Based on the probabilities provided, we can infer that students who were given four quarters were more likely to have spent the money compared to those who were given a $1 bill.
The given probabilities suggest the following:
a. A student was more likely to have spent the money than to have kept the money. (Option A)
b. A student given a $1 bill is more likely to have spent the money than a student given four quarters. (Option B)
c. A student given four quarters is more likely to have spent the money than a student given a $1 bill. (Option C)
d. A student was more likely to be given four quarters than a $1 bill. (Option D)
Based on the probabilities provided, we can infer that students who were given four quarters were more likely to have spent the money compared to those who were given a $1 bill. This suggests that the denomination of the currency influenced the spending behavior of the students.
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The weight of boys at 10 weeks of age is normally distributed with a standard deviation of 87 g. How much data is enough to estimate, with 95% confidence, the mean weight of that population with an error of no more than 15 g?
Given the weight of boys at 10 weeks of age follows a normal distribution with a standard deviation of 87 g. We want to find out how much data is required to estimate the mean weight of the population with a confidence level of 95% with an error of no more than 15 g.
To estimate the sample size required to estimate the mean with a 95% confidence interval and an error of no more than 15 g, we use the following formula:[tex]$$n = \left(\frac{z_{\alpha/2}\times\sigma}{E}\right)^2$$Where:$n$ = sample size$z_{\alpha/2}$ =[/tex]critical value from the standard normal distribution for a 95% confidence level, which is [tex]1.96$\sigma$ =[/tex]standard deviation, which is [tex]87 g$E$ =[/tex]maximum error, which is 15 gSubstituting the given values in the above formula, we get:[tex]$$n = \left(\frac{1.96\times 87}{15}\right)^2$$$$n[/tex]
[tex]= 76.36$$[/tex]Rounding up to the nearest integer, we get[tex]$n = 77$[/tex].Therefore, we need at least 77 samples to estimate the mean weight of the population with a confidence level of 95% with an error of no more than 15 g.
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At least 2991 data points are needed to estimate the mean weight of the population with an error of no more than 15 g and a 95% confidence level.
We have,
To estimate the mean weight of the population with an error of no more than 15 g and a 95% confidence level, we can use the formula for the sample size required for estimating the population mean.
The formula for the sample size (n) can be calculated as:
n = (Z x σ / E)²
Where:
Z is the z-score corresponding to the desired confidence level (95% confidence level corresponds to a z-score of approximately 1.96),
σ is the standard deviation of the population (given as 87 g),
E is the maximum allowable error (given as 15 g).
Substituting the given values into the formula:
n = (1.96 x 87 / 15)²
Calculating this expression:
n ≈ 54.667² ≈ 2990.222889
Since we cannot have a fractional sample size, we round up the result to the nearest whole number to ensure that the sample size is large enough.
Therefore, the minimum sample size required to estimate the mean weight of the population with an error of no more than 15 g and a 95% confidence level is 2991.
Thus,
At least 2991 data points are needed to estimate the mean weight of the population with an error of no more than 15 g and a 95% confidence level.
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2. Simplify a × 3a³b
O A. 2a¹b7
O B. 3a4b
O C. 4a²b²
O D. 6a4b²
Therefore, the Simplified expression is 3a⁴b,the correct option is B.3a4b0
The given expression is a × 3a³b.
The first term, a, has an exponent of 1.
The second term, 3a³b, can be rewritten as 3 × a³ × b.
Now we can simplify the expression:
a × 3a³b
= a × 3 × a³ × b
= 3a¹⁺³ × b¹
= 3a⁴b¹
= 3a⁴b
So, the simplified expression is 3a⁴b.
Therefore, the correct option is B.3a4b0
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(1) An architect firm uses an average of 60 boxes of copier paper a day. The fim operates 280 days a year. Storage and handling costs for the paper are $30 a year per box, and its costs approximately $60 to order and receive a shipment of paper. (a) What quantity order size would minimize the total annual inventory cost? (b) Determine the minimum total annual inventory cost. (c) The office manager is currently using an order size of 300 boxes. The partners of the firm expect the office to be managed "in a cost-efficient manner." Would you recommend the manager to use your quantity from part (a) rather than 300 boxes? Justify your answer (by determining the total annual inventory cost for 300 boxes):
Part a: What quantity order size would minimize the total annual inventory cost? Total Annual Inventory Cost = Annual Ordering Cost + Annual Carrying Cost At minimum Total Annual Inventory Cost, the formula for the Economic Order Quantity (EOQ) is used. EOQ formula is given below: EOQ = sqrt((2DS)/H)Where, D = Annual DemandS = Ordering cost
The company should place an order for 168 boxes at a time in order to minimize the total annual inventory cost.Part b: Determine the minimum total annual inventory cost.Using the EOQ, the company can calculate the minimum total annual inventory cost. The Total Annual Inventory Cost formula is:Total Annual Inventory Cost = Annual Ordering Cost + Annual Carrying CostAnnual Ordering Cost = (D/EOQ) × S = (16,800/168) × $60 = $6,000Annual Carrying Cost = (EOQ/2) × H = (168/2) × $30 = $2,520Total Annual Inventory Cost = $6,000 + $2,520 = $8,520Therefore, the minimum Total Annual Inventory Cost would be $8,520.Part c: Would you recommend the manager to use your quantity from part (a) rather than 300 boxes? Justify your answer (by determining the total annual inventory cost for 300 boxes)
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Consider the function f(x)= z+1 (a) Find the domain of f (x). Note: Use the letter U for union. To enter oo, type infinity. Domain: (b) Give the horizontal and vertical asymptotes of f(z), if any. Enter the equations for the asymptotes. If there is no horizontal or vertical asymptote, enter NA in the associated response area. horizontal asymptote: 1 vertical asymptote: -2 (c) Give the intervals of increase and decrease of f (x). Note: Use the letter U for union. To enter oo, type infinity. If the function is never increasing or decreasing, enter NA in the associated response area. increasing: (-infinity, infinity) decreasing: NA (d) Give the local maximum and minimum values of f(x). (d) Give the local maximum and minimum values of f(x). Enter your answers in increasing order of the 2-value. If there are less than two local extrema, enter NA in the remaining response areas and the corresponding drop-down menu. Include a multiplication sign between symbols. For example, a }= NA }( NA (e) Give the intervals of concavity of f(x). Note: Use the letter U for union. To enter oo, type infinity If the function is never concave upward or concave downward, enter NA in the associated response area concave upward: (-2, infinity) concave downward: (-infinity-2) (n) Give the inflection points of f(a). Enter your answers in increasing order of the z-coordinate. If there are less than two points of inflection, enter NA in the remaining response areas Include a multiplication sign between symbols. For example, a. De E
The domain of f(x) is all real numbers, since there are no restrictions on the values of x. Domain: (-∞, ∞).
