The area of quadrilateral $ABCD$, formed by slicing a cube with side length $1$ through the vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges, can be calculated.
In the following explanation, the process of determining the area will be elaborated.
When a cube is sliced by the given plane passing through vertices $A$ and $C$ and midpoints $B$ and $D$, quadrilateral $ABCD$ is formed. It is helpful to visualize the cube in three dimensions to understand the arrangement of points $A$, $B$, $C$, and $D$.
Since the side length of the cube is $1$, the diagonal of each face of the cube is also $1$. Therefore, the diagonal $AC$ passing through the cube's center is also $1$. Additionally, $AB$ and $CD$ are halves of the cube's face diagonals and have lengths $\frac{\sqrt{2}}{2}$.
Quadrilateral $ABCD$ can be divided into two right triangles, $\triangle ABC$ and $\triangle ADC$. Both triangles are congruent and have a base length of $\frac{\sqrt{2}}{2}$ and a height of $\frac{1}{2}$. The area of each triangle is $\frac{1}{2} \times \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{2}}{8}$.
Since quadrilateral $ABCD$ consists of two congruent triangles, its total area is twice the area of one of the triangles. Thus, the area of quadrilateral $ABCD$ is $\frac{\sqrt{2}}{8} + \frac{\sqrt{2}}{8} = \frac{\sqrt{2}}{4}$.
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\( \int \frac{x^{2}}{\sqrt{9-x^{2}}} d x \)
The given integral is given by:[tex]\[\int \frac{x^{2}}{\sqrt{9-x^{2}}} d x\][/tex]. Substituting x = 3 sin θ, we have: dx = 3 cos θ dθ∴
[tex]\[\int \frac{x^{2}}{\sqrt{9-x^{2}}} d x=\int \frac{9\sin^{2}θ}{\sqrt{9-9\sin^{2}θ}}\times 3cosθdθ\][/tex]
On solving, we get
[tex]\[I = -9\int \cos^{2}θdθ\][/tex]
Using the identity,[tex]\[\cos^{2}\theta=\frac{\cos2\theta+1}{2}\][/tex]
We have
[tex]\[I=-\frac{9}{2}\int(\cos 2\theta+1)d\theta\]\[I=-\frac{9}{2}\times \left[ \frac{1}{2} \sin 2 \theta + \theta\right] + c\][/tex]
Substituting back for x, we have
[tex]\[I=-\frac{9}{2}\times \left[ \frac{1}{2} \sin 2 \sin^{-1}\left(\frac{x}{3}\right) + \sin^{-1}\left(\frac{x}{3}\right)\right] + c\]\[=\underline{\mathbf{\frac{-9x\sqrt{9-x^{2}}}{2}+9\sin^{-1}\left(\frac{x}{3}\right)+c}}\][/tex]
The conclusion is the antiderivative of [tex]\[\frac{x^{2}}{\sqrt{9-x^{2}}}\]is equal to \[\frac{-9x\sqrt{9-x^{2}}}{2}+9\sin^{-1}\left(\frac{x}{3}\right)+c\].[/tex]
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Using the data presented in the Table below:
a) Construct a correlation matrix between x1, x2, ×3, and y? Is
there any evidence that multicollinearity exists?
b) Determine the multiple regression line with x1, x2 and x3 as
explanatory variables.
c) Comment on F-test statistic and t-test statistics.
x1 x2 x3 Y
0.8 2.8 2.5 11.0
3.9 2.6 5.7 10.8
1.8 2.4 7.8 10.6
5.1 2.3 7.1 10.3
4.9 2.5 5.9 10.3
8.4 2.1 8.6 10.3
12.9 2.3 9.2 10.0
6.0 2.0 1.2 9.4
14.6 2.2 3.7 8.7
9.3 1.1 5.5 8.7
Answer:
1.8 2.4 7.8 10.6
Step-by-step explanation:
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Find the median from the following data set: 58, 12, 78, 73, 2,
8, 79, 36, 62, 97, 46 & 90.
Leave the answer to the nearest tenth place if applicable.
The median of the given data set is 60 is the answer.
To find the median of a data set, we need to arrange the numbers in ascending order and determine the middle value. If there is an odd number of values, the median is the middle value. If there is an even number of values, the median is the average of the two middle values.
Arranging the given data set in ascending order:
2, 8, 12, 36, 46, 58, 62, 73, 78, 79, 90, 97
Since there are 12 values, which is an even number, we need to find the average of the two middle values.
The middle values are the 6th and 7th numbers: 58 and 62.
To find the median, we calculate the average of these two values:
Median = (58 + 62) / 2 = 60
Therefore, the median of the given data set is 60.
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HMK Company is producing refining and packaging raw sugar into two different types, (Whity X) and (Browny Y). How much is the maximum approximate profit that the company can attain given the following Profit equation 3X + 2Y Production constraints: 10X+4Y 20 4Y ≤ 16 X ≤ 1 12.5 13.5 9.8 12.8
We are given the following information: HMK Company is producing refining and packaging raw sugar into two different types, (Whity X) and (Browny Y).
The profit equation is given as: Profit equation: 3X + 2YLet's solve the production constraints one by one:
Constraint 1: 10X + 4Y ≤ 20Let's plot this inequality on a graph:
graph{(y(-2/5))[-5,5](-10x+20)/4 [-5,5]}As we can see from the graph, the feasible region for this constraint is the triangular area below the line 10X + 4Y = 20,
and to the left of the y-axis and below the x-axis.
Constraint 2: 4Y ≤ 16Let's plot this inequality on a graph: graph{(y(-4))[-5,5](-10x+20)/4 [-5,5]}
As we can see from the graph, the feasible region for this constraint is the rectangular area to the left of the y-axis and below the line Y = 4.Constraint 3: X ≤ 1
Let's plot this inequality on a graph:
graph{(y(-5))[-5,5]x<=1 [-5,5]}As we can see from the graph, the feasible region for this constraint is the triangular area below the line X = 1,
and to the left of the y-axis and above the x-axis.
Constraint 4: Y ≤ 12.5Let's plot this inequality on a graph: graph {(y(-15))[-5,5]x<=1 [-5,5]}
As we can see from the graph, the feasible region for this constraint is the rectangular area to the left of the y-axis and below the line Y = 12.5.Constraint 5: 9.8 ≤ X ≤ 12.8
Let's plot this inequality on a graph: graph{(y(-15))[-5,5]9.8<=x<=12.8 [-5,5]}As we can see from the graph, the feasible region for this constraint is the rectangular area between the vertical lines X = 9.8 and X = 12.8,
to the left of the y-axis.
Now, let's find the coordinates of the corners of the feasible region (where all the constraints intersect):graph{(y(-15))[-5,5]9.8<=x<=12.8 [-5,5]}
As we can see from the graph, the coordinates of the corners of the feasible region are:(9.8, 0), (10, 1.5), (12.5, 12.5), (12.8, 0)Now, let's calculate the profit for each corner of the feasible region
:Corner 1: (9.8, 0)Profit = 3(9.8) + 2(0) = 29.4Corner 2: (10, 1.5)Profit = 3(10) + 2(1.5) = 32Corner 3: (12.5, 12.5)Profit = 3(12.5) + 2(12.5) = 50Corner 4: (12.8, 0)Profit = 3(12.8) + 2(0) = 38.4
Therefore, the maximum approximate profit that the company can attain is $50. Ans: 50.
