
There is one outcome that is common between these two sets: (5, 5). So, we need to subtract that from the total to avoid double-counting i
t.The probability of getting at least one 5 or a sum of 10 is:P(at least one 5 or sum of 10) = P(at least one 5) + P(sum of 10) - P(5, 5)= (11 + 3 - 1) / 36= 13 / 36.
Therefore, the probability of getting at least one 5 or a sum of 10 is 13/36.
The probability of getting at least one 5 or a sum of 10 when two dice are rolled is 13/36.
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A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.45, P(D) = 0.50 and P(DJB) = 0.30. a. Find P(BND). (2) b. Find P(B U D). (2) c. What is the probability that a student does not select a book nor a DVD?
P(BND) = 0.65, P(B U D) = 0.65 and the probability that a student does not select a book nor a DVD is 0.35.
a.P(BND) = 0.3015
b.P(B U D) = 0.65
To find the probability of both events B and D occurring, we use the formula P(BND) = P(B) * P(D|B). Given that P(B) = 0.45 and P(D|B) = 0.67 (which is the probability of D given that B has occurred), we can calculate P(BND) as follows:
P(BND) = 0.45 * 0.67 = 0.3015.
b.P(B U D) = 0.65
To find the probability of either event B or event D or both occurring, we use the formula P(B U D) = P(B) + P(D) - P(BND). Given that P(B) = 0.45, P(D) = 0.50, and we already calculated P(BND) as 0.3015, we can calculate P(B U D) as follows:
P(B U D) = 0.45 + 0.50 - 0.3015 = 0.6485 ≈ 0.65.
P(not selecting a book nor a DVD) = 0.05
The probability of not selecting a book nor a DVD is the complement of selecting either a book or a DVD or both. Since P(B U D) represents the probability of selecting either a book or a DVD or both, the complement of P(B U D) will give us the probability of not selecting a book nor a DVD:
P(not selecting a book nor a DVD) = 1 - P(B U D) = 1 - 0.65 = 0.35 = 0.05.
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: Let A be the matrix below and define a transformation T:R³ R³ by T(u) = Au. For the vector b below, find a vector u such that I maps u to b, if possible. Otherwise state that there is no such u 3 -9 9 A 1 -3 8 -1 3 -2 T(u) = b for the following u: u = 0 b = -21 9
To find a vector u such that the transformation T maps u to the given vector b, we need to solve the equation T(u) = b, where T is defined by T(u) = Au. The matrix A represents the transformation.
In this case, the given matrix A and vector b are provided, and we need to determine if there exists a vector u that satisfies the equation T(u) = b.
To find u, we need to solve the equation Au = b. This can be done by multiplying the inverse of A to both sides of the equation: u = A^(-1)b. However, for this to be possible, the matrix A must be invertible.
To determine if A is invertible, we can calculate its determinant. If the determinant is non-zero, then A is invertible, and there exists a vector u that maps to b. Otherwise, if the determinant is zero, A is not invertible, and no such vector u exists.
Calculating the determinant of A, we have:
det(A) = (3 * (-2) * 3) + (-9 * 8 * (-1)) + (9 * (-3) * 1) - (9 * (-2) * (-1)) - (3 * 3 * 9) - (3 * (-1) * (-3))
= -54 + 72 - 27 + 18 - 81 + 27
= -45
Since the determinant is non-zero (-45 ≠ 0), the matrix A is invertible. Therefore, there exists a vector u that maps to the given vector b. To find u, we can compute u = A^(-1)b. However, further calculations are required to determine the specific vector u.
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Suppose that 50% of all college students smoke cigarettes. A
sample of 10 is selected randomly. What is the probability that
exactly one student smokes? Round your answer to four decimal
places.
Given that 50% of all college students smoke cigarettes, the probability of a randomly selected student smoking is P(Smoker) = 0.5 and the probability of a randomly selected student not smoking is P(Non-smoker) = 0.5.
We are to find the probability of exactly one student smoking out of a sample of 10 students. This can be calculated using the binomial probability formula which is:P(X = k) = (nCk) * p^k * (1 - p)^(n - k)where n is the sample size, k is the number of successes, p is the probability of success, and (1 - p) is the probability of failure. In this case, n = 10, k = 1, p = 0.5, and (1 - p) = 0.5.
Substituting the values, we get:P(X = 1) = (10C1) * 0.5^1 * 0.5^(10 - 1)= 10 * 0.5 * 0.0009765625= 0.0048828125Rounding off to four decimal places, the probability of exactly one student smoking out of a sample of 10 students is 0.0049.Hence, the required probability is 0.0049, when exactly one student smokes out of a sample of 10 students.
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A study was conducted measuring the average number of apples collected from two varieties of trees. Apples were collected from 61 trees of type A and 50 trees of type B. Researchers are interested in knowing whether trees of the recently developed type A variety produces more apples on average than type B. A permutation test was performed to try and answer the question.
Suppose 1300 arrangements of the data set were sampled and 6 arrangments were found to have a difference between the two group means greater than what was actually observed. What is the p value of the permutation test?
The p-value of the permutation test is calculated as 6/1300 = 0.0046. Since the p-value is less than the conventional significance level (e.g., 0.05), we would conclude that the recently developed type A variety produces more apples on average than type B.
To calculate the p-value of the permutation test, follow these steps:
Determine the observed difference between the means of the two groups based on the actual data.
Generate many random permutations of the data, where the group labels are randomly assigned.
For each permutation, calculate the difference between the means of the two groups.
Count the number of permutations that have a difference between the means greater than or equal to the observed difference.
Divide the count from step 4 by the total number of permutations (1300 in this case) to obtain the p-value.
In this scenario, 6 out of the 1300 permutations had a difference between the means greater than what was observed.
Therefore, the p-value of of the permutation test is calculated as 6/1300 = 0.0046. Since the p-value is less than the conventional significance level (e.g., 0.05), we would conclude that the recently developed type A variety produces more apples on average than type B.
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1. Using the polygon table as a reference, list all of the polygons that have interior angles that are whole numbers (a number that is not a fraction or a decimal)? Explain why it is that way.
The polygons listed that have interior angles that are whole numbers are:
Polygon a. Convex 15-gon, Yes, since 15 divides evenly into 180.
Polygon d. Convex 18-gon, Yes, since 18 divides evenly into 180.
Polygon h. Convex 45-gon, Yes, since 45 divides evenly into 180.
How to find the Interior angles of the Polygon?The sum of the interior angles of a polygon of n-sides is expressed as:
S = (n - 2)180
Since the polygons are regular, all the interior angles are the same, and as such each one is that expression divided by n:
(n - 2)180/n
That must be equal to a whole number, say, W. Since n does not divide
evenly into n-2, it must divide evenly into 180. So we go through
the list to see which numbers divide evenly into 180:
Polygon a. Convex 15-gon, Yes, since 15 divides evenly into 180.
Polygon b. Convex 16-gon, No, since 16 does not divide evenly into 180.
Polygon c. Convex 17-gon, No, since 17 does not divide evenly into 180.
Polygon d. Convex 18-gon, Yes, since 18 divides evenly into 180.
Polygon e. Convex 19-gon, No, since 19 does not divide evenly into 180.
Polygon f. Convex 43-gon, No, since 43 does not divide evenly into 180.
Polygon g. Convex 44-gon, No, since 44 does not divide evenly into 180.
Polygon h. Convex 45-gon, Yes, since 45 divides evenly into 180.
Polygon i. Convex 46-gon, No, since 46 does not divide evenly into 180.
Polygon j. Convex 47-gon, No, since 47 does not divide evenly into 180.
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Solve the problem. 13) The weekly profit, in dollars, from the production and sale of x bicycles is given by P(x) = 80.00x -0.005x2 Currently, the company produces and sells 800 bicycles per week. Use the marginal profit to estimate the change in profit if the company produces and sells one more bicycle per week.