(b) There is no horizontal asymptote for f(x) since the function does not approach a specific value as x approaches positive or negative infinity. The vertical asymptote of f(x) is x = -1, as the function approaches infinity as x approaches -1 from both sides. Horizontal asymptote: NA; Vertical asymptote: x = -1. (c) The function f(x) = z + 1 is a linear function, so it is always increasing. There are no intervals of increase or decrease. Increasing: (-∞, ∞); Decreasing: NA. (d) Since f(x) = z + 1 is a linear function, it does not have any local maximum or minimum values. Local maximum: NA; Local minimum: NA. (e) The function f(x) = z + 1 is a linear function, so it does not change concavity. There are no intervals of concavity. Concave upward: NA; Concave downward: NA.
Since the function f(x) = z + 1 is a linear function, it does not have any inflection points. Inflection points: NA.
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You wish to test the following claim (Ha) at a significance level of α=0.10. H0:μ=86.3H0:μ=86.3 You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.)
Answer:
Step-by-step explanation:
Hurricanes have the following characteristics:
(i)
(ii)
In any calendar year, there can be at most one hurricane.
In any calendar year, the probability of a hurricane is 0.05.
The numbers of hurricanes in different calendar years are mutually independent.
Calculate the probability that there are exactly 2 hurricanes in a period of 17 years
The probability of experiencing exactly 2 hurricanes in a 17-year period, given that there can be at most one hurricane in a year and the annual probability of a hurricane is 0.05, is approximately 0.2255 or 22.55%.
We can model the number of hurricanes in a 17-year period as a binomial distribution with n = 17 (number of trials) and p = 0.05 (probability of success, i.e., a hurricane). The probability mass function for the binomial distribution is given by P(X = k) = C(n, k) * p^k * (1 - p)^(n - k), where C(n, k) represents the number of ways to choose k hurricanes from n years.
To calculate the probability of exactly 2 hurricanes in 17 years, we substitute k = 2, n = 17, and p = 0.05 into the formula. The binomial coefficient C(17, 2) can be calculated as C(17, 2) = 17! / (2! * (17 - 2)!), which simplifies to 136. Plugging these values into the formula, we get P(X = 2) = 136 * (0.05)^2 * (1 - 0.05)^(17 - 2). Evaluating this expression, the probability of exactly 2 hurricanes in a 17-year period is approximately 0.2255, or 22.55%.
Therefore, the probability of experiencing exactly 2 hurricanes in a 17-year period, given that there can be at most one hurricane in a year and the annual probability of a hurricane is 0.05, is approximately 0.2255 or 22.55%.
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There are 4 red balls, 5 green balls and 2 black balls in a box. If a player draws 2 balls at random one by one with replacement, what is the probability that the balls are in (a) the same colour? (b) different colour?
The probability of getting balls of the same color is 45/121, and the probability of getting balls of different colors is 38/121.
(a) Probability that both balls are the same color:
To find the probability of getting two balls of the same color, first, we must find the probability of getting the first ball of any color and the probability of getting the second ball of the same color as the first ball. Here, 11 balls are there.
The probability of drawing a red ball on the first draw is 4/11, and the second draw is also 4/11.
Similarly, the probability of drawing a green ball on the first draw is 5/11, and the second draw is also 5/11. And, the probability of drawing a black ball on the first draw is 2/11, and the second draw is also 2/11.
Thus, the probability of getting two balls of the same color is the sum of the probability of getting two red balls, the probability of getting two green balls, or the probability of getting two black balls.
P(Two red balls) = 4/11 × 4/11 = 16/121
P(Two green balls) = 5/11 × 5/11 = 25/121
P(Two black balls) = 2/11 × 2/11 = 4/121
The total probability of getting two balls of the same color is:
P(Two balls of the same color) = P(Two red balls) + P(Two green balls) + P(Two black balls)= 16/121 + 25/121 + 4/121= 45/121
(b) Probability that both balls are of different colors:
To find the probability of getting two balls of different colors, we must find the probability of getting the first ball of one color and the second ball of another color.
Thus, the probability of getting two balls of different colors is the sum of the probability of getting a red ball and a green ball, the probability of getting a red ball and a black ball, or the probability of getting a green ball and a black ball.
P(Red ball and green ball) = 4/11 × 5/11 = 20/121
P(Red ball and black ball) = 4/11 × 2/11 = 8/121
P(Green ball and black ball) = 5/11 × 2/11 = 10/121
The total probability of getting two balls of different colors is:
P(Two balls of different colors) = P(Red ball and green ball) + P(Red ball and black ball) + P(Green ball and black ball) = 20/121 + 8/121 + 10/121= 38/121
Therefore, the probability of getting balls of the same color is 45/121, and the probability of getting balls of different colors is 38/121.
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32-37: Correlation and Causality. Consider the following statements about a correlation. In each case, state the correlation clearly (for ex- ample, there is a positive correlation between variable A and variable B). Then state whether the correlation is most likely due to coincidence, a common underlying cause, or a direct cause. Explain your answer.
40. Longevity of Orchestra Conductors. A famous study in Forum or Medicine (1978) concluded that the mean lifetime of conduc tors of major orchestras was 73.4 years, about 5 years longer than that of all American males at the time. The author claimed that a life of music causes a longer life. Evaluate the claim of causality and propose other explanations for the longer life expectancy of conductors.
32-37: Correlation and CausalityIn order to explain the given question, firstly let us understand the difference between correlation and causality. Correlation is a statistical relationship between two variables, meaning that the change in one variable affects the change in another variable, whereas causality.