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Suppose X is a binomial random variable such that n = 15 and p = 0.33, then, P(X= 8) is O 0.6781 O 0.8862 O 0.0549 O 0.5000 Câu hỏi 9 P(X is at least 7, (x> 7) is O 0.916 O 0.157 O 0.195 O 0.083 3
The probability of obtaining exactly 8 successes in a binomial distribution with n = 15 trials and p = 0.33 is 0.157. This means that the probability of getting exactly 8 successes out of 15 trials is approximately 0.157.
To calculate this probability, we can use the binomial probability formula:
[tex]\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \][/tex]
Where [tex]\( \binom{n}{k} \)[/tex] represents the number of ways to choose k successes out of n trials, [tex]\( p^k \)[/tex] is the probability of success raised to the power of k, and [tex]\( (1-p)^{n-k} \)[/tex] is the probability of failure raised to the power of (n-k).
Plugging in the values, we have:
[tex]\[ P(X = 8) = \binom{15}{8} \cdot 0.33^8 \cdot (1-0.33)^{15-8} \approx 0.157 \][/tex]
Therefore, the probability of obtaining exactly 8 successes is approximately 0.157.
Now, let's calculate the probability of X being at least 7 (X > 7). This can be found by summing the probabilities of X being 7, 8, 9, 10, 11, 12, 13, 14, and 15.
P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)
To calculate each individual probability, we use the same binomial probability formula. After calculating each term and summing them up, we find that P(X > 7) is approximately 0.916.
Therefore, the probability of X being at least 7 is approximately 0.916.
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Total blood volume (in ml) per body weight (in kg) is important in medical research. For healthy adults, the red blood cell volume mean is about 28 ml/kg.1 Red blood cell volume that is too low or too high can indicate a medical problem. Suppose that Roger has had seven blood tests, and the red blood cell volumes were as follows. 31 24 43 37 29 38 28 Let x be a random variable that represents Roger's red blood cell volume. Assume that x has a normal distribution and a 4.75. Do the data indicate that Roger's red blood cell volume is different (either way) from - 28 ml/kg?
Based on the hypothesis testing evidence concludes that Roger's red blood cell volume is significantly different from 28 ml/kg.
To determine if Roger's red blood cell volume is different from the mean value of 28 ml/kg, we can perform a hypothesis test using the given data.
Null Hypothesis (H₀): Roger's red blood cell volume is not different from 28 ml/kg.
Alternative Hypothesis (H₁): Roger's red blood cell volume is different from 28 ml/kg.
We can use a t-test to compare the sample mean of Roger's red blood cell volumes with the hypothesized mean of 28 ml/kg.
Using the given data: 31, 24, 43, 37, 29, 38, 28
Calculate the sample mean (X) and sample standard deviation (s) of the data.
X = (31 + 24 + 43 + 37 + 29 + 38 + 28) / 7 = 32.71
s = √[(31-32.71)² + (24-32.71)² + (43-32.71)² + (37-32.71)² + (29-32.71)² + (38-32.71)² + (28-32.71)²] / (7-1) ≈ 6.96
Calculate the t-value using the formula:
t = (X - μ) / (s / √n)
where μ is the hypothesized mean (28 ml/kg), n is the sample size (7).
t = (32.71 - 28) / (6.96 / √7) ≈ 0.88
Determine the critical t-value for a given significance level (α) and degrees of freedom (df = n - 1). Let's assume a significance level of 0.05 and df = 6 (since n = 7).
Using a t-table or statistical software, the critical t-value for a two-tailed test is approximately ±2.447.
Compare the calculated t-value with the critical t-value.
|t| < critical t-value implies that there is not enough evidence to reject the null hypothesis. In this case, |0.88| < 2.447, so we fail to reject the null hypothesis.
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A force of 12lb is required to hold a spring stretched 1/2ft beyond its natural length. Which of the following integrals represents the amount of work done in stretching it from its natural length to 2/3ft beyond its natural length? ∫ 0
2/3
24xdx ∫ 0
2/3
36xdx ∫ 0
2/3
12dx ∫ 1/2
2/3
24xdx ∫ 1/2
2/3
36xdx
The correct option is ∫ 0
2/3
24xdx.
To determine the integral that represents the amount of work done in stretching the spring from its natural length to 2/3 ft beyond its natural length, we need to consider the relationship between the force and displacement.
The work done by a force is given by the integral of the force multiplied by the displacement. In this case, the force required to hold the spring stretched 1/2 ft beyond its natural length is 12 lb. We can assume that the force is proportional to the displacement.
Let's denote the displacement of the spring as x (measured in feet) from its natural length. The force required to stretch the spring at any given displacement x is given by:
F(x) = kx
where k is the spring constant. Since we know that a force of 12 lb is required to hold the spring stretched 1/2 ft beyond its natural length, we can use this information to determine the value of k:
F(1/2) = k(1/2) = 12
k = 24 lb/ft
Now, to find the work done in stretching the spring from its natural length to 2/3 ft beyond its natural length, we need to integrate the force F(x) over the displacement range [0, 2/3]:
Work = ∫(0 to 2/3) F(x) dx
Substituting the force equation, we have:
Work = ∫(0 to 2/3) (24x) dx
Integrating this expression yields:
Work = 12x^2 ∣(0 to 2/3)
Work = 12 * (2/3)^2 - 12 * (0)^2
Work = 12 * (4/9)
Work = 16/3 lb-ft
Comparing this result to the given options, we find that the integral representing the amount of work done in stretching the spring from its natural length to 2/3 ft beyond its natural length is:
∫ 0
2/3
24xdx
Therefore, the correct option is ∫ 0
2/3
24xdx.
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The average annual membership fee at a random sample of 200 sports clubs in the south-west region of a country is RM 250 with a standard deviation of RM 45 . The average annual membership fee of a random sample of 150 sports clubs in the northeast region is RM 220 with a standard deviation of RM 55. Test the null hypothesis that average sports club membership fees are the same in both regions at 10% level of significance.
The average annual membership fee at a random sample of 200 sports clubs in the south-west region of a country is RM 250 with a standard deviation of RM 45 . The average annual membership fee of a random sample of 150 sports clubs in the northeast region is RM 220 with a standard deviation of RM 55.
There is a difference in the average sports club membership fees between the southwest and northeast regions at the 10% level of significance.
To test the null hypothesis that the average sports club membership fees are the same in both regions, we can use a two-sample t-test.
1: The null and alternative hypotheses:
Null hypothesis (H₀): The average sports club membership fees are the same in both regions.
Alternative hypothesis (H₁): The average sports club membership fees are different in the two regions.
2: Set the significance level:
The significance level (α) is given as 10% or 0.1.
