Statistics question of mean and standard deviation
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Statistics question of mean and standard deviation

[From: ] [author: ] [Date: 12-02-16] [Hit: ]
056/0.04682=-1.P(z ≤ -1.2)= 0.......
The mean is 0.973 and the standard deviation is 0.307 . The sample size is 43 with a mean amount of 0.917.

Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly selecting 43 with a mean of 0.917 or less.

The answer is .1158 but I don't know how to get that

-
mu=0.973, sigma=0.307, Xbar=0.917, n=43

Use the formula

z=(Xbar-mu)/(sigma/sqrt(n))

=(0.917-0.973)/(0.307/sqrt(43))

=-0.056/0.04682=-1.20


Use that z-score table

P(z ≤ -1.2)= 0.1158

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