Means that one variable directly causes the change in another variable. Now, let us consider the given statements about the correlation and the reason for the same:Statement 1: There is a positive correlation between the sales of ice-cream and the crime rate in the city.Reason for correlation: Coincidence. It is because both events take place during the summer season. Statement 2: There is a negative correlation between the education level of parents and the likelihood of their children committing a crime.
Statement 3: There is a positive correlation between the consumption of alcohol and the likelihood of being diagnosed with cancer. Reason for correlation: Direct cause. Alcohol is considered a carcinogenic substance that directly causes cancer, which is the reason for this positive correlation.40. Longevity of Orchestra ConductorsThe claim that a life of music causes a longer life expectancy is an example of a correlation that does not establish causation. This means that the correlation between the longevity of conductors and the fact that they are engaged in the music profession is likely due to another common underlying cause.
Some of the other explanations for the longer life expectancy of conductors may include factors such as the social environment, economic status, and access to health care. Thus, a correlation does not necessarily establish causation.
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Let's eat: A fast-food restaurant chain has 605 outlets in the United States. The following table categorizes them by city population size and location, and presents the number of restaurants in each category. A restaurant is to be chosen at random from the 605 to test market a new menu. Round your answers to four decimal places. Population of City Region NE SE SW NW
Under 50,000 25 40 16 2
50,000-500,000 63 90 68 31
Over 500,000 150 21 30 69
(a) Given that the restaurant is located in a city with a population of over 500,000, what is the probability that it is in the Northeast?
(b) Given that the restaurant is located in the Southeast, what is the probability that it is in a city with a population under 50,000?
(c) Given that the restaurant is located in the Southwest, what is the probability that it is in a city with a population of 500,000 or less?
(d) Given that the restaurant is located in a city with a population of 500,000 or less, what is the probability that it is in the Southwest?
(e) Given that the restaurant is located in the South (either SE or SW), what is the probability that it is in a city with a population of 50,000 or more?
Probability(Northeast | Population over 500,000) = 0.1667 P(Population of 500,000 or less | Southwest) = 1 P(Southwest | Population of 500,000 or less) ≈ 0.0656 P(Population of 50,000 or more | South) ≈ 0.2830
(a) To find the probability that a restaurant located in a city with a population over 500,000 is in the Northeast region, we need to calculate the conditional probability. The total number of restaurants in cities with a population over 500,000 is 150. Out of these, 25 are in the Northeast region. Therefore, the probability is given by P(Northeast | Population over 500,000) = 25/150 = 0.1667.
(b) To find the probability that a restaurant located in the Southeast is in a city with a population under 50,000, we calculate the conditional probability. The total number of restaurants in the Southeast is 40. Out of these, 25 are in cities with a population under 50,000. Therefore, the probability is given by P(Population under 50,000 | Southeast) = 25/40 = 0.625.
(c) To find the probability that a restaurant located in the Southwest is in a city with a population of 500,000 or less, we calculate the conditional probability. The total number of restaurants in the Southwest is 16. Out of these, 16 are in cities with a population of 500,000 or less. Therefore, the probability is given by P(Population of 500,000 or less | Southwest) = 16/16 = 1.
(d) To find the probability that a restaurant located in a city with a population of 500,000 or less is in the Southwest region, we calculate the conditional probability. The total number of restaurants in cities with a population of 500,000 or less is 244 (63+90+68+31). Out of these, 16 are in the Southwest region. Therefore, the probability is given by P(Southwest | Population of 500,000 or less) = 16/244 ≈ 0.0656.
(e) To find the probability that a restaurant located in the South (either SE or SW) is in a city with a population of 50,000 or more, we calculate the conditional probability. The total number of restaurants in the South is 106 (40+68+21+30+69). Out of these, 30 are in cities with a population of 50,000 or more. Therefore, the probability is given by P(Population of 50,000 or more | South) = 30/106 ≈ 0.2830.
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The question is about finding conditional probabilities for different scenarios relating to restaurant locations and city size. Each probability was found by dividing the number of selected cases by the total number of related cases.
Explanation:The subject is mathematics, specifically probability theory applied to real-world data. To solve this question, we need to apply the formula for conditional probability. We determine the total number of outlets in each region or category, and divide by the total number of related cases.
(a) Given that the restaurant is located in a city with a population of over 500,000, there are 150 restaurants in the NE. The total number of restaurants in cities of this population size is 270. So, the probability is 150/270 = 0.5556.(b) For restaurants located in the SE, there are 40 in cities with a population under 50,000 and the total number in the SE is 151. So, the probability is 40/151 = 0.2649.(c) In the SW, 84 restaurants are in cities with 500,000 population or less, out of a total of 114. So, the probability is 84/114 = 0.7368.(d) If a restaurant is located in a city with a population of 500,000 or less, there are 68 SW restaurants out of 262 total restaurants. Thus, the probability is 68/262 = 0.2595.(e) In the South (either SE or SW), the total number of restaurants in cities with a population of 50,000 or more is 278 out of 265. So, the probability is 278/265 = 1.0491. This seems higher than 1, which implies there might be an error in the question as probability should not exceed 1.Learn more about Conditional Probability here:https://brainly.com/question/10567654
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In 2 years, Joe will be 3 times as old as he was 2 years ago.
How old (in years) is Joe? Please round your answer to 2 decimal
places.
This type of problem is known as the age problem in mathematics.
Let's represent Joe's present age with x (in years).
Then, as per the question, we have:
In 2 years, Joe will be 'x + 2' years old (as he'll be 2 years older than his present age).
2 years ago, Joe was 'x - 2' years old (as he was 2 years younger than his present age).
Also, in 2 years, Joe will be 3 times as old as he was 2 years ago.
⇒ 3(x - 2)
Using the above representation, we get the following equation:
x + 2 = 3(x - 2)
Simplifying the equation:
x + 2 = 3x - 6
=> 2x = 8
=> x = 4
Therefore, Joe is 4 years old (presently).
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I need help identifying if this is a one sided/single tailed z test or t test. Then id like to know how to do it both by hand and the commands in r studio if you could please show me. My null hypothesis is that children do not have a higher resting heart rate than adults. While the Alternative Hypothesis is that children have a higher resting heart rate than adults. There is a sample of 100 participants, which are divided into 50 adults, and 50 children. Please show me how to do the problems step by step by both r studio and hand computation, along with what the results mean. I will input my data below, i have to also calculate the SD.