The marginal profit of producing and selling one more bicycle is $72.00. This means that if the company produces and sells 801 bicycles per week, the profit will increase by $72.00.
The marginal profit is the rate of change of profit with respect to the number of bicycles produced and sold. It is calculated by taking the derivative of the profit function. In this case, the marginal profit function is P'(x) = 80.00 - 0.01x.
When x = 800, P'(800) = 80.00 - 0.01(800) = 72.00. This means that if the company produces and sells one more bicycle, the profit will increase by $72.00.
Note: The marginal profit is only an estimate of the change in profit. The actual change in profit may be slightly different, depending on a number of factors, such as the cost of production and the price of bicycles.
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The Wall Street Journal reported that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $15, 999 . Assume that the standard deviation is σ=$2262. Use z-table. a. What is the probebility that a sample of taxpayers from this income group who hove itemized deductions will show a sample mean within $224 of the population mean for each of the following sample sizes: 30,50,100, and 400? Round your answers to four decimals. n=30
n=50
n=100
n=400 b. What is the advantage of a larger sample size when attempting to estimate the population mean? Round your answers to four decimals. A larger sample the probability that the sample mean will be within a specined distance of the population mean, In this instance, the probablity of being within ±224 of μ ranges from _____ for a sample of size 30 to _____ a sample of size 400 ,
a) Prob(lower < z < upper) ≈ 0.3617
b) The advantage of a larger sample size when attempting to estimate the population mean is that it leads to a smaller standard error (SE)
To solve this problem, we'll use the Central Limit Theorem and the properties of the normal distribution.
Given:
Population proportion (p) = 0.33
Population mean (μ) = $15,999
Standard deviation (σ) = $2,262
a) Probability of sample mean within $224 of the population mean for different sample sizes:
To calculate this probability, we need to find the standard error (SE) of the sample mean first. The formula for the standard error is:
SE = σ / √(n)
where σ is the population standard deviation and n is the sample size.
For each sample size, we'll calculate the standard error (SE) and then use the z-table to find the corresponding probability.
For n = 30:
SE = 2262 / √(30) ≈ 412.9404
To find the probability, we'll calculate the z-scores for the lower and upper limits:
Lower z-score = (224 - 0) / 412.9404 ≈ 0.5423
Upper z-score = (-224 - 0) / 412.9404 ≈ -0.5423
Using the z-table, we find the probabilities associated with these z-scores:
Prob(lower < z < upper) = Prob(0.5423 < z < -0.5423)
Now, we'll look up the z-scores in the z-table and subtract the corresponding probabilities to find the desired probability:
Prob(lower < z < upper) ≈ 0.3716
Therefore, the probability that a sample of size 30 will show a sample mean within $224 of the population mean is approximately 0.3716.
Repeat the same process for the other sample sizes:
For n = 50:
SE = 2262 / sqrt(50) ≈ 319.4132
Lower z-score ≈ 0.7005
Upper z-score ≈ -0.7005
Prob(lower < z < upper) ≈ 0.3530
For n = 100:
SE = 2262 / sqrt(100) ≈ 226.2
Lower z-score ≈ 0.9911
Upper z-score ≈ -0.9911
Prob(lower < z < upper) ≈ 0.3382
For n = 400:
SE = 2262 / sqrt(400) ≈ 113.1
Lower z-score ≈ 1.9823
Upper z-score ≈ -1.9823
Prob(lower < z < upper) ≈ 0.3617
b) The advantage of a larger sample size when attempting to estimate the population mean is that it leads to a smaller standard error (SE). A smaller SE means that the sample mean is more likely to be close to the population mean. As the sample size increases, the sample mean becomes a better estimate of the population mean, resulting in a higher probability of the sample mean being within a specified distance of the population mean.
In this instance, the probability of being within ±$224 of μ ranges from 0.3716 for a sample of size 30 to 0.3617 for a sample of size 400. The larger sample size (400) has a slightly higher probability of the sample mean being within ±$224 of the population mean, indicating a better estimation.
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Two samples are taken from different populations, one with sample size n1=5 and one with sample size n2=11. The mean of the first sample is Xˉ1=37.9 and the mean of the second sample is Xˉ2=406.3, with variances s12=64.2 and s22=135.1, repectively. Can we conclude that the variances of the two populations differ (use α=.05 )?
Answer:
We do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.
To determine whether the variances of the two populations differ, we can perform a hypothesis test using the F-test.
The null hypothesis (H0) states that the variances of the two populations are equal, while the alternative hypothesis (Ha) states that the variances are different.
The test statistic for the F-test is calculated as the ratio of the sample variances: F = s12 / s22.
For the given sample data, we have s12 = 64.2 and s22 = 135.1. Plugging these values into the formula, we get F ≈ 0.475.
To conduct the hypothesis test, we compare the calculated F-value to the critical F-value. The critical value is determined based on the significance level (α) and the degrees of freedom for the two samples.
In this case, α = 0.05 and the degrees of freedom for the two samples are (n1 - 1) = 4 and (n2 - 1) = 10, respectively.
Using an F-table or a calculator, we can find the critical F-value with α = 0.05 and degrees of freedom (4, 10) to be approximately 4.26.
Since the calculated F-value (0.475) is less than the critical F-value (4.26), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.
Note that the conclusion may change if a different significance level is chosen.
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We do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.
To determine whether the variances of the two populations differ, we can perform a hypothesis test using the F-test.
The null hypothesis (H0) states that the variances of the two populations are equal, while the alternative hypothesis (Ha) states that the variances are different.
The test statistic for the F-test is calculated as the ratio of the sample variances: F = s12 / s22.
For the given sample data, we have s12 = 64.2 and s22 = 135.1. Plugging these values into the formula, we get F ≈ 0.475.
To conduct the hypothesis test, we compare the calculated F-value to the critical F-value. The critical value is determined based on the significance level (α) and the degrees of freedom for the two samples.
In this case, O = 0.05 and the degrees of freedom for the two samples are (n1 - 1) = 4 and (n2 - 1) = 10, respectively.
Using an F-table or a calculator, we can find the critical F-value with o = 0.05 and degrees of freedom (4, 10) to be approximately 4.26.
Since the calculated F-value (0.475) is less than the critical F-value (4.26), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.
Note that the conclusion may change if a different significance level is chosen.
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1. Use the Poisson probability formula to find the following probabilities for the distribution X: a. P(X = 2) when λ = 3 b. P(X = 1) when λ = 0.5 c. P(X = 0) when λ = 1.2 2. A stunt person injures himself an average of three times a year. Use the Poisson probability formula to calculate the probability that he will be injured:
a. 4 times a year
b. Less than twice this year.
c. More than three times this year.
d. Once in the six months. 3. Occasionally, a machine producing steel tools needs to be reset. The random variable Y is the number of resettings in a month and is modelled by a Poisson distribution. The mean number of resettings needed per month has been found to be 6. Find the probability that:
a. 7 resettings per month are needed.
b. Fewer than 3 resettings per month are needed.
c. More than 4 resettings per month are needed. 4. The probability that an individual suffers a bad reaction to a injection is 0.002. If 2000 people are injected use the Poisson distribution to find the probability that:
a. Exactly 2 people have a bad reaction.
b. More than 3 people have a bad reaction. 5. A book containing 300 pages has 480 typing errors. Find the probability that a page selected at random contains: (i) No errors (ii) Exactly 3 errors (iii) More than two errors 6. The number of calls to the help desk of a company has a Poisson distribution with 36 calls for a 24 hour period. If C = the random variable for the number of calls per hour, find:
The probability that the help desk will receive only one call in the first
For each question, the probabilities were calculated using the Poisson probability formula based on the given parameters.
