**Transcription:**The average amount of money spent for lunch per person in the college cafeteria is $6.96 and the standard deviation is $2.49. Suppose that 7 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.a. What is the distribution of $ X $? $ X \sim N(\ \_\_\_\_\_\_ , \ \_\_\_\_\_\_ $ )b. What is the distribution of $ \bar{x} $? $ \bar{x} \sim N(\ \_\_\_\_\_\_ , \ \_\_\_\_\_\_ $ )c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.8484 and $6.599.\_\_\_\_\_\_d. For the group of 7 patrons, find the probability that the average lunch cost is between $5.8484 and $6.599.\_\_\_\_\_\_e. For part d), is the assumption that the distribution is normal necessary? ○ No ○ Yes

Sampling distribution of x¯ : If simple random sample of size n is drawn from Normal population…

Answered: The average amount of money spent for…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Private nonprofit four-year colleges charge, on average, $27,741 per year in tuition and fees. The standard deviation is $7,012. Assume the distribution is normal. Let X be the cost for a randomly selected college. Round all answers to 4 decimal places where possible. b. Find the probability that a randomly selected Private nonprofit four-year college will cost less than 28,487 per year. Incorrect (0.5438)c. Find the 78th percentile for this distribution. $Incorrect(33140) (Round to the nearest dollar.)
It is commonly believed that the mean body temperature of a healthy adult is 98.6° F. You are not entirely convinced. You believe that it is higher than 98.6° F. The temperatures for 12 randomly selected healthy adults are shown below. Assume that the distribution of the population is normal. 100.1, 96.8, 98.7, 100, 99.1, 98.3, 97.5, 100.9, 99.1, 100, 98.1, 100.1 What can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ? Ha: Select an answer Select an answer c. The test statistic O Question Help: (please show your answer to 3 decimal places.) (Please show your answer to 4 decimal places.) d. The p-value = e. The p-value is a f. Based on this, we should Select an answer the null hypothesis. g. Thus, the final conclusion is that ... There is sufficient evidence to conclude that the population mean body temperature for healthy adults is higher than 98.6° F. There is not sufficient…
Please answer B, C, & D.
a. The mean of this distribution is b. The standard deviation is c. The probability that the the child will be older than 5 years old?
Please answer B, C, & D.
Refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Find the mean and standard deviation for the number of girls in 8 births. Click the icon to view the data table. The mean is u= girl(s). (Round to one decimal place as needed.) The standard deviation is o = e# girl(s). (Round to one decimal place as needed.) Table of numbers of girls and probabilities Number of Girls x 0 1 2 3 4 5 6 7 8 átv P(x) 0.004 0.034 0.117 0.218 0.278 0.217 0.102 0.027 0.003 D - X NiZ Next
Please answer B, C, and D.
A normal population has a mean of 61 and a standard deviation of 22. You select a random sample of 35. Use Appendix. for the values Compute the probability that the sample mean is (Round the final answers to 4 decimal places.) a. Greater than 64 Probability b. Less than 57. Probability c. Between 57 and 64. Probability
please send correct answer for Q7
Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees and standard deviation of 1.00 degrees C. A thermometer is randomly selected and tested. Draw a sketch and find the temperature reading corresponding to Upper P 84, the 84 th percentile. This is the temperature reading separating the bottom 84 % from the top 16 %. A. Which graph represents Upper P 84? B. The temperature for Upper P 84 is approximately (Round to two decimal places as needed.)
Please answer B & C.
Please answer A, B, & C.

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