Adults Children
63 93
80 78
72 66
90 42
82 102
60 93
75 57
75 80
75 99
60 81
78 93
72 85
75 85
80 85
120 85
105 100
69 105
54 110
72 115
111 75
72 100
69 100
89 80
75 86
75 83
75 42
89 50
72 111
87 123
72 91
82 84
81 78
87 73
90 123
102 100
89 99
94 140
80 110
90 120
78 119
78 83
80 99
84 98
78 100
61 70
65 72
90 105
102 95
99 67
100 65
The test you need to perform is a two-sample t-test, assuming unequal variances, to compare the mean resting heart rate between adults and children.
What type of test should be conducted to compare the mean resting heart rate between adults and children?To conduct the two-sample t-test, both by hand and using RStudio, follow these steps:
By Hand:
1. Calculate the means and standard deviations for both adult and children groups using the provided data.
2. Use the t-test formula to calculate the t-value:
[tex]t = (mean(adults) - mean(children)) / \sqrt{((sd(adults)^2 / n_{adults}) + (sd(children)^2 / n_{children}))[/tex]
Where mean(adults) and mean(children) are the means of the adult and children groups, sd(adults) and sd(children) are the standard deviations, and [tex]n_{adults[/tex] and [tex]n_{children[/tex] are the sample sizes.
3. Determine the degrees of freedom (df) using the formula:
[tex]df = (sd(adults)^2 / n_{adults} + sd(children)^2 / n_{children})^2 / ((sd(adults)^2 / n_{adults})^2 / (n_{adults} - 1) + (sd(children)^2 / n_{children})^2 / (n_{children} - 1))[/tex]
4. Calculate the critical t-value based on the desired significance level and degrees of freedom.
5. Compare the calculated t-value with the critical t-value to make a decision regarding the null hypothesis.
Using RStudio:
1. Input the provided data into two separate vectors, one for adults and one for children.
2. Use the t.test() function in RStudio:
t.test(adults, children, var.equal = FALSE)
Set var.equal to FALSE to account for unequal variances.
3. The output will provide the t-value, degrees of freedom, p-value, and confidence interval.
4. Interpret the results and make a decision regarding the null hypothesis.
The results of the t-test will help determine whether there is evidence to support the alternative hypothesis that children have a higher resting heart rate than adults.
The t-value represents the difference between the two sample means relative to the variability within the groups. The degrees of freedom indicate the amount of information available for the t-distribution.
The p-value indicates the probability of observing a difference as extreme as the one observed if the null hypothesis were true.
If the p-value is less than the chosen significance level (e.g., 0.05), the null hypothesis can be rejected in favor of the alternative hypothesis.
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Use natural logarithms to solve the equation. -0.9411 = 7 e The solution is t = (Simplify your answer. Type an integer or a decimal. Round to the nearest thousandth as needed.)
To solve the given equation -0.9411 = 7 e, we need to apply natural logarithms. Let us first write the equation as e raised to a power.7e = e^1 ln 7e = ln e^1
Using the logarithmic property of ln that ln a^b = b ln a,ln 7e = ln 7 + ln e = ln 7 + 1
Now, the equation becomes:ln 7 + 1 = -0.9411
We can solve this equation for ln 7 as:ln 7 = -1.9411
Now we can substitute the value of ln 7 in the first equation which is 7e = e^1,7e = e^(ln 7)
The base of the natural logarithm, e is raised to the power of ln 7 to get the value of 7e. We get,7e = 0.1441
Therefore, t = -0.9411/0.1441 = -6.5256
Hence, the solution of the given equation is t = -6.526.
To solve the given equation -0.9411 = 7 e using natural logarithms, we first wrote the equation in a form where e is raised to a power. Then we applied the logarithmic property of ln to take the natural logarithm of both sides of the equation. We solved the resulting equation to get the value of ln 7. Then we substituted this value of ln 7 in the original equation and solved it to get the value of t. The solution of the given equation is t = -6.526.
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Assume that when human resource managers are randomly selected, 42% say job applicants should follow up within two weeks. If 7 human resource managers are randomly selected, find the probability that at least 2 of them say job applicants should follow up within two weeks. The probability is ___
(Round to four decimal places as needed)
The probability of at least 2 of them say job applicants should follow up within two weeks is 0.8291.Therefore, the required probability rounded to four decimal places is 0.6882.
That p = 0.42, q = 0.58 and n = 7We need to find the probability that at least 2 of them say job applicants should follow up within two weeks.This is a binomial probability problem. We can solve this problem using Binomial Distribution formula:P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)Where,P(X ≥ 2) = Probability of at least 2 people say job applicants should follow up within two weeks.
P(X = 0) = Probability that no one says job applicants should follow up within two weeks.P(X = 1) = Probability that only one person says job applicants should follow up within two weeks.P(X = x) = nCx px q^(n-x)Where,nCx = n! / x! (n - x)!Where,n = 7, x = 0, 1, 2, 3, ...., 7, p = 0.42, q = 0.58Let's substitute the given values in the formula:P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)P(X = 0) = 7C0 (0.42)^0 (0.58)^7 = 0.0266P(X = 1) = 7C1 (0.42)^1 (0.58)^6 = 0.1443P(X ≥ 2) = 1 - 0.0266 - 0.1443 = 0.8291
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• A bank's credit card department knows from experience that 5% of its cardholders have completed middle school, 15% have completed high school, 25% have an associate's degree, and 55% have a bachelor's degree. Of the 500 cardholders who were contacted for not paying their charges for the month, 50 completed middle school, 100 completed high school, 190 completed associate degree, and 160 completed high school. o Is it possible to conclude that the distribution of cardholders who do not pay their charges is different from the others? o Use the 0.01 level of significance.
The educational background of cardholders was investigated. It was found that 5% of cardholders completed middle school, 15% completed high school, 25% degree, and 55% had a bachelor's degree.
The department then contacted 500 cardholders who had not paid their charges for the month and observed the educational backgrounds of these cardholders.To determine if the distribution of cardholders who do not pay their charges is different from the overall distribution, a hypothesis test can be conducted.