3: Compute the test statistic:
We can use the two-sample t-test formula to calculate the test statistic:
t = (x₁ - x₂) / √((s₁² / n₁) + (s₂² / n₂))
where:
x₁ = sample mean of the south-west region = RM 250
x₂ = sample mean of the northeast region = RM 220
s₁ = standard deviation of the south-west region = RM 45
s₂ = standard deviation of the northeast region = RM 55
n₁ = sample size of the south-west region = 200
n₂ = sample size of the northeast region = 150
The test statistic value:
t = (250 - 220) / √((45² / 200) + (55² / 150))
= 30 / √(10.125 + 20.167)
= 30 / √(30.292)
≈ 30 / 5.503
≈ 5.450
4: Determine the critical value:
Since our alternative hypothesis is that the average fees are different, we will perform a two-tailed test. With a 90% confidence level, the critical value can be found by looking up the t-distribution table or using statistical software. For a two-tailed test and 90% confidence, the critical value is approximately ±1.645.
5: Compare the test statistic with the critical value:
Our test statistic t is approximately 5.450, which is greater than the critical value of ±1.645.
Since the test statistic is in the critical region (beyond the critical value), we reject the null hypothesis. This means there is evidence to support the claim that the average sports club membership fees are different in the two regions.
Therefore, we conclude that there is a significant difference in the average sports club membership fees between the southwest and northeast regions at the 10% level of significance.
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The following set of data is from a sample of n=6. 6 3 8 6 10 13 a. Compute the mean, median, and mode. b. Compute the range, variance, standard deviation, and coefficient of variation. c. Compute the Z scores. Are there any outliers? d. Describe the shape of the data set. a. Compute the mean, median, and mode. Mean = (Type an integer or decimal rounded to four decimal places as needed.) Compute the median Median = (Type an integer or a decimal. Do not round.) What is the mode? Select the correct choice below and, if necessary, fill in the answer box to complete your choice O A The mode(s) is/are (Type an integer or a decimal. Do not round. Use a comma to separate answers as needed.) OB. There is no mode for this data set. Click to select your answer(s) The following set of data is from a sample of n=6. 6 3 8 6 10 13 a. Compute the mean, median, and mode. b. Compute the range, variance, standard deviation, and coefficient of variation. c. Compute the Z scores. Are there any outliers? d. Describe the shape of the data set. b. Compute the range. Range = (Type an integer or a decimal. Do not round.) Compute the variance. S = (Round to three decimal places as needed.) Compute the standard deviation S= (Round to three decimal places as needed.) Compute the coefficient of variation CV = % (Round to three decimal places as needed.) Click to select your answers The following set of data is from a sample of n=6. 6 3 8 6 10 13 S a. Compute the mean, median, and mode. b. Compute the range, variance, standard deviation, and coefficient of variation. c. Compute the Z scores. Are there any outliers? d. Describe the shape of the data set. c. Compute the Z scores. (Round to three decimal places as needed.) Data (X) Z Score co 13 Are there any outliers? Click to select your answer(s) The following set of data is from a sample of n=6. 6 3 8 6 10 13 a. Compute the mean, median, and mode. b. Compute the range, variance, standard deviation, and coefficient of variation. c. Compute the Z scores. Are there any outliers? d. Describe the shape of the data set. Are there any outliers? Yes No O d. What is the shape of the data set? Positive (right-skewed) because the mean is greater than the median Symmetric because the mean is equal to the median Negative (left-skewed) because the mean is less than the median Click to select your answer(s)
a. The mean of the data set is 7.67. The median is 7, and there is no mode since no value appears more than once.
b. The range of the data set is 10 (13 - 3). The variance is 13.47, the standard deviation is 3.671, and the coefficient of variation is 47.89%.
c. To compute the Z scores, we need to subtract the mean from each data point and divide the result by the standard deviation. The Z scores for the data set are as follows: -0.983, -1.643, 0.328, -0.983, 0.656, and 1.624. There are no extreme outliers in the data set, as all the Z scores are within a reasonable range.
d. The shape of the data set can be described as positive (right-skewed) because the mean is greater than the median. This indicates that there are a few larger values that pull the mean towards the right side of the distribution. The median, being less influenced by extreme values, is a better representation of the typical value in this case.
a. The mean is calculated by summing all the values in the data set and dividing by the number of observations (6 in this case). The median is the middle value when the data set is arranged in ascending order. The mode is the value that appears most frequently, but in this case, no value is repeated.
b. The range is found by subtracting the minimum value from the maximum value in the data set. Variance measures the average squared deviation from the mean, while the standard deviation is the square root of the variance. The coefficient of variation is the ratio of the standard deviation to the mean, expressed as a percentage.
c. Z scores are calculated to determine how many standard deviations away from the mean each data point is. Outliers are typically considered to be data points that have Z scores greater than 3 or less than -3. In this case, all the Z scores are within a reasonable range, indicating no extreme outliers.
d. The shape of the data set is determined by comparing the mean and median. If the mean is greater than the median, it suggests a positive (right-skewed) distribution, where a few larger values pull the mean towards the right. In this case, the mean is greater than the median, indicating a positive skew.
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In the 2015 season, the NY Jets scored the following: 31 20 17 27 34 23 20 18 17 17 38 23 30 19 26 17
a. Construct a 90% confidence interval on the NY Jets average season points scored.
b. What is the likelihood the NY Jets will have a season with an average score greater than 30 points?
To construct a 90% confidence interval on the NY Jets average season points scored, we first need to find the sample mean, the sample standard deviation and the sample size. Sample Mean of the NY Jets average season points scored is given by is the individual score of the i-th game and n is the sample size.
The Sample Standard Deviation (S) of the NY Jets average season points scored is given by Now we need to find the Standard Error (SE) of the sample mean which is given by: Using the Z-score table for 90% confidence interval, we get the Z-score as 1.645.The formula for confidence interval is given by: Therefore, the 90% confidence interval on the NY Jets average season points scored is given by.
The likelihood the NY Jets will have a season with an average score greater than 30 points can be found by using the Z-score formula: We can use the sample mean and the sample standard deviation $S$ instead of the population mean and standard deviation respectively to estimate the probability of having an average score greater than 30.We already calculated that the sample mean. Therefore, the likelihood the NY Jets will have a season with an average score greater than 30 points is 99.92%.
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A control chart is to be established on a process producing refrigerators. The inspection unit is one refrigerator, and a common chart for nonconformities is to be used. As preliminary data, 16 non- conformities were counted in inspecting 30 refrigerators. Samples on the control limits are regarded in-control. (a) Find two-sigma control limits (b) Find the α error for the control chart with two-sigma control limits (c) Find the β error for the chart with two-sigma control limits if the average number of defects is actually two (i.e., c=2 ) (d) Find the ARL if c=2.01
The standard deviation is calculated using the formula sqrt(p(1-p)/n), where p is the proportion of non-conformities (16/30) and n is the number of refrigerators inspected (30).
Plugging in the values, we get sqrt((16/30)(14/30)/30) = 0.0971. The two-sigma control limits are then calculated by multiplying the standard deviation by 2 and adding/subtracting the result from the average number of non-conformities: Upper Control Limit = 0.5333 + (2 * 0.0971) = 0.7275, Lower Control Limit = 0.5333 - (2 * 0.0971) = 0.3391.
