(h)Zhelong claimed she needs different to 30 minutes on average to repair brokencustomers' computers. A total of 17 computers left by customers were randomlyselected. Through the record from the sample, the average time taken was 28.5 minuteswith a standard deviation of 2.5 minutes. Assume the time is normally distributed.(i)Determine is the population standard deviation given for the above case study?

Answered: (b) Zhelong claimed she needs different…

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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The heights of adult men in America are normally distributed, with a mean of 69.6 inches and a standa deviation of 2.61 inches. The heights of adult women in America are also normally distributed, but with mean of 64.7 inches and a standard deviation of 2.56 inches. If a man is 6 feet 3 inches tall, what is his Z-score (to two decimal places)?
The average length of a phone call for cell phone users in the United States is 8.4 minutes, with standard deviation 2.4 minutes. If you were to take a random sample of 40 cell phone users in US, what is the standard deviation of the sampling distribution?
In a certain country the heights of adult men are normally distributed with a mean of 70.2 inches and a standard deviation of 2.7 inches. The country's military requires that men have heights between 66 inches and 79 inches. Determine what percentage of this country's men are eligible for the military based on height. The percentage of men that are eligible for the military based on height is nothing%. (Round to two decimal places as needed.)
In a recent awards ceremony, the age of the winner for best actor was 36 and the age of the winner for best actress was 51. For all best actors, the mean age is 43.2 years and the standard deviation is 5.3 years. For all best actresses, the mean age is 38.6 years and the standard deviation is 11.4 years. (All ages are determined at the time of the awards ceremony.) Relative to Use z scores to compare the given values. their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. ..... had the more extreme age. and the z score for the actress is z = the Since the z score for the actor is z= Score T (Round to two decimal places.) U.S. adu lect five sing Table %00% 4. two of the 4. %6 23.08% 3 e fewer th. 98.61% 3. 96.67% (1) More Next %00 P Pearson Always Learning Copyright © 2022, 2018, 2014 Pearson Educati 95.24% 23min 10/18/21 10/18/21 L. 9:30am 9:53am OL %00 5min 10/08/21 10/08/21 E Notes 065 Comments States) 11:50am 11:53am %00% 26min…
In a study of facial behavior, people in a control group are timed for eye contact for a 5 minute period. They are normally distributed with a mean of 177 seconds and a standard deviation of 53 seconds. I need to find the percentage of times within 53 seconds of the mean of 177 seconds and round it one decimal place if needed.
The ages of a group of 138 randomly selected adult females have a standard deviation of 17.5 years. Assume that the ages of female statistics students have less variation than ages of females in the general population, so leto= 17.5 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 95% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The required sample size is (Round up to the nearest whole number as needed.)
In 16 women with diabetes, the morning blood sugar mean was 49.4 with a standard deviation of 24.0. In normal healthy women, the mean was 47.0. From the sample information, can we conclude that the blood sugar mean in women with rheumatoid arthritis is greater than that in normal women? Assume a significance level of 0.05.
Consider population data with 5 being the standard deviation, 5 being the mean. Compute an 88.9% chebyshev interval around the population mean
1) A half-century ago, the mean height of women in a particular country in their 20s was 63.7 inches. Assume that the heights of today's women in their 20s are approximately normally distributed with a standard deviation of 2.31 inches. If the mean height today is the same as that of a half-century ago, what percentage of all samples of 26 of today's women in their 20s have mean heights of at least 65.04 inches?
Almost all medical schools in the United States require applicants to take the Medical College Admission Test (MCAT). On one exam, the scores of all applicants on the biological sciences part of the MCAT were approximately Normal with mean 9.4 and standard deviation 2.6. For applicants who actually entered medical school, the mean score was 10.6 and the standard deviation was 1.7. (a) What percent of all applicants had scores higher than 11? (b) What percent of those who entered medical school had scores between 8 and 12?

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