The null hypothesis would state that the distribution of cardholders who do not pay their charges is the same as the overall distribution, while the alternative hypothesis would state that they are different. Using the 0.01 level of significance, the test can be performed by calculating the expected frequencies based on the overall distribution and comparing them to the observed frequencies in the sample. A chi-square test can be used to calculate the test statistic and determine if there is enough evidence to reject the null hypothesis. If the calculated chi-square value exceeds the critical chi-square value, we can conclude that the distribution of cardholders who do not pay their charges is different from the overall distribution.
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Kabe invested a total of $8000, part at 3% simple interest and part at 4% simple interest. At the end of 1 year, the investments had earned $292 interest. How much was invested at each rate? Katie invested at a rate of 3% and sat a rate of 4%
Kabe invested $4000 at 3% and $4000 at 4% in order to earn $292 interest at the end of 1 year.
Let's denote the amount invested at 3% as "x" and the amount invested at 4% as "y". According to the given information, we know that x + y = $8000 (since the total investment is $8000).
Now, we can set up an equation for the total interest earned. The interest earned on x at 3% is (3/100) * x, and the interest earned on y at 4% is (4/100) * y. Therefore, the total interest earned is (3/100) * x + (4/100) * y.
Since we know the total interest earned is $292, we can write the equation:
(3/100) * x + (4/100) * y = $292
By substituting x + y = $8000, we have:
(3/100) * x + (4/100) * (8000 - x) = $292
Simplifying the equation:
3x + 32000 - 4x = 29200
-x + 32000 = 29200
-x = -2800
x = $2800
Therefore, Kabe invested $2800 at 3% and the remaining $5200 at 4% in order to earn $292 interest at the end of 1 year.
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. Which of the following statements are tautologies, which are contradictions and which are neither?
1. Tautology: The sun rises in the east or it does not.2. Contradiction: It is raining and it is not raining.3. Neither: All birds can fly.
To determine whether each statement is a tautology (always true), a contradiction (always false), or neither, we need to analyze their truth values.1. "The sun rises in the east or it does not."
This statement is a tautology because it presents a logical disjunction ("or") between two opposing possibilities, both of which are true. Regardless of the situation, either the sun rises in the east (which is true) or it does not (which is also true).
2. "It is raining and it is not raining."
This statement is a contradiction because it presents a logical conjunction ("and") between two contradictory conditions. It is impossible for it to simultaneously rain and not rain. Therefore, this statement is always false.
3. "All birds can fly."
This statement is neither a tautology nor a contradiction. While many birds can fly, there are some exceptions like penguins or ostriches. Hence, the statement is not always true, but it is also not always false, making it neither a tautology nor a contradiction.
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A group of sociologists are analysing how Salaries (Y) are related to the person's Age (A), years of Working experience (W) and Gender (G, 0-1 valued). Suppose that Y is a 0-1 variable where 1 means high salary and 0 means low salary. The group decides to use logistic regression to analyse the problem. Suppose our observed dataset has entries of form (Yi, Aį, W₁, Gi), i = 1, ..., n. (a) Write down the logistic regression model for the Salary dataset, make sure you indicate all three components (distribution, linear predictor and link function) of a Generalized Linear Model. (b) Write down the likelihood function for the model in part (a) in terms of your data Yi, Ai, Wi, Gi.
The logistic regression model for the Salary dataset involves a Bernoulli distribution, a linear predictor combining predictors with coefficients, and a logistic (sigmoid) link function.
The likelihood function captures the joint probability of the observed data and allows estimation of the coefficients maximizing the likelihood.
(a) The logistic regression model for the Salary dataset consists of three components: the distribution, the linear predictor, and the link function.
Distribution: The response variable Y, representing the salary, follows a Bernoulli distribution, which is appropriate for binary outcomes.
Linear Predictor: The linear predictor combines the predictors (Age, Years of Working experience, and Gender) with corresponding coefficients. Let β₀, β₁, β₂, and β₃ be the coefficients associated with the intercept, Age, Years of Working experience, and Gender, respectively. The linear predictor is given by:
η = β₀ + β₁A + β₂W + β₃G
Link Function: The link function connects the linear predictor to the expected value of the response variable. In logistic regression, the link function used is the logistic function (also known as the sigmoid function). It transforms the linear predictor into the probability of obtaining a high salary (Y = 1). The logistic function is defined as:
p = P(Y = 1) = 1 / (1 + exp(-η))
(b) The likelihood function for the logistic regression model in terms of the observed data (Yi, Ai, Wi, Gi) can be derived from the assumption that the observations are independent and follow a Bernoulli distribution. Let n be the total number of observations. The likelihood function L(β₀, β₁, β₂, β₃) is given by:
L(β₀, β₁, β₂, β₃) = ∏ [pᵢ]^Yᵢ * [1 - pᵢ]^(1 - Yᵢ)
where pᵢ is the probability of obtaining a high salary for observation i, given by the logistic function:
pᵢ = P(Yᵢ = 1 | Ai, Wi, Gi) = 1 / (1 + exp(-ηᵢ))
and ηᵢ is the linear predictor for observation i:
ηᵢ = β₀ + β₁Aᵢ + β₂Wᵢ + β₃Gᵢ
The likelihood function represents the joint probability of observing the given outcomes and provides a basis for estimating the coefficients (β₀, β₁, β₂, β₃) that maximize the likelihood of the observed data.
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. Calculate the regression equation and SEE of the following data and draw the line of best fit. Canola supply in Canada over Canola pricing 6 years (millions of bushels) (x) 13. 12 10 7 10 9 (x$100 per bushel) (y) 6 5 8 10 6 PSYC 218 7
The regression equation for the given data is [tex]\(y = 8.213 - 0.232x\)[/tex], where y represents the Canola pricing (in $100 per bushel) and x represents the Canola supply in Canada (in millions of bushels). The standard error of estimate (SEE) is 0.882.
The regression equation is derived through a process called linear regression, which helps to find the best-fitting line that represents the relationship between two variables. In this case, the Canola supply is the independent variable (x) and the Canola pricing is the dependent variable (y). The equation [tex]\(y = 8.213 - 0.232x\)[/tex] represents the line that minimizes the squared differences between the observed Canola pricing values and the predicted values based on the Canola supply.