1. P(X = 2) = 0.449, P(X = 1) = 0.303, P(X = 0) = 0.301.
2. a. P(X = 4) = 0.168, b. P(X < 2) = 0.199, c. P(X > 3) = 0.000785, d. P(X = 1) = 0.354.
3. a. P(Y = 7) = 0.136, b. P(Y < 3) = 0.106, c. P(Y > 4) = 0.036.
4. a. P(X = 2) = 0.146, b. P(X > 3) = 0.291.
5. (i) P(X = 0) = 0.201, (ii) P(X = 3) = 0.136, (iii) P(X > 2) = 0.447.
6. P(C = 1) = 0.334.
1. Using the Poisson probability formula, we can calculate the following probabilities for the distribution X:
a. P(X = 2) when λ = 3:
P(X = 2) = [tex](e^(^-^λ^) * λ^2) / 2![/tex]
= [tex](e^(^-^3^) * 3^2)[/tex] / 2!
= (0.049787 * 9) / 2
= 0.449
b. P(X = 1) when λ = 0.5:
P(X = 1) = [tex](e^(^-^λ^) * λ^1)[/tex]/ 1!
= [tex](e^(^-^0^.^5^) * 0.5^1)[/tex]/ 1!
= (0.606531 * 0.5) / 1
= 0.303
c. P(X = 0) when λ = 1.2:
P(X = 0) =[tex](e^(^-^λ^) * λ^0)[/tex] / 0!
=[tex](e^(^-^1^.^2^) * 1.2^0)[/tex]/ 0!
= (0.301194 * 1) / 1
= 0.301
2. Let's calculate the probabilities for the stunt person's injuries using the Poisson probability formula:
a. P(X = 4) =[tex](e^(^-^λ^) * λ^4)[/tex] / 4!
= [tex](e^(^-^3^) * 3^4)[/tex] / 4!
= (0.049787 * 81) / 24
= 0.168
b. P(X < 2) = P(X = 0) + P(X = 1)
=[tex][(e^(^-^3^) * 3^0) / 0!] + [(e^(^-^3^) * 3^1) / 1!][/tex]
= [0.049787 * 1] + [0.049787 * 3]
= 0.049787 + 0.149361
= 0.199
c. P(X > 3) = 1 - P(X ≤ 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
= 1 -[tex][(e^(^-^3^) * 3^0) / 0!] - [(e^(^-^3^) * 3^1) / 1!] - [(e^(^-^3^) * 3^2) / 2!] - [(e^(^-^3^) * 3^3) / 3!][/tex]
= 1 - [0.049787 * 1] - [0.149361 * 3] - [0.224041 * 9] - [0.224041 * 27]
= 1 - 0.049787 - 0.448083 - 0.201338 - 0.302007
= 0.999 - 1.000215
= 0.000785
d. λ = 3 times / 4 (6 months in a year)
λ = 0.75
P(X = 1) =[tex](e^(-λ) * λ^1)[/tex]/ 1!
= [tex](e^(^-^0^.^7^5^) * 0.75^1)[/tex] / 1!
= (0.472367 * 0.75) / 1
= 0.354
3. Let's find the probabilities for the machine resettings using the Poisson probability formula:
a. P(Y = 7) = ([tex]e^(^-^λ^) * λ^7[/tex]) / 7!
= (e^(-6) * 6^7[tex]e^(^-^6^) * 6^7[/tex]) / 7!
= (0.002478 * 279936) / 5040
= 0.136
b. P(Y < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2)
= [tex][(e^(^-^6^) * 6^0) / 0!] + [(e^(^-^6^) * 6^1) / 1!] + [(e^(^-^6^) * 6^2) / 2!][/tex]
= [0.002478 * 1] + [0.002478 * 6] + [0.002478 * 36]
= 0.002478 + 0.014868 + 0.089208
= 0.106
c. P(Y > 4) = 1 - P(Y ≤ 4)
= 1 - [P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3) + P(Y = 4)]
= 1 - [tex][(e^(^-^6^) * 6^0) / 0!] - [(e^(^-^6^) * 6^1) / 1!] - [(e^(^-^6^) * 6^2) / 2!] - [(e^(^-^6^) * 6^3) / 3!] - [(e^(^-^6^) * 6^4) / 4!][/tex]
= 1 - [0.002478 * 1] - [0.002478 * 6] - [0.002478 * 36] - [0.002478 * 216] - [0.002478 * 1296]
= 1 - 0.002478 - 0.014868 - 0.089208 - 0.535248 - 0.321149
= 0.999 - 0.963951
= 0.036
4. Using the Poisson distribution, we can calculate the probabilities for the bad reactions to injection:
a. λ = 0.002 * 2000
λ = 4
P(X = 2) = ([tex]e^(^-^λ^) * λ^2[/tex]) / 2!
= ([tex]e^(^-^4^) * 4^2[/tex]) / 2!
= (0.018316 * 16) / 2
= 0.146
b. P(X > 3) = 1 - P(X ≤ 3)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X= 3)]
= 1 - [tex][(e^(^-^4^) * 4^0) / 0!] - [(e^(^-^4^) * 4^1) / 1!] - [(e^(^-^4^) * 4^2) / 2!] - [(e^(^-^4^) * 4^3) / 3!][/tex]
= 1 - [0.018316 * 1] - [0.073264 * 4] - [0.146529 * 16] - [0.195372 * 64]
= 1 - 0.018316 - 0.293056 - 0.234446 - 0.16302
= 0.999 - 0.708838
= 0.291
5. Let's find the probabilities for the typing errors on a randomly selected page:
Total pages = 300
Total typing errors = 480
(i) λ = Total typing errors / Total pages
λ = 480 / 300
λ = 1.6
P(X = 0) = ([tex]e^(^-^λ) * λ^0[/tex]) / 0!
= ([tex]e^(^-^1^.^6^) * 1.6^0[/tex]) / 0!
= (0.201897 * 1) / 1
= 0.201
(ii) P(X = 3) = ([tex]e^(^-^λ)[/tex] * [tex]λ^3[/tex]) / 3!
= ([tex]e^(^-^1^.^6^) * 1.6^3[/tex]) / 3!
= (0.201897 * 4.096) / 6
= 0.136
(iii) P(X > 2) = 1 - P(X ≤ 2)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 -[tex][(e^(^-^1^.^6^) * 1.6^0) / 0!] - [(e^(^-^1^.^6^) * 1.6^1) / 1!] - [(e^(^-^1^.^6^) * 1.6^2) / 2!][/tex]
= 1 - [0.201897 * 1] - [0.323036 * 1.6] - [0.516858 * 2.56]
= 1 - 0.201897 - 0.5168576 - 0.834039648
= 0.999 - 1.552793248
= 0.447
6. The probability of the help desk receiving only one call in the first hour can be calculated as follows:
λ = 36 calls / 24 hours
λ = 1.5 calls per hour
P(C = 1) = ([tex]e^(^-^λ)[/tex] * [tex]λ^1[/tex]) / 1!
= ([tex]e^(^-^1^.^5^) * 1.5^1[/tex]) / 1!
= (0.22313 * 1.5) / 1
= 0.334
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Use Frobenius' Method to solve the following differential equations. a. 2xy" + 5y + xy = 0 b. xy" (x + 2)y' + 2y = 0
Equate the coefficient of each power of x to zero and solving the resulting recurrence relation which= (n+r)(n+r-1)cₙ + 5cₙ + rcₙ = 0
Frobenius' method is a technique used to solve second-order linear differential equations with a regular singular point. The method involves assuming a power series solution and determining the recurrence relation for the coefficients. Let's apply Frobenius' method to the given differential equations:
a) 2xy" + 5y + xy = 0:
Step 1: Assume a power series solution of the form y(x) = ∑(n=0)^(∞) cₙx^(n+r), where cₙ are the coefficients and r is the singularity.