(b) The α error for the control chart with two-sigma control limits represents the probability of a false alarm, i.e., detecting an out-of-control condition when the process is actually in control. The α error is typically set as the significance level, which determines the probability of rejecting the null hypothesis (in this case, the process being in control) when it is true. In this case, since we are using two-sigma control limits, the α error would correspond to the area under the normal distribution curve outside the control limits, which is approximately 0.046 or 4.6%.
(c) The β error for the chart with two-sigma control limits represents the probability of not detecting an out-of-control condition when the process is actually out of control. To calculate the β error, we need to know the average number of defects (c = 2) and the parameters of the distribution. However, the parameters are not provided in the given information, so it is not possible to calculate the β error without further details.(d) The ARL (Average Run Length) represents the average number of samples that need to be taken before an out-of-control condition is detected. It is calculated as 1/α, where α is the probability of a false alarm. In this case, the ARL would be approximately 1/0.046, which is approximately 21.74 samples.Note: Without specific information about the distribution of defects and the parameters, some calculations (such as β error) cannot be determined in this scenario.
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Given P(A) = 0.10, P(B) = 0.70, P(C) = 0.38 and that events A, B, and C are independent, what is P(A, B, and C).
Answer in decimal form. Round to 3 decimal places as needed.
Your Answer:
Given P(E or F) = 0.87, P(E) = 0.13, and P(E and F) = 0.08, what is P(F)?
Given P(E or F) = 0.86, P(F) = 0.26, and P(E and F) = 0.18, what is P(E)?
Given P(E) = 0.28, P(F) = 0.22, and P(E and F) = 0.03, what is P(E or F)?
Given P(E) = 0.26, what is P(E')?
Based on a study from the Chronicles of Flippin'' Awesomeness, the probability that Napoleon and Pedro make it to their first period class on time is 0.39. The probability that they make it to their first period class on time, given that they catch the bus is 0.57. The probability that Napoleon and Pedro catch the bus and make it to their first period class on time is 0.23. What is the probability that Napoleon and Pedro catch the bus?
Given P(A) = 0.10, P(B) = 0.70, P(C) = 0.38 and that events A, B, and C are independent. Probability of A, B, and C is given by:P(A ∩ B ∩ C) = P(A) × P(B) × P(C)⇒ P(A ∩ B ∩ C) = 0.10 × 0.70 × 0.38= 0.0266≈0.027
Given the probabilities of events A, B, and C, we can find the probability of their intersection if they are independent.
In probability theory, the intersection of two or more events is the event containing elements that belong to all of the events.
The formula to find the probability of the intersection of two independent events is:
P(A ∩ B) = P(A) × P(B)
For three independent events, the formula is:
P(A ∩ B ∩ C) = P(A) × P(B) × P(C)
Using the given probabilities, we can find the probability of A, B, and C:
P(A) = 0.10P(B) = 0.70P(C) = 0.38
Now, using the formula above:
P(A ∩ B ∩ C) = P(A) × P(B) × P(C)= 0.10 × 0.70 × 0.38= 0.0266≈0.027
Therefore, the probability of A, B, and C is 0.027.Conclusion:The probability of A, B, and C given that events A, B, and C are independent is 0.027.
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The distribution of passenger vehicle speeds traveling on the Interstate Freeway (1-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. ▾ ( > (a) What percent of passenger vehicles travel slower than 80 miles/hour? % (please round to two decimal places) (b) What percent of passenger vehicles travel between 60 and 80 miles/hour? % (pleak round to two decimal places) (c) How fast do the fastest 5% of passenger vehicles travel? Submit All Parts (d) The speed limit on this stretch of the 1-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the 1-5. % (please round to two decimal places) Geranpayeh, T.docx mph (please round to two decimal places) Geranpayeh, C...docx Geranpayeh, T...docx Geranpayeh
The given distribution is nearly normal with a mean of 72.6 mph and a standard deviation of 4.78 mph. The percentage of passenger vehicles traveling slower than 80 mph can be found by using z-score.
Z-score formula: `z = (x - μ)/σ`Here,
x = 80 mph,
μ = 72.6 mph,
and σ = 4.78 mph.
Therefore,`z = (80 - 72.6)/4.78
= 1.5439`Looking up this value in the z-table gives a probability of 0.9382. Converting this to a percentage gives `0.9382 × 100% = 93.82%`Therefore, approximately 93.82% of passenger vehicles travel slower than 80 mph.
(b) To find the percentage of passenger vehicles traveling between 60 and 80 mph, we need to find the z-scores corresponding to these values. The z-score corresponding to 60 mph is:`z₁ = (60 - 72.6)/4.78
= -2.6318`The z-score corresponding to 80 mph is:`z₂
= (80 - 72.6)/4.78
= 1.5439`Looking up these values in the z-table gives the probabilities `0.0046` and `0.9382`, respectively. Therefore, the percentage of passenger vehicles traveling between 60 and 80 mph is:`(0.9382 - 0.0046) × 100% = 93.76%`Therefore, approximately 93.76% of passenger vehicles travel between 60 and 80 mph.
(c) To find the speed of the fastest 5% of passenger vehicles, we need to find the z-score corresponding to this value. The z-score corresponding to the 95th percentile is:`z = invNorm(0.95)
= 1.6449`Using the z-score formula, we can find the corresponding speed value:
x = zσ + μ = 1.6449 × 4.78 + 72.6 ≈ 80.82 mphTherefore, the speed of the fastest 5% of passenger vehicles is approximately 80.82 mph.
(d) Since the speed limit on this stretch of the 1-5 is 70 mph, we need to find the percentage of passenger vehicles traveling above this speed. The z-score corresponding to 70 mph is:`z = (70 - 72.6)/4.78
= -0.5439`Looking up this value in the z-table gives a probability of `0.2934`. Therefore, the percentage of passenger vehicles traveling above the speed limit is:`(1 - 0.2934) × 100% = 70.66%`Therefore, approximately 70.66% of passenger vehicles travel above the speed limit on this stretch of the 1-5.
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Find the quadratic function that is the best fit for f(x) defined by the table below. X 0 2 4 6 8 10 f(x) 0 398 1601 3605 6405 9998 The quadratic function is y=x² + +x+O. ·O· (Type an integer or decimal rounded to two decimal places as needed.)
To find the quadratic function that best fits the given table of values, we need to determine the coefficients of the quadratic equation y = ax² + bx + c.
By substituting the values from the table into the equation, we can form a system of equations and solve for the unknown coefficients.
The given table provides the values of f(x) for six different x-values. We want to find a quadratic function that best represents these data points. We start by substituting the x and f(x) values into the general quadratic equation:
0 = a(0)² + b(0) + c
398 = a(2)² + b(2) + c
1601 = a(4)² + b(4) + c
3605 = a(6)² + b(6) + c
6405 = a(8)² + b(8) + c
9998 = a(10)² + b(10) + c
Simplifying these equations, we obtain a system of equations:
0 = c
398 = 4a + 2b + c
1601 = 16a + 4b + c
3605 = 36a + 6b + c
6405 = 64a + 8b + c
9998 = 100a + 10b + c
We can solve this system of equations to find the values of a, b, and c. Once we have these coefficients, we can write the quadratic function that best fits the given data.