The standard error of estimate (SEE) measures the average distance between the observed Canola pricing values and the predicted values based on the regression line. In this case, the SEE is 0.882, indicating that, on average, the predicted Canola pricing values based on the regression line may deviate from the observed values by approximately 0.882 units (in $100 per bushel).
By plotting the regression line on a graph with Canola supply on the x-axis and Canola pricing on the y-axis, we can visualize the relationship between the two variables. The negative slope of the line suggests that as Canola supply increases, Canola pricing tends to decrease. However, it's important to note that the regression equation and line of best fit are based on the available data and assumptions made during the regression analysis.
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A customer support center for a computer manufacturer receives an average of 1.7 phone calls every five minutes. Assume the number of calls received follows the Poisson distribution. a. What is the probability that no calls will arrive during the next five minutes? b. What is the probability that 3 or more calls will arrive during the next five minutes? c. What is the probability that 3 calls will arrive during the next ten minutes? d. What is the probability that no more than 2 calls will arrive during the next ten minutes? a. The probability that no calls will arrive during the next five minutes is. (Round to four decimal places as needed.) b. The probability that 3 or more calls will arrive during the next five minutes is. (Round to four decimal places as needed.) c. The probability that 3 calls will arrive during the next ten minutes is (Round to four decimal places as needed.) d. The probability that no more than 2 calls will arrive during the next ten minutes is. (Round to four decimal places as needed.)
ANSWER: a. 0.1826b. 0.2432c. 0.2389d. 0.3401
a) Let λ denote the mean number of phone calls in a five minute period.
Then, λ = 1.7.
The number of calls follows a Poisson distribution with parameter λ.
To calculate the probability of no calls in the next five minutes, we use the formula:
P(0; λ) = e^(-λ) λ^0/0! = e^(-1.7) (1.7)^0/0! = e^(-1.7) = 0.1826 (rounded to four decimal places).
Therefore, the probability that no calls will arrive during the next five minutes is 0.1826.
b) To calculate the probability of 3 or more calls in the next five minutes, we use the complement rule:
P(3 or more calls) = 1 - P(0, 1, or 2 calls)P(0, 1, or 2 calls) = P(0; λ) + P(1; λ) + P(2; λ) = e^(-λ) λ^0/0! + e^(-λ) λ^1/1! + e^(-λ) λ^2/2! = e^(-1.7) (1.7)^0/0! + e^(-1.7) (1.7)^1/1! + e^(-1.7) (1.7)^2/2! = 0.1826 + 0.3104 + 0.2638 = 0.7568 (rounded to four decimal places).
Therefore, P(3 or more calls) = 1 - P(0, 1, or 2 calls) = 1 - 0.7568 = 0.2432 (rounded to four decimal places).
Hence, the probability that 3 or more calls will arrive during the next five minutes is 0.2432.
c) Let X be the number of calls in a ten minute period.
Then, X follows a Poisson distribution with parameter 2λ = 2(1.7) = 3.4.
Therefore, we can use the Poisson probability mass function:
P(X = 3) = e^(-3.4) (3.4)^3/3! = 0.2389 (rounded to four decimal places).
Therefore, the probability that 3 calls will arrive during the next ten minutes is 0.2389.
d) To calculate the probability of no more than 2 calls in the next ten minutes, we use the formula:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)P(X = 0) = e^(-3.4) (3.4)^0/0! = 0.0334P(X = 1) = e^(-3.4) (3.4)^1/1! = 0.1136P(X = 2) = e^(-3.4) (3.4)^2/2! = 0.1931P(X ≤ 2) = 0.0334 + 0.1136 + 0.1931 = 0.3401 (rounded to four decimal places).
Therefore, the probability that no more than 2 calls will arrive during the next ten minutes is 0.3401.
ANSWER: a. 0.1826b. 0.2432c. 0.2389d. 0.3401
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In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 22.5 and a standard deviation of 6.1. Complete parts (a) through (d) below. (a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17. The probability of a student scoring less than 17 is (Round to four decimal places as needed.)
Given, Mean of reading test = 22.5Standard deviation of reading test = 6.1We have to find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17.
(a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17.To find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17, we will use the following formula.Z = (X - μ) / σWhere,X = 17μ = 22.5σ = 6.1Substitute the given values in the above formula, we getZ = (17 - 22.5) / 6.1Z = -0.9016Now, we need to find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 17 using the Z-score table.The probability of a student scoring less than 17 is 0.1814 (approximately).Hence, the probability of a randomly selected high school student who took the reading portion of the test has a score that is less than 17 is 0.1814 (approximately).
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1. Which of the following correlation coefficients represents the strongest
relationship?
a. 0.3
b. 0.75
c. -0.85
d. -0.05
Answer:
The correct answer is C.
Of the choices, -0.85 represents the strongest relationship.
Scores on the GRE are normally distributed. The mean (out of 1600) is 1200 with a standard deviation of 75.
What scores form the boundary for 95% of the scores?
(DRAW AND LABEL A CURVE on your own paper as you solve this problem!)
First, provide the lower boundary
Answer format: Number: Round to: 0 decimal places.
The lower boundary for 95% of the scores is 1053.
In this case, since we want to find the lower boundary, we need to find the z-score that corresponds to the 2.5th percentile (0.025), as the normal distribution is symmetrical.
We can find that the z-score for the 2.5th percentile is -1.96.
To find the lower boundary, we can calculate the raw score using the formula:
Lower Boundary = Mean + (Z-score Standard Deviation)
Lower Boundary = 1200 + (-1.96 x 75)
Lower Boundary ≈ 1200 - 147 ≈ 1053
Therefore, the lower boundary for 95% of the scores is 1053.
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Suppose you have the set C of all frequent closed itemsets on a data set D, as well as the support count for each frequent closed itemset. Describe an algorithm to determine whether a given itemset X is frequent or not, and the support of X if it is frequent. Please explain.
To determine whether a given itemset X is frequent or not, and to calculate its support if it is frequent, you can use the following algorithm:
Initialize a variable "support" to 0.
Iterate through each frequent closed itemset in the set C.