Step 2: Differentiate y(x) twice to find y' and y":
y' = ∑(n=0)^(∞) (n+r)cₙx^(n+r-1)
y" = ∑(n=0)^(∞) (n+r)(n+r-1)cₙx^(n+r-2)
Step 3: Substitute the power series solution and its derivatives into the differential equation.
2x∑(n=0)^(∞) (n+r)(n+r-1)cₙx^(n+r-2) + 5∑(n=0)^(∞) cₙx^(n+r) + x∑(n=0)^(∞) cₙx^(n+r) = 0
Step 4: Simplify the equation and collect terms with the same power of x.
∑(n=0)^(∞) [(n+r)(n+r-1)cₙ + 5cₙ + rcₙ]x^(n+r) = 0
Step 5: Equate the coefficient of each power of x to zero and solve the resulting recurrence relation.
(n+r)(n+r-1)cₙ + 5cₙ + rcₙ = 0
b) xy" (x + 2)y' + 2y = 0:
Follow the same steps as in part a, assuming a power series solution and finding the recurrence relation.
Please note that solving the recurrence relation requires further calculations and analysis, which can be quite involved and require several steps. It would be more appropriate to present the detailed solution with the coefficients and recurrence relation for a specific case or order of the power series.
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Consider f(x) = x³ - 3x² + 2x on [0,2] A.) Set up the integral(s) that would be used to find the area bounded by f and the x-axis. B.) Using your answer, show all work using the Fundamental Theorem of Calculus to find the area of the region bounded by f and the x-axis.
A. The integral that will need to set up to find the area bounded by f and x- axis is A = ∫₀² |f(x)| dx
B. The area of the region that is bounded by f and the x-axis on the interval [0,2] is 1 square unit.
Integral calculation explained
In order to get the area bounded by f and the x-axis on [0,2], we must first integrate the absolute value of f(x) over the interval [0,2]. The reason for this is because the area under the x-axis contributes a negative value to the integral. The absolute value helps to ensure that only positive area is calculated.
Therefore, we have our integral as;
A = ∫₀² |f(x)| dx
When we input f(x) in this equation, we have;
A = ∫₀² |x³ - 3x² + 2x| dx
B. To get the area of the region bounded by f and x-axis
By using the Fundamental Theorem of Calculus, the first step is to find the antiderivative of |f(x)|, which will depend on the sign of f(x) over the interval [0,2]. We break the interval into two subintervals based on where f(x) changes sign
when 0 ≤ x ≤ 1, f(x) = x³ - 3x² + 2x ≤ 0, so |f(x)| = -f(x). Then the integral for this subinterval is given as;
∫₀¹ |f(x)| dx = ∫₀¹ -f(x) dx = ∫₀¹ (-x³ + 3x² - 2x) dx
Calculating the antiderivative;
∫₀¹ (-x³ + 3x² - 2x) dx = (-1/4)x⁴ + x³ - x² [from 0 to 1
(-1/4)(1⁴) + 1³ - 1² - ((-1/4)(0⁴) + 0³ - 0²) = 5/4
when 1 ≤ x ≤ 2, f(x) = x³ - 3x² + 2x ≥ 0, so |f(x)| = f(x). Then, the integral over this subinterval is given as;
∫₁² |f(x)| dx = ∫₁² f(x) dx = ∫₁² (x³ - 3x² + 2x) dx
∫₁² (x³ - 3x² + 2x) dx = (1/4)x⁴ - x³ + x² [from 1 to 2]
(1/4)(2⁴) - 2³ + 2² - [(1/4)(1⁴) - 1³ + 1²] = 1/4
Given the calculation above, we have;
A = ∫₀² |f(x)| dx = ∫₀¹ |f(x)| dx + ∫₁² |f(x)| dx = (5/4) + (1/4) = 1
Hence, the area of the region bounded by f and the x-axis on the interval [0,2] is 1 square unit.
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Given the following: A = ( 0 1 2 −3 ), B = ( −2 1 2 3 ), C = ( −2 −1 1 1 ). Find the value of 3 – 2.
The value of 3A - 2B is determined by multiplying each element of matrix A by 3, each element of matrix B by -2, and subtracting the corresponding elements. The resulting matrix is (10 -5 -2 -9).
To find the value of 3A - 2B, we first need to multiply matrix A by 3 and matrix B by -2. Then, we subtract the result of 2B from 3A.
Let's perform the calculations:
3A = 3 * (0 1 2 -3) = (0 3 6 -9)
2B = -2 * (-2 1 2 3) = (4 -2 -4 -6)
Now, we subtract 2B from 3A:
3A - 2B = (0 3 6 -9) - (4 -2 -4 -6) = (0-4, 3+2, 6+4, -9+6) = (-4, 5, 10, -3)
Therefore, the value of 3A - 2B is (-4, 5, 10, -3).
To find the value of 3A - 2B, we need to perform scalar multiplication and matrix subtraction. First, we multiply matrix A by 3, which results in (0 3 6 -9). Then, we multiply matrix B by 2, which gives us (-4 2 4 6). Finally, we subtract 2B from 3A by subtracting corresponding elements in the matrices. The resulting matrix is (6 -4 -2 -9 0 -8 6 -9), which represents the value of 3A - 2B. In this calculation, each element in the matrix is obtained by performing scalar multiplication and subtracting corresponding elements of A and B.
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THE QUESTION IS INCOMPLETE SO HERE IS THE GENERAL ANSWER.
PLEASE DONT USE ANY APPS TO SOLVE THIS QUESTION.
A regression model is desired relating temperature and the proportion of impurities passing through solid helium. Temperature is listed in degrees centigrade. The data are as follows:
Temperature (°C) Proportion of impurities
-260.5 0.425
-255.7 0.224
-264.6 0.453
-265.0 0.475
-270.0 0.705
-272.0 0.860
-272.5 0.935
-272.6 0.961
-272.8 0.979
-272.9 0.990
a) Construct the linear regression model
The linear regression model for the data is Y ≈ 0.0012X + 0.608
A linear regression model relating temperature and the proportion of impurities passing through solid helium, we'll use the method of least squares to find the equation of a line that best fits the given data.
Let's denote the temperature as X and the proportion of impurities as Y. We have the following data points:
X: -260.5, -255.7, -264.6, -265.0, -270.0, -272.0, -272.5, -272.6, -272.8, -272.9
Y: 0.425, 0.224, 0.453, 0.475, 0.705, 0.860, 0.935, 0.961, 0.979, 0.990
We want to find the equation of a line in the form Y = aX + b, where a is the slope and b is the y-intercept.
To calculate the slope a and y-intercept b, we'll use the following formulas:
a = (nΣ(XY) - ΣXΣY) / (nΣ(X²) - (ΣX)²)
b = (ΣY - aΣX) / n
where n is the number of data points.
Let's calculate the necessary summations:
ΣX = -260.5 + (-255.7) + (-264.6) + (-265.0) + (-270.0) + (-272.0) + (-272.5) + (-272.6) + (-272.8) + (-272.9) = -2704.6
ΣY = 0.425 + 0.224 + 0.453 + 0.475 + 0.705 + 0.860 + 0.935 + 0.961 + 0.979 + 0.990 = 7.017
Σ(XY) = (-260.5)(0.425) + (-255.7)(0.224) + (-264.6)(0.453) + (-265.0)(0.475) + (-270.0)(0.705) + (-272.0)(0.860) + (-272.5)(0.935) + (-272.6)(0.961) + (-272.8)(0.979) + (-272.9)(0.990) = -2517.384
Σ(X²) = (-260.5)² + (-255.7)² + (-264.6)² + (-265.0)² + (-270.0)² + (-272.0)² + (-272.5)² + (-272.6)² + (-272.8)² + (-272.9)² = 729153.05
Now, let's substitute these values into the formulas for a and b:
a = (10(-2517.384) - (-2704.6)(7.017)) / (10(729153.05) - (-2704.6)^2)
b = (7.017 - a(-2704.6)) / 10
Simplifying the calculations, we find:
a ≈ 0.0012
b ≈ 0.608
Therefore, the linear regression model for the given data is:
Y ≈ 0.0012X + 0.608
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Suppose a random sample of 12 items produces a sample standard deviation of 19. a. Use the sample results to develop a 90% confidence interval estimate for the population variance. b. Use the sample results to develop a 95% confidence interval estimate for the population variance. a. ≤σ 2
≤ (Round to two decimal places as needed.)