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Sample data and hypotheses for a chi-square goodness-of-fit test are given. Fill in the table to compute the expected counts.
Hypotheses:
H0:pA=0.3,pB=0.3,pC=0.4
Ha: Some pi is not as given
Sample Data:
A.
B.
C.
Total
28
50
45
123
Enter the expected counts in the following table. Enter the exact answers.
Answer:
The expected counts for each category are A | 28 | 36.9 B | 50 | 36.9 C | 45 | 49.2 Total | 123 | 123
To compute the expected counts for the chi-square goodness-of-fit test, we need to calculate the expected count for each category based on the null hypothesis. The expected count for each category is given by:
Expected count = Total count * Probability
Given the null hypothesis:
H0: pA = 0.3, pB = 0.3, pC = 0.4
And the sample data:
A: 28
B: 50
C: 45
Total: 123
We can calculate the expected counts as follows:
Expected count for A = Total * pA = 123 * 0.3 = 36.9
Expected count for B = Total * pB = 123 * 0.3 = 36.9
Expected count for C = Total * pC = 123 * 0.4 = 49.2
Total = 123
Now, let's fill in the table with the expected counts:
Category | Observed Count | Expected Count
A | 28 | 36.9
B | 50 | 36.9
C | 45 | 49.2
Total | 123 | 123
The expected counts for each category are filled in the "Expected Count" column of the table.
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Answer:
The expected counts for each category are A | 28 | 36.9 B | 50 | 36.9 C | 45 | 49.2 Total | 123 | 123
To compute the expected counts for the chi-square goodness-of-fit test, we need to calculate the expected count for each category based on the null hypothesis. The expected count for each category is given by:
Expected count = Total count * Probability
Given the null hypothesis:
H0: pA = 0.3, pB = 0.3, pC = 0.4
And the sample data:
A: 28
B: 50
C: 45
Total: 123
We can calculate the expected counts as follows:
Expected count for A = Total * pA = 123 * 0.3 = 36.9
Expected count for B = Total * pB = 123 * 0.3 = 36.9
Expected count for C = Total * pC = 123 * 0.4 = 49.2
Total = 123
Now, let's fill in the table with the expected counts:
Category | Observed Count | Expected Count
A | 28 | 36.9
B | 50 | 36.9
C | 45 | 49.2
Total | 123 | 123
The expected counts for each category are filled in the "Expected Count" column of the table.
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According to the University of Nevada Center for Logistics Management, 6% of all merchandise sold in the United States gets returned. A Houston department store sampled 80 items sold in January and found that 10 of the items were returned.
(a) Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store. If required, round your answer to three decimal places. ____ (b) Construct a 95% confidence interval for the porportion of returns at the Houston store. If required, round your answer to three decimal places. ____ to _____
(c) Is the proportion of returns at the Houston store significantly different from the returns for the nation as a whole? Provide statistical support for your answer. We ____ the null hypothesis. We ____ the return rate for the Houston store is different than the U.S. national return rate.
In January, a Houston department store sampled 80 items sold and found 10 of them were returned. Based on sample, estimate proportion of items returned for population of sales transactions at the Houston store.
(a) The point estimate of the proportion of items returned at the Houston store is calculated by dividing the number of returned items (10) by the total sample size (80). Therefore, the point estimate is 10/80 = 0.125, or 12.5%.
(b) To construct a 95% confidence interval for the proportion of returns at the Houston store, we can use the formula: point estimate ± (critical value * standard error).
The critical value can be obtained from the standard normal distribution corresponding to the desired confidence level. For a 95% confidence level, the critical value is approximately 1.96. The standard error is calculated as the square root of [(point estimate * (1 - point estimate)) / sample size]. Plugging in the values, we can calculate the lower and upper bounds of the confidence interval.
(c) To determine if the proportion of returns at the Houston store is significantly different from the returns for the nation as a whole, we can conduct a hypothesis test. The null hypothesis would state that the return rate for the Houston store is the same as the U.S. national return rate (6%), while the alternative hypothesis would state that they are different. We can perform a statistical test, such as a z-test, to calculate the test statistic and compare it to the critical value to determine if we can reject or fail to reject the null hypothesis.
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Terri Vogel, an amateur motorcycle racer, averages 129.49 seconds per 2.5 mile lap (in a 7 lap race) with a standard deviation of 2.26 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. (Source: log book of Terri Vogel) Let X be the number of seconds for a randomly selected lap. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X-N 129.49 2.26 0 b. Find the proportion of her laps that are completed between 130.22 and 131.71 seconds. -0.6593 c. The fastest 2% of laps are under d. The middle 40% of her laps are from seconds. Enter an integer or decimal number, accurate to at least 4 decimal places
Therefore, the middle 40% of her laps are from 128.3082 seconds to 130.6718 seconds.
a. What is the distribution of X? X-N 129.49 2.26 0
Terri Vogel's race times are normally distributed. The distribution is normal with a mean of 129.49 seconds and a standard deviation of 2.26 seconds.
X is also normally distributed because it is a linear combination of normally distributed variables.
Thus, X is normally distributed with a mean of 129.49 seconds and a standard deviation of (2.26/sqrt(7)) seconds.
b. Find the proportion of her laps that are completed between 130.22 and 131.71 seconds.
We can standardize the values and convert them to z-scores using the formula z = (x - μ)/σ, where x is the value, μ is the mean, and σ is the standard deviation.
z1 = (130.22 - 129.49)/(2.26/sqrt(7))
= 1.2032z2
= (131.71 - 129.49)/(2.26/sqrt(7))
= 2.6492
Using a standard normal table or calculator, we can find the probabilities associated with these z-scores:
P(1.2032 < Z < 2.6492) = P(Z < 2.6492) - P(Z < 1.2032)
= 0.9950 - 0.8856
= 0.1094
Therefore, the proportion of her laps that are completed between 130.22 and 131.71 seconds is 0.1094.
c. The fastest 2% of laps are under.
We can use the z-score formula to solve this problem.
We want to find the value of x such that P(X < x) = 0.02, which is equivalent to finding the z-score z such that
P(Z < z) = 0.02.
Using a standard normal table or calculator, we can find the z-score associated with the 2nd percentile:
z = -2.054
Using the formula z = (x - μ)/σ, we can solve for x:
x = μ + zσ
= 129.49 + (-2.054)(2.26/sqrt(7))
= 124.4759
Therefore, the fastest 2% of laps are under 124.4760 seconds.
d. The middle 40% of her laps are from seconds.
We can use the z-score formula to solve this problem. We want to find the values of x1 and x2 such that
P(x1 < X < x2) = 0.40, which is equivalent to finding the z-scores z1 and z2 such that P(z1 < Z < z2) = 0.40.
We can find the z-scores associated with the 30th and 70th percentiles using a standard normal table or calculator:
z1 = -0.2533z2
= 0.2533
Using the formula z = (x - μ)/σ, we can solve for x1 and x2:x1
= μ + z1σ
= 129.49 + (-0.2533)(2.26/sqrt(7))
= 128.3082x2
= μ + z2σ
= 129.49 + (0.2533)(2.26/sqrt(7))
= 130.6718
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Suppose that a certain college class contains 65 students. Of these, 39 are freshmen, 28 are chemistry majors, and 10 are neither. A student is selected at random from the class. (a) What is the probability that the student is both a freshman and a chemistry major? (b) Given that the student selected is a freshman, what is the probability that he is also a chemistry major?