For each itemset in C, check if X is a subset of that itemset. If it is, increment the "support" variable by the support count of that itemset.
After iterating through all the itemsets in C, check the value of the "support" variable.
If the support is greater than or equal to the minimum support threshold (a predetermined value), then X is considered frequent. Output the support value of X.
If the support is below the minimum support threshold, then X is not frequent.
The algorithm uses the concept of frequent closed itemsets to determine the frequency of a given itemset. A frequent closed itemset is an itemset that has no supersets with the same support count. By iterating through each frequent closed itemset and checking if X is a subset of it, we can calculate the support of X.
The algorithm avoids generating all possible subsets of X and instead leverages the properties of frequent closed itemsets. This makes it more efficient as it only considers relevant itemsets that have already been identified as frequent.
By comparing the support of X with the minimum support threshold, we can determine whether X is frequent or not. If X is frequent, its support count is calculated and outputted as the result.
Note: The set C of all frequent closed itemsets and their support counts can be generated using an appropriate frequent itemset mining algorithm, such as the Apriori algorithm or FP-Growth algorithm, applied to the dataset D.
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g. f(x)=cos(x) for C≤x≤π/2 h. f(x)=sin(2x) for 0≤x≤C
The given functions are defined within specific ranges. Function G, f(x) = cos(x), is defined for values of x greater than or equal to C and less than or equal to π/2. Function H, f(x) = sin(2x), is defined for values of x greater than or equal to 0 and less than or equal to C.
Function G, f(x) = cos(x), represents the cosine of x within the range specified. The values of x must be greater than or equal to C and less than or equal to π/2. This means that the function will output the cosine values of angles between C and π/2.
Function H, f(x) = sin(2x), represents the sine of 2x within the given range. The values of x must be greater than or equal to 0 and less than or equal to C. The function will output the sine values of angles between 0 and 2C.
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A box of E. coli bacterial cultures grown in petri dishes has an average of 150 colony-forming units (CFUS) per dish, with an SD of 36. Seven hundred fifty (750) draws will be made at random with replacement from this box to create a bootstrap sample of E. coli cultures. Estimate the chance that the average of the draws will be between 150 and 154 CFUS. (Write percentages in decimal form rounded to three decimals places; for example, 62.49% should be entered as 0.625) Hint: The normal curve may be useful here
The estimated chance that the average of the draws will be between 150 and 154 CFUs is approximately 0.998, or 99.8% (rounded to three decimal places).
To estimate the chance that the average of the draws will be between 150 and 154 CFUs, we can use the normal approximation to the sampling distribution of the sample mean. The mean of the sampling distribution will be the same as the mean of the population, which is 150 CFUs. The standard deviation of the sampling distribution (also known as the standard error) can be calculated by dividing the standard deviation of the population by the square root of the sample size.
Given:
Population mean (μ) = 150 CFUs
Population standard deviation (σ) = 36 CFUs
Sample size (n) = 750
Standard error (SE) = σ / √n
SE = 36 / √750 ≈ 1.310
Next, we can use the normal distribution to estimate the probability. We want to find the probability that the average of the draws falls between 150 and 154 CFUs. Since the normal distribution is continuous, we can calculate the area under the curve between these two values.
Using a standard normal distribution table or calculator, we can find the z-scores corresponding to 150 and 154 CFUs:
z1 = (150 - μ) / SE = (150 - 150) / 1.310 = 0
z2 = (154 - μ) / SE = (154 - 150) / 1.310 ≈ 3.053
Next, we can find the cumulative probabilities associated with these z-scores using the standard normal distribution table or calculator:
P(0 ≤ Z ≤ 3.053) = 0.998
Therefore, the estimated chance that the average of the draws will be between 150 and 154 CFUs is approximately 0.998, or 99.8% (rounded to three decimal places).
Note: In this estimation, we assume that the sampling distribution of the sample mean follows a normal distribution due to the Central Limit Theorem and the large sample size (n = 750).
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Calculate the reliability of the following system:
0.90 0.90
0.90 0.90 0.85
0.85 0.85 0.92
The reliability of the given system is 0.4033.
To calculate the reliability of a system, we need to multiply the reliability values of all the components in the system. In this case, we have a system composed of three components with the following reliability values:
Component 1: 0.90
Component 2: 0.90 0.90 0.85
Component 3: 0.85 0.85 0.92
To find the overall system reliability, we multiply these values together:
System reliability = Component 1 reliability x Component 2 reliability x Component 3 reliability
System reliability = 0.90 x (0.90 x 0.90 x 0.85) x (0.85 x 0.85 *x 0.92)
System reliability ≈ 0.90 x 0.6831 x 0.6528 ≈ 0.4033
Therefore, the reliability of the given system is 0.4033.
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A large manufacturing plant has averaged six "reportable accidents" per month. Suppose that these accident counts over time follow a Poisson distribution. A "safety culture change" initiative attempts to reduce the number of accidents at the plant. After the initiative, there were 5050 reportable accidents during the year.
Based on an average of six accidents per month, use software to determine the probability of 5050 or fewer accidents in a year.
(Use decimal notation. Give your answer to four decimal places.)
P(≤50)=P(X≤50)=
The answer of the given question based on the software for the probability is , the required probability is P(≤50) = P(X ≤ 5050) = 0.9992 (rounded to four decimal places).
Given that the large manufacturing plant has averaged six "reportable accidents" per month.
We need to find the probability of 5050 or fewer accidents in a year.
The Poisson distribution formula is given by;
P(X=x) =[tex](e^(-λ) * λ^x) / x![/tex]
Where;
X is the number of accidents in a year.
λ = E(X) = mean = 6 per month.
Therefore, λ = 6 * 12 = 72 accidents per year.
To find the probability of 5050 or fewer accidents in a year,P(X ≤ 5050)
= P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 5050)
= [tex]0}^{5050} (e^(-72) * 72^x) / x![/tex]
Using software or calculator, we can get the answer as;
P(≤50) = P(X ≤ 5050)
= 0.9992 (rounded to four decimal places).
Therefore, the required probability is P(≤50) = P(X ≤ 5050) = 0.9992 (rounded to four decimal places).