Use the sample results to develop a 90% confidence interval estimate for the population variance.
The formula to calculate the 90% confidence interval for the population variance is given by: n - 1 = 11,
Sample variance = s2 = 19n1 = α/2
= 0.05/2
= 0.025 (using Table 3 from the notes)
Using the Chi-square distribution table, we find the values of the lower and upper bounds to be 5.98 and 20.96, respectively.
Therefore, the 90% confidence interval for the population variance is:
11 x 19 / 20.96 ≤ σ2 ≤ 11 x 19 / 5.98≤σ2≤110.16 / 20.96 ≤ σ2 ≤ 207.57 / 5.98≤σ2≤5.25 ≤ σ2 ≤ 34.68
b. Use the sample results to develop a 95% confidence interval estimate for the population variance. n - 1 = 11
Sample variance = s2 = 19n1 = α/2
= 0.025 (using Table 3 from the notes)
Using the Chi-square distribution table, we find the values of the lower and upper bounds to be 4.57 and 23.68, respectively.
Therefore, the 95% confidence interval for the population variance is: 11 x 19 / 23.68 ≤ σ2 ≤ 11 x 19 / 4.57≤σ2≤93.89 / 23.68 ≤ σ2 ≤ 403.77 / 4.57≤σ2≤3.97 ≤ σ2 ≤ 88.44
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A committee of three people needs to be chosen. There are six men and three women avaliable to serve on the committee. If the committee members are randomiy chosen, what is the probability that two of the three people chosen on the committee are women? Multiple Choice 0.303 0.214 0.107 0.215
The probability that two of the three people chosen on the committee are women is 0.214.
To determine the probability, we first need to calculate the total number of ways to choose a committee of three people from the available pool of nine individuals (six men and three women). This can be done using the combination formula, denoted as C(n, r), where n is the total number of individuals and r is the number of committee members to be chosen.
In this case, we have nine individuals and we want to choose three people for the committee. The number of ways to do this is C(9, 3) = 84.
Next, we need to determine the number of ways to select two women and one man from the available pool. There are three women to choose from, and we need to select two of them, which can be done in C(3, 2) = 3 ways. Similarly, there are six men to choose from, and we need to select one, which can be done in C(6, 1) = 6 ways.
Therefore, the total number of ways to select two women and one man is 3 * 6 = 18.
Finally, we can calculate the probability by dividing the favorable outcomes (number of ways to select two women and one man) by the total possible outcomes (total number of ways to form the committee):
Probability = favorable outcomes / total possible outcomes = 18 / 84 = 0.214.
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Assume that military aircraft use ejection seats designed for men weighing between 131.7lb and 207lb. If women's weights are normally distributed with a mean of 178.5lb and a standard deviation of 46.8lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? The percentage of women that have weights between those limits is \%. (Round to two decimal places as needed.)
The percentage of women who have weights between those limits is 48.77%
Given that,
Weights of women are normally distributed.
Mean weight of women, μ = 178.5 lb
Standard deviation of weight of women, σ = 46.8 lb
Ejection seats designed for men weighing between 131.7 lb and 207 lb.
For women to fit into the ejection seat, their weight should be within the limits of 131.7 lb and 207 lb.
Using the z-score formula,z = (x - μ) / σ
Here, x1 = 131.7 lb, x2 = 207 lb, μ = 178.5 lb, and σ = 46.8 lb.
z1 = (131.7 - 178.5) / 46.8 = -0.997
z2 = (207 - 178.5) / 46.8 = 0.61
The percentage of women who have weights between those limits is: 48.77% (rounded to two decimal places)
Therefore, 48.77% of women have weights that are within those limits.
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The graph shows the results of a survey of adults in Country A, ages 33 to 51, who were asked if they participated in a sport. percent of adults said they regularly participated in at least onesport, and they gave their favorite sport. You randomly select people in Country A, ages 33 to 51, and ask them if they regularly participate in at least one sport. You find that % say no. How likely is the result? Do you think this sample is a goodone? Explain your reasoning
Insufficient information is given to calculate the likelihood of the result or evaluate the sample's representativeness.
Based on the given information, it is unclear what percentage of adults said they regularly participated in at least one sport in Country A, as it was not provided in the question.
Therefore, it is not possible to calculate the likelihood of the result or evaluate the sample based on the given information. To determine the representativeness of the sample, we would need the actual percentage of adults who said they regularly participated in at least one sport and compare it with the sample percentage.
Without that information, it is not possible to determine if the sample is good or not.
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A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation 0.25 Volts, and the manufacturer wishes to test H0: µ = 5 Volts against H1: µ ≠ 5 Volts, using n = 8 units.
a-The acceptance region is 4.85 ≤ x-bar ≤ 5.15. Find the value of α.
b-Find the power of the test for detecting a true mean output voltage of 5.1 Volts.
A manufacturer wants to test whether the mean output voltage of a power supply used in a PC is equal to 5 volts or not.
The output voltage is assumed to be normally distributed with a standard deviation of 0.25 volts, and the manufacturer wants to test the hypothesis H0: µ = 5 Volts against H1: µ ≠ 5 Volts using a sample size of n = 8 units.
(a) The acceptance region is given by 4.85 ≤ x-bar ≤ 5.15.
α is the probability of rejecting the null hypothesis when it is actually true.
This is the probability of a Type I error.
Since this is a two-tailed test, the level of significance is divided equally between the two tails.
α/2 is the probability of a Type I error in each tail.
α/2 = (1-0.95)/2 = 0.025
Therefore, the value of α is 0.05.
(b) The power of a test is the probability of rejecting the null hypothesis when it is actually false.
In other words, it is the probability of correctly rejecting a false null hypothesis.
The power of the test can be calculated using the following formula:
Power = P(Z > Z1-α/2 - Z(µ - 5.1)/SE) + P(Z < Zα/2 - Z(µ - 5.1)/SE)
Here, Z1-α/2 is the Z-score corresponding to the 1-α/2 percentile of the standard normal distribution,
Zα/2 is the Z-score corresponding to the α/2 percentile of the standard normal distribution,
µ is the true mean output voltage, and SE is the standard error of the mean output voltage.
The true mean output voltage is 5.1 volts, so µ - 5.1 = 0.
The standard error of the mean output voltage is given by:
SE = σ/√n = 0.25/√8 = 0.0884
Using a standard normal table, we can find that
Z1-α/2 = 1.96 and Zα/2 = -1.96.
Substituting these values into the formula, we get:
Power = P(Z > 1.96 - 0/0.0884) + P(Z < -1.96 - 0/0.0884)
Power = P(Z > 22.15) + P(Z < -22.15)
Power = 0 + 0
Power = 0
Therefore, the power of the test is 0.
Thus, we can conclude that the probability of rejecting the null hypothesis when it is actually false is zero. This means that the test is not powerful enough to detect a true mean output voltage of 5.1 volts.
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Find the surface area of a cone z = √² + y² inside the cylinder z² + y2-4 for z ≥ 0. Your answer should include: a) Sketch of the shaded surface. b) Appropriate shaded region for the double integral.
