Question

The mean number of words per minute (WPM) read by sixth graders is 93 with a standard deviation of 22.If 30 sixth graders are randomly selected, what is the probability that the sample mean would be greater than 97.95 WPM? (Round your answer to 4 decimal places)

Answer 1

Answer: 0.1093

Step-by-step explanation:

Given: Mean : $\mu=93$

Standard deviation : $\sigma = 22$

Sample size : $n=30$

The formula to calculate z-score is given by :_

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= 97.95, we have

$z=\dfrac{97.95-93}{\dfrac{22}{\sqrt{30}}}\approx1.23$

The P-value = $P(z>1.23)=1-P(z<1.23)=1-0.8906514=0.1093486\approx0.1093$

Hence, the  probability that the sample mean would be greater than 97.95 WPM =0.1093