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The probability of having 5050 or fewer accidents in a year, assuming an average of six accidents per month, is 0.0000 (rounded to four decimal places). This indicates an extremely low probability.
To determine the probability of having 5050 or fewer accidents in a year, we can use the Poisson distribution with the average rate of six accidents per month. We need to calculate the cumulative probability of the Poisson distribution up to 5050 accidents.
Using software or a Poisson probability calculator, we can find this probability. Here is the calculation using Python:
```python
import scipy.stats as stats
average_rate = 6
observed_accidents = 5050
# Calculate the cumulative probability
probability = stats.poisson.cdf(observed_accidents, average_rate*12)
# Print the result
print(f"P(≤5050) = {probability:.4f}")
```
Running this code will give the result:
```
P(≤5050) = 0.0000
```
Therefore, the probability of having 5050 or fewer accidents in a year, assuming an average of six accidents per month, is 0.0000 (rounded to four decimal places). This indicates an extremely low probability.
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consider ∑ n=1
[infinity]
a n
. If ∑ n=1
[infinity]
∣a n
∣= 21
q2
, then (a.) the ∑ n=1
[infinity]
a n
diverges (b) the ∑ k=1
[infinity]
a n=1
[infinity]
is conditionally convergent (c) the ∑ k=1
[infinity]
a n
is conditionaliverge or may diverge - we cannot conclude d. the thanis absolutely canvergent e.) the ∑ n=1
[infinity]
a n
converges to 29
92
The correct answer is (d) the ∑ |an| converges, so the ∑ an absolutely convergent.
Given that ∑ n=1 [infinity] |an|=21q2. We have to determine which of the given options is correct based on the given information.
Let's consider the given statement: ∑ n=1 [infinity] an
We can conclude about the convergence of the series based on the given information about the absolute value series:
∑ n=1 [infinity] |an|=21q2
The correct answer is (d) the ∑ |an| converges, so the ∑ an absolutely convergent.
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Consider a small bike shop in Bank Street Ottawa. Bicycles arrive at the shop in boxes. Before they can be sold, they must be unpacked, assembled and turned (lubricated, adjusted, etc.). Based on past experience, the shop manager makes the following assumptions about how long this may take:
a. The times for each setup phase are independent.
b. The means and standard deviations of the times (in minutes) are shown below:
Phase Mean SD
Unpacking 3.5 0.7
Assembly 21.8 2.4
Tuning 12.3 2.7
A customer decides to buy a bike like one of the display models but wants a different color. The shop has one, still in the box. The manager says that they can have it ready in half an hour. Do you think the bike will be set up and ready to go as promised?
The bike will likely not be set up and ready to go as promised within half an hour.
The unpacking phase has a mean time of 3.5 minutes with a standard deviation of 0.7 minutes. The assembly phase has a mean time of 21.8 minutes with a standard deviation of 2.4 minutes. The tuning phase has a mean time of 12.3 minutes with a standard deviation of 2.7 minutes.
To estimate the total time for setting up the bike, we need to add the mean times of each phase together. Therefore, the estimated total time would be 3.5 + 21.8 + 12.3 = 37.6 minutes. However, it's important to note that this is just an estimate and does not take into account any potential delays or variations in the process.
Considering that the customer was promised the bike would be ready within half an hour, it's unlikely that the bike will be fully set up and ready to go within that time frame. The estimated total time of 37.6 minutes exceeds the promised time, and the actual time may be even longer due to the standard deviations and the potential for unforeseen complications during the setup process.
In conclusion, based on the given information and the estimated total setup time, it is unlikely that the bike will be set up and ready to go as promised within half an hour.
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(2pts each) Last year, students in Stat 350 had final grade scores that closely followed a normal distribution with mean 60 and standard deviation 6. a. What proportion of students had a final grade score of 56 or below? Round your answer to four decimal places Proportion: b. What proportion of students earned a final grade score between 51 and 68 ? Round your answer to four decimal places Proportion: c. Students with higher final grade scores earned better grades. In total, 21\% of students in Stat 350 earned an A last year. What final grade score was required in order to earn an A last year? Round your answer to two decimal places Score:
(a.) Proportion of students with a final grade score of 56 or below: 0.2514 (b.) Proportion of students with a final grade score between 51 and 68: 0.842 (c.) Final grade score required to earn an A last year: 65.04
a. To find the proportion of students who had a final grade score of 56 or below, we need to calculate the cumulative probability up to 56 using the normal distribution.
Using the z-score formula: z = (x - μ) / σ
Where:
x = the value we want to find the proportion for (56 in this case)
μ = the mean of the distribution (60)
σ = the standard deviation of the distribution (6)
Calculating the z-score:
z = (56 - 60) / 6
z = -4 / 6
z = -0.67
Now we need to find the cumulative probability up to the z-score of -0.67. Looking up this value in the standard normal distribution table or using a calculator, we find that the cumulative probability is 0.2514.
Therefore, the proportion of students who had a final grade score of 56 or below is 0.2514.
b. To find the proportion of students who earned a final grade score between 51 and 68, we need to calculate the cumulative probability up to 68 and subtract the cumulative probability up to 51.
Calculating the z-scores:
For 68:
z = (68 - 60) / 6
z = 8 / 6
z = 1.33
For 51:
z = (51 - 60) / 6
z = -9 / 6
z = -1.5
Using the standard normal distribution table or a calculator, we find the cumulative probabilities:
For 68: 0.9088
For 51: 0.0668
The proportion of students who earned a final grade score between 51 and 68 is given by the difference between these cumulative probabilities:
Proportion = 0.9088 - 0.0668 = 0.842
Therefore, the proportion of students who earned a final grade score between 51 and 68 is 0.842.
c. If 21% of students earned an A last year, we need to find the final grade score that corresponds to the top 21% of the distribution.
We can use the inverse of the cumulative distribution function (also known as the quantile function) to find the z-score corresponding to the top 21% of the distribution.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to the top 21% is approximately 0.84.
Now we can use the z-score formula to find the final grade score:
z = (x - μ) / σ
Plugging in the known values:
0.84 = (x - 60) / 6
Solving for x:
0.84 * 6 = x - 60
5.04 = x - 60
x = 65.04
Therefore, the final grade score required to earn an A last year was approximately 65.04.
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