EP INTRODUCTION TO PROBABILITY+STAT.
14th Edition
ISBN: 2810019974203
Author: Mendenhall
Publisher: CENGAGE L
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The population mean and standard deviation are given below. Find the required probability and determine whether the
given sample mean would be considered unusual.
For a sample of n = 65, find the probability of a sample mean being greater than 225 if μ = 224 and σ = 3.5.
For a sample of n = 65, the probability of a sample mean being greater than 225 if μ=224 and σ = 3.5 is 0.0102
(Round to four decimal places as needed.)
***Please do not just simply copy and paste the other solution for this problem posted on bartleby as that solution does not have all of the parts completed for this problem. Please answer this I will leave a like on the problem. The data needed to answer this question is given in the following link (file is on view only so if you would like to make a copy to make it easier for yourself feel free to do so)
https://docs.google.com/spreadsheets/d/1aV5rsxdNjHnkeTkm5VqHzBXZgW-Ptbs3vqwk0SYiQPo/edit?usp=sharing
The data needed to answer this question is given in the following link (file is on view only so if you would like to make a copy to make it easier for yourself feel free to do so)
https://docs.google.com/spreadsheets/d/1aV5rsxdNjHnkeTkm5VqHzBXZgW-Ptbs3vqwk0SYiQPo/edit?usp=sharing
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- Business Discussarrow_forwardYou want to obtain a sample to estimate the proportion of a population that possess a particular genetic marker. Based on previous evidence, you believe approximately p∗=11% of the population have the genetic marker. You would like to be 90% confident that your estimate is within 0.5% of the true population proportion. How large of a sample size is required?n = (Wrong: 10,603) Do not round mid-calculation. However, you may use a critical value accurate to three decimal places.arrow_forward2. [20] Let {X1,..., Xn} be a random sample from Ber(p), where p = (0, 1). Consider two estimators of the parameter p: 1 p=X_and_p= n+2 (x+1). For each of p and p, find the bias and MSE.arrow_forward
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