**Title: Understanding the Distribution of Movie Production Costs**The average production cost for major movies is 69 million dollars and the standard deviation is 21 million dollars. Assume the production cost distribution is normal. Suppose that 18 randomly selected major movies are researched. Answer the following questions. Give your answers in millions of dollars, not dollars. Round all answers to 4 decimal places where possible.---**a. What is the distribution of \( X \)?**\[ X \sim N(69, 21) \]**b. What is the distribution of \( \bar{x} \)?**\[ \bar{x} \sim N \left(69, \frac{21}{\sqrt{18}}\right) \]\[ \bar{x} \sim N(69, 4.9497) \]---**c. For a single randomly selected movie, find the probability that this movie's production cost is between 65 and 69 million dollars.**\[ \Pr(65 < X < 69) = \text{(Answer required)} \]---**d. For the group of 18 movies, find the probability that the average production cost is between 65 and 69 million dollars.**\[ \Pr(65 < \bar{x} < 69) = \text{(Answer required)} \]---**e. For part d), is the assumption of normal necessary?**\[ \text{Yes} \]---In the equations above:- \( X \) represents the production cost of a single movie.- \( \bar{x} \) represents the average production cost of the 18 selected movies.- \( N(\mu, \sigma) \) denotes a normal distribution with mean \( \mu \) and standard deviation \( \sigma \).This problem involves understanding key concepts of normal distribution and the impact of sample size on the standard deviation of the sample mean.**Graphs and Diagrams:**There are no graphs or diagrams in the provided text.

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Answered: The average production cost for major…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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The standard deviation of the given the data (14, 13, 16, 9, 3} is ♦ If the average American male is 5'10" and the standard deviation is 3", then Yao Ming who is 7'6" is within ◆ standard deviations of the mean? An outlier is an extreme value that is significantly less or more than the rest of the data. In the data below take out the outlier, then calculate using a statistics calculator or spreadsheet, the standard deviation of the heights of 15 soda cans. {92.8, 92.8, 92.9, 92.9, 92.9, 92.8, 92.7, 92.9, 92.1, 92.7, 92.8, 92.9, 92.9, 92.7, 92.8} Round to the nearest 100th. Answer:
The following frequency table shows the number of fires per month in a city. Complete the table entries to find the mean and standard deviation for this distribution. E Click the icon to view the table. Complete the table below. More Info (Type integers or decimals rounded to three decimal places as needed.) 21 Product, Number, Frequency, Product, Deviation, Deviation 2 2 (x-x) (x-x)²•f 21 f (x•f) (x-x)² Number, Frequency, Product, f Deviation, Deviation (x•f) (x-x) (x-x)² X-X 2 X-X 4. 3 1 6. 0. Ex-73D Print Done Help me solve this View an example Get more help - 99+
Q.9. Calculate missing frequencies from the following when mean is 13-75. No. of students : 4 8 10 12 18 24 22 frequency : 13 11 9 8 10
Population 1 Population 2 Use the given statistics to complete parts (a) and (b). Assume that the populations are normally distributed. (a) Test whether H, > H, at the a = 0.01 level of significance for the given sample data. (b) Construct a 90% confidence interval about u, -H2. 28 16 45.3 43.2 9.7 6.7 (a) Identify the null and alternative hypotheses for this test. O A. Ho: H1 > H2 O B. H: H1 =H2 OC. Hg: H1 = H2 H;: H1 # H2 O F. Ho: H1 H2. O B. Do not reject Hg. There is not sufficient evidence at the a = 0.01 level of significance to conclude that u, > H2. O C. Do not reject H,. There is sufficient evidence at the a= 0.01 level of significance to conclude that u, > H2- O D. Reject Hn. There is not sufficient evidence at the a= 0.01 level of significance to conclude that u, > H2- (b) The 90% confidence interval about u, - H, is the range from a lower bound of to an upper bound of (Round to three decimal places as needed.)
D, e, f and h
The graph illustrates the distribution of test scores taken by College Algebra students. The maximum possible score on the test was 130, while the mean score was 71 and the standard deviation was 13. 32 45 97 110 58 71 84 Distribution of Test Scores Q What is the approximate percentage of students who scored less than 45 on the test? What is the approximate percentage of students who scored between 71 and 84 on the test? What is the approximate percentage students who scored between 58 and 84 on the test? What is the approximate percentage of students who scored between 45 and 97 on the test?
The average production cost for major movies is 57 million dollars and the standard deviation is 20 million dollars. Assume the production cost distribution is normal. Suppose that 46 randomly selected major movies are researched. Answer the following questions. Give your answers in millions of dollars, not dollars. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X - N( 57 20 b. What is the distribution of ? - N c. For a single randomly selected movie, find the probability that this movie's production cost is between 52 and 55 million dollars. 58 d. For the group of 46 movies, find the probability that the average production cost is between 52 and 55 million dollars. e. For part d), is the assumption of normal necessary? Yes No
Please answer 7-8 only, show your complete solution and dont use excel
After 4 points are added to every score in a distribution, the mean is calculated and found to be μ = 26. What was the value of the mean for the original distribution? A. 30 B. Cannot be determined from the information given C. 34 D. 22
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