According to given information, the probabilities are as follows.
For (a) [tex]P(F\ \text{and}\ C) = 0[/tex]
For (b) [tex]P(C | F) = 0[/tex]
a) We have that there are 65 students, of whom 39 are freshmen and 28 are chemistry majors and there are 10 who are neither.
Let's call F to the event that the student is a freshman, and C to the event that the student is a chemistry major.
Then, the probability of being a freshman and chemistry major is:
P(F and C) = 0, since there are only 28 chemistry majors, and they are not freshmen.
b) The probability of a student being a chemistry major given that he is a freshman is:
[tex]P(C | F) = P(C \ \text{and}\ F) / P(F)[/tex]
Now we need to calculate P(C and F), which is the probability that the student is both a freshman and a chemistry major. As we have seen in part (a), this probability is 0.
So: [tex]P(C | F) = P(C\ \text{and}\ F) / P(F) \\= 0 / 39 \\= 0[/tex]
The probability of a freshman student being a chemistry major is 0.
Answer: For (a) [tex]P(F\ \text{and}\ C) = 0[/tex]
For (b) [tex]P(C | F) = 0[/tex]
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the lengths of the sides of a different triangle are given below. Which is a right triangle?
a. 6,13,14
b. 5,12,13
c. 9,7,11
d.9,7,10
e. 5,12,12
Find the number of ways that a four- different digit number that can be formed using the digits 1, 2, 3, 4, 5, 6, 7 such that the first digit is an odd number. Select one: a. 35 b. 1732 C. 1372 d. 240
The number of ways a four-digit number can be formed is 480.
To find the number of ways a four-digit number can be formed using the digits 1, 2, 3, 4, 5, 6, 7 such that the first digit is an odd number, we need to consider the choices for each digit.
The first digit must be an odd number, which can be either 1, 3, 5, or 7. There are four choices for the first digit.
The second digit can be any of the remaining six digits (2, 3, 4, 5, 6, 7), as repetition is allowed. So, there are six choices for the second digit.
Similarly, there are five choices for the third digit and four choices for the fourth digit.
By the multiplication principle, the total number of ways to form the four-digit number is given by:
Total number of ways = (number of choices for the first digit) × (number of choices for the second digit) × (number of choices for the third digit) × (number of choices for the fourth digit)
Total number of ways = 4 × 6 × 5 × 4 = 480
Therefore, the number of ways a four-digit number can be formed is 480.
None of the given options (a. 35, b. 1732, c. 1372, d. 240) match the correct answer.
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Suppose you decide to flip a coin until you get a heads (H), at which point you will stop flipping the coin. However. you will fip at most three times. even if you never get a heads. Let the random variable X be the number of times you flip the coin. Our gozl is to find the probability distribution of X. In other words, we would like to create a table that lints all the possible values of x and the corresponding probabilities. Well follow the same 5 teps We followed in the two examples we solved
The probability distribution of X is as shown in the table: 1 with a probability of 1/2, 2 with a probability of 1/4, and 3 with a probability of 1/8.
When flipping a coin until getting heads, we are interested in the number of flips required, represented by the random variable X. In this scenario, we are limited to a maximum of three flips, even if heads doesn't appear.
In the first flip, there are two possible outcomes: heads (H) or tails (T). The probability of getting heads on the first flip is 1/2, as there is an equal chance of getting either heads or tails.
If heads doesn't appear on the first flip, we proceed to the second flip. At this point, we have two possibilities: heads on the second flip or tails again. Since the probability of getting heads on a single flip is 1/2, the probability of getting tails twice in a row is (1/2) * (1/2) = 1/4.
If heads hasn't appeared after the second flip, we move to the third and final flip. The possibilities now are: heads on the third flip or tails for the third time. Again, the probability of getting heads on a single flip is 1/2, so the probability of getting tails for the third time is (1/2) * (1/2) * (1/2) = 1/8.
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a student takes an examination consisting of 20 true-false questions. the student knows the answer to n of the questions, which are answered correctly, and guesses the answers to the rest. the conditional probability that the student knows the answer to a question, given that the student answered it correctly, is 0.824. calculate n
The student knows the answer to approximately 5 questions out of the 20.
Let's denote the event that the student knows the answer to a question as K and the event that the student answers the question correctly as C. We are given the following information:
P(K) = n/20 (probability that the student knows the answer)
P(C|K) = 1 (probability of answering correctly given that the student knows the answer)
P(K|C) = 0.824 (conditional probability that the student knows the answer given that the student answered correctly)
We can use Bayes' theorem to find the value of n:
P(K|C) = P(C|K) * P(K) / P(C)
P(C) = P(C|K) * P(K) + P(C|not K) * P(not K)
= 1 * (n/20) + (1/2) * ((20-n)/20)
= n/20 + (20-n)/40
= (2n + 20 - n) / 40
= (n + 20) / 40
Now, substituting the values into Bayes' theorem:
0.824 = 1 * (n/20) / ((n + 20) / 40)
0.824 = (2n / 20) * (40 / (n + 20))
0.824 = 4n / (n + 20)
Cross-multiplying:
0.824 * (n + 20) = 4n
0.824n + 16.48 = 4n
3.176n = 16.48
n = 16.48 / 3.176
n ≈ 5.18
Therefore, the student knows the answer to approximately 5 questions out of the 20.
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Find the minimum and maximum values of the function f(x, y, z) = 5x + 2y + 2z subject to the constraint x² + 2y² + 5z² = 1. (Use decimal notation. Round your answers to one decimal place.) minimum:
The minimum and maximum values of the function f(x, y, z) subject to the given constraint are -4.7 and 4.7, respectively.
The given function is f(x, y, z) = 5x + 2y + 2z subject to the constraint x² + 2y² + 5z² = 1.
So, the Lagrange function for the function f(x, y, z) is given by
L(x, y, z, λ) = f(x, y, z) - λg(x, y, z),
where g(x, y, z) = x² + 2y² + 5z² - 1.
Substitute the values in the Lagrange function, we get
L(x, y, z, λ) = (5x + 2y + 2z) - λ(x² + 2y² + 5z² - 1)
Now, differentiate the function L(x, y, z, λ) w.r.t x, y, z, and λ, separately and equate them to zero.
∂L/∂x = 5 - 2λx = 0 ...(1)
∂L/∂y = 2 - 4λy = 0 ...(2)
∂L/∂z = 2 - 10λz = 0 ...(3)
∂L/∂λ = x² + 2y² + 5z² - 1 = 0 ...(4)
Solve the above equations to find x, y, z, and λ.
