Question

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 = 70 ), find the probability of a sample mean being less than 21.6 if ( mu = 22 ) and ( sigma = 1.31 ).
click the icon to view page 1 of the standard normal table.
click the icon to view page 2 of the standard normal table.
for a sample of ( n = 70 ), the probability of a sample mean being less than 21.6 if ( mu = 22 ) and ( sigma = 1.31 ) is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \).
Given \( \sigma = 1.31 \) and \( n = 70 \), then \( \sigma_{\bar{x}}=\frac{1.31}{\sqrt{70}}\approx\frac{1.31}{8.3666}\approx0.1566 \).

Step2: Calculate the z - score

The formula for the z - score is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \).
Here, \( \bar{x}=21.6 \), \( \mu = 22 \), and \( \sigma_{\bar{x}}\approx0.1566 \).
So \( z=\frac{21.6 - 22}{0.1566}=\frac{- 0.4}{0.1566}\approx - 2.55 \).

Step3: Find the probability

We want to find \( P(\bar{X}<21.6) \), which is equivalent to \( P(Z < - 2.55) \).
Looking up the value in the standard - normal table, \( P(Z < - 2.55)=0.0054 \).

Answer:

\(0.0054\)