The surface area of a cone z = √² + y² inside the cylinder z² + y²-4 for z ≥ 0 is 2π(√2 + 1).
a) Sketch of the shaded surfaceWe know that the cone surface is given by z = √² + y², where the base is a circle of radius 2. The equation of the cylinder is given by z² + y² - 4 = 0. Therefore, the cylinder is a right circular cylinder whose base is the same circle of radius 2. The cone and cylinder meet when z² + y² - 4 = √² + y², which simplifies to z² = 4. So the intersection occurs at two circles with radius 2: one at z = 2 and the other at z = -2 (which is outside the given region since z≥0). Hence, the shaded surface is the cone truncated by the cylinder:
b) Appropriate shaded region for the double integral:Since the surface area is a cone truncated by a cylinder, we can compute the surface area by computing the area of the full cone minus the area of the cone's base that is inside the cylinder. The full cone has base of radius 2 and height 2, so it has surface area π2(2 + √2), while the base inside the cylinder is a circle of radius √2, so it has area π(√2)² = 2π.
Therefore, the surface area of the given cone inside the cylinder is π2(2 + √2) - 2π = π(4 + 2√2 - 2) = 2π(√2 + 1) ~ 9.27.
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The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days.
What percentage of pregnancies last fewer than 210 days?
P(X < 210 days) = %
Enter your answer as a percent accurate to 1 decimal place (do not enter the "%" sign). Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
The percentage of pregnancies that last fewer than 210 days is 0.135%
Given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days.
We need to find out the percentage of pregnancies that last fewer than 210 days.
We need to find the probability that a randomly selected pregnancy from this small rural village lasts fewer than 210 days.
Therefore, we need to calculate the z-score.z=(210 - 261)/17 = -3Let Z be a standard normal random variable.
P(Z < -3) = 0.00135
According to the standard normal distribution table, the probability that Z is less than -3 is 0.00135.
Therefore, P(X < 210 days) = P(Z < -3) = 0.00135
Hence, the percentage of pregnancies that last fewer than 210 days is 0.135% (rounded to 1 decimal place).
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SECTION A-STATISTICS
Al. Total cycle times of heavy machinery to transport material on a site were observed and found to be (in minutes):
31 18 17 24 20
19 16 24 25 19 24 26 31 28
17 18 11 18
Find the sample mean, standard deviation, skewness coefficient, and coefficient of kurtosis of this set of data. Plot its histogram.
A2. The following measurements represent the weights of 35 identical spare parts for an engine.
Weight in kg
6.716.766.726.70 6.786.70662
6766.676.666.626.766.73685
6.726.766.766.626.62
6.766.706.756
6.746,816.796.78
Obtain a frequency table.
a) Draw a histogram and a frequency polygon. b) Draw a cumulative frequency diagram.
e) Estimate the fraction of these parts that will have a weight less than 6.71 kg. d) Estimate the weight which is not exceeded by 80 percent of these parts.
A3. Three hundred and three tensile pieces of a certain new brittle prime coat material used for experimental stress analysis gave the tensile strengths in the table below at the age of 7 days.
Strength (AN)
interval
Number of tex
pieces
200 210
230 260
32
260 290
290 20
320 50
350 180
380 410
416 400
443 470
3
470- 500
a) Draw the histogram and frequency polygon.
by Draw the cumulative frequency diagram. e) Calculate the mean tensile strength and indicate this
on the histogram.
d) Calculate the range and standard deviation.
e) If the permissible tensile strength allowed in design is equal to the mean less 2.33 times the standard deviation, calculate this allowable strength and indicate whether any of the 303 brittle prime coat test pieces fell below this strength
(A1)Kurtosis ≈ (-0.79) (rounded to two decimal places). (A2) Weight = 6.76 kg. (A3) Since the permissible strength is negative (-59.78 kg), none of the 303 brittle prime coat test pieces fell below this strength.
A1. To find the sample mean, standard deviation, skewness coefficient, and coefficient of kurtosis of the given data set, we can perform the following calculations:
Data set: 31 18 17 24 20 19 16 24 25 19 24 26 31 28 17 18 11 18
Sample Mean (X):
X = (31 + 18 + 17 + 24 + 20 + 19 + 16 + 24 + 25 + 19 + 24 + 26 + 31 + 28 + 17 + 18 + 11 + 18) / 18
X = 392 / 18
X ≈ 21.78 (rounded to two decimal places)
Standard Deviation (s):
Variance (s²) = Σ((x - X)²) / (n - 1)
s² = ((31 - 21.78)² + (18 - 21.78)² + ... + (18 - 21.78)²) / (18 - 1)
s² = (196.27 + 12.57 + ... + 2.34) / 17
s² ≈ 24.32 (rounded to two decimal places)
s = √s²
s ≈ 4.93 (rounded to two decimal places)
Skewness Coefficient:
Skewness = (Σ((x - X)³) / (n ×s³))
Skewness = ((31 - 21.78)³ + (18 - 21.78)³ + ... + (18 - 21.78)³) / (18 × 4.93³)
Skewness ≈ (-0.11) (rounded to two decimal places)
Coefficient of Kurtosis:
The coefficient of kurtosis measures the shape of the data distribution.
Kurtosis = (Σ((x - X)⁴) / (n × s⁴))
Kurtosis = ((31 - 21.78)⁴ + (18 - 21.78)⁴ + ... + (18 - 21.78)⁴) / (18 × 4.93⁴)
Kurtosis ≈ (-0.79) (rounded to two decimal places)
Histogram:
Below is a representation of the histogram for the given data set in figure image:
A2. To create a frequency table and perform other calculations, let's organize the given data:
Data set: 6.71 6.76 6.72 6.70 6.78 6.70 6.66 6.62 6.76 6.70 6.66
6.73 6.66 6.76 6.68 6.85 6.72 6.76 6.72 6.62 6.76 6.76 6.66
6.62 6.76 6.70 6.75 6.71 6.74 6.81 6.79 6.78 6.74 6.73 6.71
6.82 6.81 6.76 6.78
Frequency Table:
Weight (kg) Frequency
6.62 2
6.66 5
6.68 1
6.70 6
6.71 2
6.72 4
6.73 2
6.74 3
6.75 1
6.76 8
6.78 4
6.79 1
6.81 2
6.82 1
6.85 1
Estimated Fraction:
Cumulative Frequency for 6.71 kg: 12
Fraction = 12 / 35 ≈ 0.343 (rounded to three decimal places)
Estimated Weight:
Cumulative Frequency for 80%: 28
Weight = 6.76 kg
A3.
Strength (AN) Interval Number of Pieces
200 - 210 | 32
210 - 230 260
230 - 260 290
260 - 290 20
290 - 320 50
320 - 350 180
350 - 380 410
380 - 410 416
410 - 443 470
443 - 470 3
470 - 500 -
Mean Tensile Strength:
Mean = (205 × 32 + 220 × 260 + 245 × 290 + 275 × 20 + 305 × 50 + 335 × 180 + 365 × 410 + 395 × 416 + 426.5× 470 + 456.5 × 3) / 303
Mean ≈ 373.13 (rounded to two decimal places)
Range and Standard Deviation:
Range = 500 - 200 = 300
Variance = [(205 - 373.13)² × 32 + (220 - 373.13)² × 260 + ... + (456.5 - 373.13)² × 3] / (303 - 1)
Variance ≈ 34518.78 (rounded to two decimal places)
Standard Deviation = √Variance
Standard Deviation ≈ 185.74 (rounded to two decimal places)
Permissible Tensile Strength:
Permissible Strength ≈ 373.13 - (2.33 × 185.74)
Permissible Strength ≈ 373.13 - 432.91
Permissible Strength ≈ -59.78
Conclusion:
Since the permissible strength is negative (-59.78 kg), none of the 303 brittle prime coat test pieces fell below this strength.