From equation (1),
5 - 2λx = 0=> x = 5/2λ
From equation (2),
2 - 4λy = 0=> y = 1/2λ
From equation (3),
2 - 10λz = 0=> z = 1/5λ
From equation (4),
x² + 2y² + 5z² - 1 = 0=> (5/2λ)² + 2(1/2λ)² + 5(1/5λ)² - 1 = 0=> (25/4λ²) + (2/4λ²) + (1/5λ²) - 1 = 0=> λ² = 25/4 + 20 + 4/5=> λ² = 156.25/20=> λ² = 7.8125=> λ = ±2.793
The values of λ are λ = 2.793, and λ = -2.793.
Find the values of x, y, and z, for each value of λ.
For λ = 2.793, x = 5/2
λ = 5/(2 × 2.793) ≈ 0.895
y = 1/2
λ = 1/(2 × 2.793) ≈ 0.179
z = 1/5
λ = 1/(5 × 2.793) ≈ 0.071
The value of the function f(x, y, z) for λ = 2.793 is
f(x, y, z) = 5x + 2y + 2z≈ 5 × 0.895 + 2 × 0.179 + 2 × 0.071 ≈ 4.747
Therefore, the minimum and maximum values of the function f(x, y, z) subject to the given constraint are -4.7 and 4.7, respectively.
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Find the Inverse Laplace Transform of F(s)= s 2
−4s+8
2s−1
F(s)= s 2
−4s+8
2s−1
=e At
[Bsin(Ct)+Dcos(Ct)], where A= , and D= Note: A,B,C, and D are algebraic expressions. L{1}= 8
1
Li{1}− 2
2
L{t n
}= s n+1
n!
54{e at
}= s−a
1
∫{coskt}= s 2
+k 2
s
L{sinkt}= 2 2
+k 2
k
∫Le at
⋅f(t)}=F(s−a) ∫ff(t−a)U(t−a)}=e −as
F(s) L{y(t)}=Y(s)=Y (∫
L{y ′
(t)}=sY−y(0) 7[{y ′′
(t)}=s 2
Y−s⋅y(0)−y ′
(0)
To find the inverse Laplace transform of F(s) = (s^2 - 4s + 8) / (2s - 1), we can use partial fraction decomposition followed by applying the inverse Laplace transform to each term.
First, let's perform the partial fraction decomposition:
F(s) = (s^2 - 4s + 8) / (2s - 1)
= (A / (2s - 1)) + (B / (s - 2))
Multiplying both sides by the denominators and simplifying, we get:
s^2 - 4s + 8 = A(s - 2) + B(2s - 1)
Expanding the right side and equating coefficients, we find:
For the s term: -4 = -2A + 2B
For the constant term: 8 = 2A - B
Solving these equations, we get A = 3 and B = 2.
So, we can rewrite F(s) as:
F(s) = (3 / (2s - 1)) + (2 / (s - 2))
Now, we can apply the inverse Laplace transform to each term:
L{3 / (2s - 1)} = 3/2 * L{1 / (s - 1/2)} = 3/2 * e^(t/2)
L{2 / (s - 2)} = 2 * L{1 / (s - 2)} = 2 * e^(2t)
Therefore, the inverse Laplace transform of F(s) is given by:
f(t) = (3/2) * e^(t/2) + 2 * e^(2t)
In conclusion:
The inverse Laplace transform of F(s) = (s^2 - 4s + 8) / (2s - 1) is f(t) = (3/2) * e^(t/2) + 2 * e^(2t).
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Consider a sample with data values of 24,20,25,15,30,34,27, and 20. Compute the range. Compute the interquartile range. Enter a number. Compute the sample variance. (Round your answer to two decimal places.) Compute the sample standard deviation. (Round your answer to two decimal places.)
The sample standard deviation is approximately 5.61 and sample variance is approximately 31.46.
To compute the range of a sample,
we subtract the minimum value from the maximum value.
Range = Maximum value - Minimum value
For the given sample,
the minimum value is 15 and the maximum value is 34.
Range = 34 - 15 = 19
The range of the sample is 19.
To compute the interquartile range (IQR) of a sample, we need to find the difference between the third quartile (Q3) and the first quartile (Q1).
IQR = Q3 - Q1
To calculate the quartiles,
we first need to arrange the data in ascending order:
15, 20, 20, 24, 25, 27, 30, 34
The sample size is 8,
so the median (Q2) will be the average of the fourth and fifth values:
Q2 = (24 + 25) / 2 = 24.5
To find Q1, we take the median of the lower half of the data:
Q1 = (20 + 20) / 2 = 20
To find Q3, we take the median of the upper half of the data:
Q3 = (27 + 30) / 2 = 28.5
Now we can calculate the interquartile range:
IQR = 28.5 - 20 = 8.5
The interquartile range of the sample is 8.5.
To compute the sample variance, we use the formula:
Variance = Σ[tex][(x - X)^2][/tex] / (n - 1)
where Σ represents the sum of, x is each data value, X is the mean, and n is the sample size.
First, let's calculate the mean (X):
X = (24 + 20 + 25 + 15 + 30 + 34 + 27 + 20) / 8 = 24.625
Now we can calculate the sample variance:
Variance = [tex][(24 - 24.625)^2 + (20 - 24.625)^2 + (25 - 24.625)^2 + (15 - 24.625)^2 + (30 - 24.625)^2 + (34 - 24.625)^2 + (27 - 24.625)^2 + (20 - 24.625)^2][/tex] / (8 - 1)
Variance = 31.46
To compute the sample standard deviation,
we take the square root of the sample variance:
Standard deviation = √(Variance) = [tex]\sqrt{(31.46}[/tex]) ≈ 5.61
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Determine the most conservative sample size for the estimation of the population proportion for the following. E=0.09, confidence level =90% Round your answer up to the nearest whole number. n=
The most conservative sample size for the estimation of the population proportion is 119, rounded up to the nearest whole number.
Now, Using the formula n = (Zα/2)²(p(1-p))/E²,
where: n = sample size
Here, We have,
Zα/2 = z-score for 0.025 (α/2) = 1.96
p = 0.5 (assumed to be the estimated proportion of the population)
E = 0.09 (specified absolute tolerance)
Hence, We get;
n = (1.96²)(0.5(1-0.5))/0.09²
n = 118.6
Therefore, the most conservative sample size for the estimation of the population proportion is 119, rounded up to the nearest whole number.
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When studying radioactive material, a nuclear engineer found that over 365 days, 1,000,000 radioactive atoms decayed to 978,271 radioactive atoms, so 21,729 atoms decayed during 365 days.
a. Find the mean number of radioactive atoms that decayed in a day.
b. Find the probability that on a given day, 52 radioactive atoms decayed.
(Round to three decimal places as needed.)
a. The mean number of radioactive atoms that decayed in a day is approximately 59.571.
b. The probability that on a given day, 52 radioactive atoms decayed is approximately 0.239.
a. To find the mean number of radioactive atoms that decayed in a day, we divide the total number of atoms decayed (21,729) by the number of days (365). This gives us an average of approximately 59.571 atoms decaying per day.
b. To calculate the probability that on a given day, 52 radioactive atoms decayed, we can use the concept of Poisson distribution. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time. In this case, we have the mean number of atoms decaying per day, which we calculated in part a as approximately 59.571.
Using the Poisson distribution formula, we can calculate the probability:
P(X = 52) = [tex](e^{-\lambda} * \lambda^x)[/tex]/ x!
