The mean number of words per WPM read by sixth graders is 94

The mean number of words per (WPM) read by sixth graders is 94 with a variance of 256.If 88 sixth graders randomly selected, what is the probability that the sample mean would differ from the population mean by more than 4.72 WPM?

Solution

Mean = 94

Variance = 256 ---> standard deviation = sqrt(256) --> SD = 16

94 - 4.72 = 89.28
94 + 4.72 = 98.72

Since it must differ by more than 4.72 wpm, we are finding the probability P(x < 89.28 or x > 98.72)

P(x < 89.28) :

z = (x - u) / (SD / sqrt(n))

z = (89.28 - 94) / (16 / sqrt(88)) = -2.767345298295823437193721766991235105547054728513

Use this link to convert z to P value ----> https://www.easycalculation.com/statistics/p-value-for-z-score.php

P(z < -2.767345298295823437193721766991235105547054728513) = 0.0028

P(x > 98.72) :

z = (x - u) / (SD / sqrt(n))

z = (98.72 - 94) / (16 / sqrt(88))

z = 2.767345298295823437193721766991235105547054728513

P(z > 2.767345298295823437193721766991235105547054728513) = 0.0028

So, P(x < 89.28 or x > 98.72) = 0.0028 + 0.0028 ----> 0.0056

Answer -----> 0.0056