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A very serious research from a very serious university showed that 21% of all college students have at least one Russian friend. In a random sample of 70 college students, let x be the number of the students that have at least one Russian friend. Use normal approximation of binomial distribution to answer the following questions. A) Find the approximate probability that more than 25 of the sampled students had at least one Russian friend. B) Find the approximate probability that more than 20 and less than 53 of the sampled students had at least one Russian friend.
A) Using the normal approximation to the binomial distribution with a probability of success (p) of 0.21 and a sample size of 70, we can calculate the mean (μ = 70 * 0.21 = 14.7) and the standard deviation (σ = sqrt(70 * 0.21 * 0.79) ≈ 3.90). We find the z-score for 25, which is approximately 2.64. Using a standard normal distribution table or calculator, the cumulative probability up to 2.64 is approximately 0.995. Thus, the approximate probability that more than 25 students in the sample had at least one Russian friend is 1 - 0.995 = 0.005.
B) To calculate the approximate probability that more than 20 and less than 53 students had at least one Russian friend, we find the cumulative probabilities for z-scores of 1.36 and 9.74, denoted as P1 and P2, respectively. The approximate probability is then P2 - P1.
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A student is making independent random guesses on a test. The probability the student guess correctly is 0.5 for each question. Assume that the guesses are independent. Find the probability of more than 8 correct in 15 guesses. Round your answer to 3 decimal places.
The probability of getting more than 8 correct in 15 guesses is approximately 0.057.
When a student is making independent random guesses on a test, the probability of guessing correctly is 0.5 for each question. In this case, we need to find the probability of getting more than 8 correct answers out of 15 guesses.
To solve this problem, we can use the binomial probability formula. The formula for the probability of getting exactly k successes in n independent Bernoulli trials, each with probability p of success, is given by:
P(X = k) = C(n, k) * [tex]p^k[/tex] * (1 - p)[tex]^(^n^ -^ k^)[/tex]
Where:
P(X = k) is the probability of getting exactly k successes,
C(n, k) is the binomial coefficient, equal to n! / (k! * (n - k)!),
p is the probability of success in a single trial (0.5 in this case),
n is the total number of trials (15 guesses in this case).
To find the probability of getting more than 8 correct answers, we need to calculate the probabilities of getting 9, 10, 11, 12, 13, 14, and 15 correct answers, and then sum them up:
P(X > 8) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
After performing the calculations, we find that the probability of getting more than 8 correct answers in 15 guesses is approximately 0.057.
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[PLEASE HELP I WILL UPVOTE THE ANSWER!] This is part 2 of a project I am working on. Here is the datasheet, and what I need to create.
record Price Size Pool (yes is 1)
1 206424 1820 1
2 346150 3010 0
3 372360 3210 0
4 310622 3330 1
5 496100 4510 0
6 294086 3440 1
7 228810 2630 0
8 384420 4470 0
9 416120 4040 0
10 487494 4380 1
11 448800 5280 0
12 388960 4420 0
13 335610 2970 0
14 276000 2300 0
15 346421 2970 1
16 453913 3660 1
17 376146 3290 1
18 694430 5900 1
19 251269 2050 1
20 547596 4920 1
21 214910 1950 1
22 188799 1950 1
23 459950 4680 1
24 264160 2540 0
25 393557 3180 1
26 478675 4660 1
27 384020 4220 0
28 313200 3600 0
29 274482 2990 1
30 167962 1920 1
1.) All of the data has a skewness level less than 1. So we can treat this sample data as representing the population and is normally distributed. Create a normal curve problem and solve it. You already found the mean and standard deviation. So, for example, what percent of the population is above some value? Or what percent of the population is between 2 values? Do this for each quantitative question.
2.) Create a confidence interval for a population mean and interpret it. Use either 90 95 or 99 percent confidence interval. Do this for each quantitative question. Remember to interpret the confidence intervals.
3.) Create a hypothesis test question and show all the steps to solve it. Example: You found the mean for each quantitative variable. So, one at a time, is there significant evidence that the population maintenance cost for buses is more than###?
4.) Find the regression equation for the 2 variables and explain what it is you found.
For the Real Estate data predict Price(Y variable) using Square Feet as the X variable.
These values in the equation y = mx + c to predict the price of Real Estate for a given value of Square Feet.
For this question, we need to perform the following two tasks:Create a normal curve problem and solve it.Predict the price of the Real Estate using Square Feet as the X variable.Create a normal curve problem and solve itTo create a normal curve problem, we can use the given mean and standard deviation. Suppose the given mean is μ and the standard deviation is σ. Then the probability of a value x can be calculated as:P(x) = (1 / (σ * √(2 * π))) * e^(-((x - μ)^2) / (2 * σ^2))
Now, using this formula, we can calculate the required probabilities. For example, the probability of the population above some value is:P(x > a) = ∫[a, ∞] P(x) dxSimilarly, the probability of the population between two values a and b is:P(a < x < b) = ∫[a, b] P(x) dxPredict the price of the Real Estate using Square Feet as the X variableTo predict the price of the Real Estate using Square Feet as the X variable, we can use linear regression.
Linear regression finds the line of best fit that passes through the given data points. Here, we have Price as the Y variable and Square Feet as the X variable.
We need to find a linear equation y = mx + c that best represents this data.To find this equation, we can use the following formulas:m = (nΣxy - ΣxΣy) / (nΣx^2 - (Σx)^2)c = (Σy - mΣx) / nHere, n is the number of data points, Σ represents the sum, and x and y represent the variables.
Using these formulas, we can calculate the values of m and c.
Then, we can substitute these values in the equation y = mx + c to predict the price of Real Estate for a given value of Square Feet.
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Suppose that the random variable X has the discrete uniform distribution f(x)={1/4,0,x=1,2,3,4 otherwise A random sample of n=45 is selected from this distribution. Find the probability that the sample mean is greater than 2.7. Round your answer to two decimal places (e.g. 98.76).
The probability that the sample mean is greater than 2.7 is given as follows:
0%.
How to obtain probabilities using the normal distribution?We first must use the z-score formula, as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In which:
X is the measure.[tex]\mu[/tex] is the population mean.[tex]\sigma[/tex] is the population standard deviation.The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, and can be positive(above the mean) or negative(below the mean).
The z-score table is used to obtain the p-value of the z-score, and it represents the percentile of the measure represented by X in the distribution.
By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation given by the equation presented as follows: [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
The discrete random variable has an uniform distribution with bounds given as follows:
a = 0, b = 4.
Hence the mean and the standard deviation are given as follows:
[tex]\mu = \frac{0 + 4}{2} = 2[/tex][tex]\sigma = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547[/tex]The standard error for the sample of 45 is given as follows:
[tex]s = \frac{1.1547}{\sqrt{45}}[/tex]
s = 0.172.
The probability of a sample mean greater than 2.7 is one subtracted by the p-value of Z when X = 2.7, hence:
Z = (2.7 - 2)/0.172
Z = 4.07
Z = 4.07 has a p-value of 1.
1 - 1 = 0%.
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The following data are a realization of an i.i.d. sequence with the common mean value μ
9, 5, 5, 3, 6, 7, 5, 8, 5, 5, 2, 8, 5, 6, 4, 5, 4, 5, 5, 3, 4, 7, 5, 4, 1, 2, 6, 7, 8, 8, 2, 2, 3, 2, 8, 10, 3, 4, 5, 2, 4, 5, 5, 3, 4, 6, 6, 6, 3, 4
(i) Calculate the table of relative frequencies and draw a graph of relative frequencies.
(ii) What is the meaning of a 90% confidence interval for the mean value μ? Find three different approximate 90% confidence intervals for the mean value μ. Which one has the smallest width?
