You survey 100 UCSC students and create arandom variable, called campus_housing, equalto 1 if they live on campus and zero otherwise.You then calculate the sample mean of thisrandom variable and find that it is equal to 0.74Answer the following questions using theinformation provided:a. True or False. The sample mean youcalculated is a parameter.b. How many students in your sample do notlive on campus?c. Choose the right answer: What share ofUCSC students live on campus?i. 26 percentii. 74 percentii. It is not possible to answer with theinformation provided.

Given that, Sample size is n= 100 Let X represents the students who lives on campus. Then we…

Answered: You survey 100 UCSC students and create…

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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About 3% of the population has a particular genetic mutation. 900 people are randomly selected.Find the mean for the number of people with the genetic mutation in such groups of 900.
There are eight patients waiting in a doctor's waiting room. They have been waiting for 9 minutes, 2 minutes, 5 minutes, 4 minutes, 12 minutes, 7 minutes, 8 minutes, and 11 minutes. Calculate the sample variance for the patients' wait times (in minutes squared). Round to the nearest hundredth as needed.
About 2% of the population has a particular genetic mutation. 800 people are randomly selected. Find the mean for the number of people with the genetic mutation in such groups of 800.
About 2% of the population has a particular genetic mutation. 900 people are randomly selected. Find the mean for the number of people with the genetic mutation in such groups of 900.
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An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 240 Fundash runners and 200 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 240 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.2 minutes. The 200 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.4 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, μ1 , of Fundash is not equal to the mean completion time, μ2 , of Coolsprint?…
In each part, give the value of the standardized statistic (z-score) for the sample mean. 100, σ = 8, n = 4 (a) x = 88, μ = Z = (b) x = 88, μ = 100, σ = 8, n = 64 Z = (c) x = 10.63, μ = 10.0, σ = 7, n = 10,000 z =
Ti A fitness instructor at a large gym would like to know if the individuals that take fitness classes in the morning tend to get in a greater number of steps than individuals that take evening fitness classes. She selects a random sample of 10 individuals that take fitness classes in the morning and 10 individuals that take fitness classes in the evening. She gives each person a step-counting At 10 device to wear for one week. At the end of the week she collects the devices and calculates the mean and standard deviation of the number of steps taken by the individuals in each group. Let uM = the true mean number of steps taken by all individuals who take fitness classes in the morning and HE = the true mean number of steps taken by all individuals who take fitness classes in the evening. What are the appropriate hypotheses for this test? O Ho: HM - HE = 0; Ho: HM - HE 0; H: HM - HE = 0 O Ho: HM HE = 0; Ha: HM - HE > 0 O Ho: HM HE < 0; H HM - HE = 0 O Ho: HM - HE - 0: H HM - HE #0
About 9% of the population has a particular genetic mutation. 600 people are randomly selected. Find the mean for the number of people with the genetic mutation in such groups of 600.
Nine observations from a work measurement study using continuous timing are shown below. Allowances are determined as personal, six percent; fatigue, five percent; delay, six percent. Use the Work Measurement Excel template to determine the standard time for this operation. Work Observation Element 1 2 3 4 5 6 7 8 9 Performance Rating A 0.11 0.14 0.09 0.07 0.13 0.11 0.15 0.14 0.15 115% B 0.23 0.21 0.21 0.22 0.27 0.23 0.27 0.27 0.27 85% C 0.47 0.51 0.47 0.46 0.50 0.50 0.51 0.47 0.49 90% D 0.64 0.68 0.63 0.61 0.69 0.69 0.69 0.68 0.73 110% E 0.73 0.77 0.74 0.68 0.80 0.83 0.79 0.80 0.84 95% F 1.01 1.02 1.01 1.02 1.11 1.09 1.02 1.08 1.11 105% Round your answer to two decimal places.
About 1% of the population has a particular genetic mutation. 200 people are randomly selected. Find the mean for the number of people with genetic mutation in such group of 200.

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