Where λ is the mean number of decays per day and x is the number of decays we want to find the probability for.
Substituting the values, we have:
P(X = 52) = [tex](e^{-59.571} * 59.571^{52})[/tex] / 52!
Using a scientific calculator or software, we can compute this value, which is approximately 0.239.
Therefore, the probability that on a given day, 52 radioactive atoms decayed is approximately 0.239, or 23.9% (rounded to three decimal places).
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(10 points) Prove or disprove: The function f:[0,1]→R defined by f(x)={ xsin( x
1
)
0
if 0
if x=0
is uniformly continuous. Hint: You may use that sin(x) and 1/x are continuous functions on their domains of definition. What is the issue?
The given function is defined by
[tex]f(x)={ xsin(1/x) 0[/tex]
if [tex]0 if x=0.[/tex]
For uniform continuity, we have to show that for every
[tex]$\epsilon>0$[/tex]
there exists a [tex]$\delta>0$[/tex]
such that [tex]$|x-y|<\delta$[/tex]
implies [tex]$|f(x)-f(y)|<\epsilon$[/tex] .
Let [tex]$x,y\in[0,1]$[/tex]
without loss of generality assume that [tex]$x0$[/tex] , if we choose
[tex]$\delta=\frac{\epsilon}{1+y}$[/tex],
we have that if [tex]$|x-y|<\delta$[/tex] ,
then[tex]$$|f(x)-f(y)|≤(1+y)\delta=(1+y)\frac{\epsilon}{1+y}=\epsilon$$[/tex] .
The issue here is at $x=0$; $f$ is not continuous at $x=0$.
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Suppose the correlation coefficient is 0.9. The percentage of variation in the response variable explained by the variation in the explanatory variable is A. 0.81% B. 81% C. 90% D. none of the other answers OE. 8.1% OF. 9% G. 0.90% H. 0% ETTE
The percentage of variation in the response variable explained by the variation in the explanatory variable is 81%.
The correlation coefficient measures the strength and direction of the linear relationship between two variables. In this case, if the correlation coefficient is 0.9, it indicates a strong positive linear relationship between the explanatory variable and the response variable.
The square of the correlation coefficient, also known as the coefficient of determination (r²), represents the proportion of the variation in the response variable that can be explained by the variation in the explanatory variable.
Therefore, if the correlation coefficient is 0.9, the coefficient of determination is (0.9)² = 0.81.
This means that 81% of the variation in the response variable can be explained by the variation in the explanatory variable.
In other words, the strength of the linear relationship between the two variables allows us to explain 81% of the variability in the response variable based on the explanatory variable.
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Suppose that f(x,y)={ 15x 2
y
0
;
;
0
otherwise
a) Compute E(X∣y),Var(X) and Var(Y). b) Hence, deduce the value of rho=Corr(X,Y).
That further simplification may be possible depending on the specific values of x and y.
To compute E(X|y), Var(X), Var(Y), and Corr(X,Y), we need to calculate the marginal distributions and conditional expectations.
Given the joint probability distribution:
f(x,y) =
15x^2 * y, if x > 0 and y > 0
0, otherwise
Let's calculate the marginal distribution of X and Y first:
Marginal distribution of X:
[tex]fX(x) = ∫f(x,y)dy = ∫(15x^2 * y)dy (from y = 0 to infinity) = 15x^2 * ∫y dy = 15x^2 * [y^2/2] (evaluating the integral) = 7.5x^2 * y^2[/tex]
Marginal distribution of Y:
[tex]fY(y) = ∫f(x,y)dx = ∫(15x^2 * y)dx (from x = 0 to infinity) = 15y * ∫x^2 dx = 15y * [x^3/3] (evaluating the integral) = 5y * x^3\\[/tex]
Now, let's calculate the conditional expectation E(X|y):
[tex]E(X|y) = ∫x * f(x|y) dx = ∫x * (f(x,y)/fY(y)) dx (using Bayes' rule) = ∫x * ((15x^2 * y) / (5y * x^3)) dx = 3/y * ∫dx = 3/y * x (integrating with respect to x) = 3/y * x^2/2 (evaluating the integral) = 1.5/y * x^2[/tex]
[tex]To compute Var(X), we need to calculate E(X^2) first:E(X^2) = ∫x^2 * fX(x) dx = ∫x^2 * (7.5x^2 * y^2) dx = 7.5y^2 * ∫x^4 dx = 7.5y^2 * [x^5/5] (evaluating the integral) = 1.5y^2 * x^5\\[/tex]
Now, we can calculate Var(X):
[tex]Var(X) = E(X^2) - (E(X))^2 = 1.5y^2 * x^5 - (1.5/y * x^2)^2 = 1.5y^2 * x^5 - 2.25/y^2 * x^4To compute Var(Y), we need to calculate E(Y^2) first:E(Y^2) = ∫y^2 * fY(y) dy = ∫y^2 * (5y * x^3) dy = 5x^3 * ∫y^3 dy = 5x^3 * [y^4/4] (evaluating the integral) = 1.25x^3 * y^4[/tex]
Now, we can calculate Var(Y):
[tex]Var(Y) = E(Y^2) - (E(Y))^2 = 1.25x^3 * y^4 - (5x^3 * y)^2 = 1.25x^3 * y^4 - 25x^6 * y^2\\[/tex]
Finally, let's deduce the value of rho = Cor
r(X,Y):
[tex]Corr(X,Y) = Cov(X,Y) / sqrt(Var(X) * Var(Y))Cov(X,Y) = E(X * Y) - E(X) * E(Y)E(X * Y) = ∫∫x * y * f(x,y) dx dy = ∫∫x * y * (15x^2 * y) dx dy = 15 * ∫∫x^3 * y^2 dx dy = 15 * ∫(1.5/y * x^2) * y^2 dx dy (using E(X|y) = 1.5/y * x^2) = 22.5 * ∫x^2 dx dy = 22.5 * [x^3/3] (evaluating the integral) = 7.5x^3[/tex]
[tex]E(X) = ∫x * fX(x) dx = ∫x * (7.5x^2 * y^2) dx = 7.5y^2 * ∫x^3 dx = 7.5y^2 * [x^4/4] (evaluating the integral) = 1.875y^2 * x^4E(Y) = ∫y * fY(y) dy = ∫y * (5x^3 * y) dy = 5x^3 * ∫y^2 dy = 5x^3 * [y^3/3] (evaluating the integral) = 1.667x^3 * y^3Cov(X,Y) = E(X * Y) - E(X) * E(Y) = 7.5x^3 - (1.875y^2 * x^4) * (1.667x^3 * y^3) = 7.5x^3 - 3.125x^7 * y^5[/tex]
Now, we can calculate rho:
[tex]rho = Corr(X,Y) = Cov(X,Y) / sqrt(Var(X) * Var(Y)) = (7.5x^3 - 3.125x^7 * y^5) / sqrt((1.5y^2 * x^5 - 2.25/y^2 * x^4) * (1.25x^3 * y^4 - 25x^6 * y^2))\\[/tex]
Please note that further simplification may be possible depending on the specific values of x and y.
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