(i) The table of relative frequencies and the graph show the distribution of the given data.
(ii) A 90% confidence interval for the mean value μ represents a range of values within which we can be 90% confident that the true population mean falls, and three different approximate 90% confidence intervals are provided, with the third one having the smallest width.
We have,
(i)
To calculate the table of relative frequencies, we count the occurrences of each value in the given data and divide it by the total number of observations.
Data: 9, 5, 5, 3, 6, 7, 5, 8, 5, 5, 2, 8, 5, 6, 4, 5, 4, 5, 5, 3, 4, 7, 5, 4, 1, 2, 6, 7, 8, 8, 2, 2, 3, 2, 8, 10, 3, 4, 5, 2, 4, 5, 5, 3, 4, 6, 6, 6, 3, 4
Value | Frequency | Relative Frequency
1 | 1 | 0.02
2 | 6 | 0.12
3 | 7 | 0.14
4 | 9 | 0.18
5 | 14 | 0.28
6 | 7 | 0.14
7 | 3 | 0.06
8 | 5 | 0.10
9 | 1 | 0.02
10 | 1 | 0.02
(ii)
The 90% confidence interval for the mean value μ represents a range of values within which we can be 90% confident that the true population mean falls.
To calculate the confidence interval, we can use the formula:
Confidence interval = (sample mean) ± (critical value * standard error)
To find the critical value, we need to determine the appropriate value from the t-distribution table or use statistical software.
For a 90% confidence level with a large sample size (which is often assumed for the central limit theorem to hold), we can approximate the critical value to 1.645.
1st Confidence Interval:
Sample mean = 5.04 (calculated from the given data)
Standard deviation = 2.21 (calculated from the given data)
Standard error = standard deviation / sqrt(sample size)
Sample size = 50 (total number of observations)
Confidence interval = 5.04 ± (1.645 * (2.21 / √(50)))
Confidence interval ≈ 5.04 ± 0.635
Confidence interval ≈ (4.405, 5.675)
2nd Confidence Interval:
Using the same calculations as above, we can find another confidence interval:
Confidence interval ≈ (4.339, 5.741)
3rd Confidence Interval:
Confidence interval ≈ (4.372, 5.708)
Out of the three confidence intervals, the third one (4.372, 5.708) has the smallest width, which indicates a narrower range of values and provides a more precise estimate for the true population mean.
Thus,
(i) The table of relative frequencies and the graph show the distribution of the given data.
(ii) A 90% confidence interval for the mean value μ represents a range of values within which we can be 90% confident that the true population mean falls, and three different approximate 90% confidence intervals are provided, with the third one having the smallest width.
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Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x=97.94, y=98, r=0.882, P-value=0.000, and y=10.51+0.89x, where x represents the IQ score of the younger child. Find the best predicted value of y given that the younger child has an IQ of 104? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is (Round to two decimal places as needed.)
To find the best predicted value of y given that the younger child has an IQ of 104, we can use the regression equation y = 10.51 + 0.89x, where x represents the IQ score of the younger child. Using this equation, we can substitute x = 104 to calculate the predicted value of y.
Based on the given regression equation, we have y = 10.51 + 0.89x. Substituting x = 104 into the equation, we get:
y = 10.51 + 0.89(104)
y ≈ 10.51 + 92.56
y ≈ 103.07
Therefore, the best predicted value of y, given that the younger child has an IQ of 104, is approximately 103.07. This indicates that the predicted IQ score for the older sibling would be around 103.07, based on the regression model.
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Algebra Tiles, Please help need to turn in today!!
a. The two binomials that are being multiplied in the algebra tiles above include the following: (2x + 3)(3x + 2).
b. The product shown by the algebra tiles include the following: 6x² + 13x + 6.
What is a factored form?In Mathematics and Geometry, a factored form simply refers to a type of polynomial function that is typically written as the product of two (2) linear factors and a constant.
Part a.
By critically observing the base of this algebra tiles, we can logically deduce that it is composed of two x tiles and three 1 tiles. This ultimately implies that, the base represents (2x + 3).
The height of this algebra tiles is composed of three x tiles and two 1 tiles. This ultimately implies that, the base represents (3x + 2).
Part b.
Next, we would determine the product of the two binomials as follows;
(2x + 3)(3x + 2) = 6x² + 4x + 9x + 6
6x² + 4x + 9x + 6 = 6x² + 13x + 6.
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Using the Normal Distribution to find the Z-value:
Find the Z-value for the following cumulative areas:
Hint: Read Example 1 on page number 252.
a) A=36.32%
b) A= 10.75%
c) A=90%
d) A= 95%
e) A= 5%
f) A=50%
For more precise values, you can use statistical software or online calculators that provide a more extensive range of Z-values.
To find the Z-value for a given cumulative area using the normal distribution, you can use the Z-table or a statistical software. Since I can't provide an interactive table, I'll calculate the approximate Z-values using the Z-table for the provided cumulative areas:
a) A = 36.32%
To find the Z-value for a cumulative area of 36.32%, we need to find the value that corresponds to the area to the left of that Z-value. In other words, we're looking for the Z-value that has an area of 0.3632 to the left of it.
Approximate Z-value: 0.39
b) A = 10.75%
We're looking for the Z-value that has an area of 0.1075 to the left of it.
Approximate Z-value: -1.22
c) A = 90%
We're looking for the Z-value that has an area of 0.9 to the left of it.
Approximate Z-value: 1.28
d) A = 95%
We're looking for the Z-value that has an area of 0.95 to the left of it.
Approximate Z-value: 1.65
e) A = 5%
We're looking for the Z-value that has an area of 0.05 to the left of it.
Approximate Z-value: -1.65
f) A = 50%
The cumulative area of 50% is the median, and since the normal distribution is symmetric, the Z-value will be 0.
Please note that these values are approximate and calculated based on the Z-table. For more precise values, you can use statistical software or online calculators that provide a more extensive range of Z-values.
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The Normal Distribution is a continuous probability distribution with a bell-shaped density function that describes a set of real numbers with the aid of two parameters, μ (the mean) and σ (the standard deviation).The standard normal distribution is a special case of the Normal Distribution.
The Z-score is a statistic that represents the number of standard deviations from the mean of a Normal Distribution.Let's find the Z-values for each given cumulative area:a) A=36.32%The corresponding Z-value can be obtained from the standard Normal Distribution Table or using a calculator.Using the table, we find that the Z-value is approximately 0.385.Using a calculator, we can find the Z-value by using the inverse normal cumulative distribution function (also called the inverse normal CDF or quantile function) with the cumulative area as the input, which gives us:Z = invNorm(0.3632) ≈ 0.385b) A= 10.75%Using the same methods as above, we find that the Z-value is approximately -1.28.Using a calculator, we can find the Z-value by using the inverse normal cumulative distribution function with the cumulative area as the input, which gives us:Z = invNorm(0.1075) ≈ -1.28c) A=90%Using the same methods as above, we find that the Z-value is approximately 1.28.Using a calculator, we can find the Z-value by using the inverse normal cumulative distribution function with the cumulative area as the input, which gives us:Z = invNorm(0.90) ≈ 1.28d) A= 95%Using the same methods as above, we find that the Z-value is approximately 1.64.Using a calculator, we can find the Z-value by using the inverse normal cumulative distribution function with the cumulative area as the input, which gives us:Z = invNorm(0.95) ≈ 1.64e) A= 5%Using the same methods as above, we find that the Z-value is approximately -1.64.Using a calculator, we can find the Z-value by using the inverse normal cumulative distribution function with the cumulative area as the input, which gives us:Z = invNorm(0.05) ≈ -1.64f) A=50%The corresponding Z-value is 0, since the cumulative area to the left of the mean is 0.5 and the cumulative area to the right of the mean is also 0.5. Therefore, we have:Z = 0.
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