Answer: 0.1093
Step-by-step explanation:
Given: Mean : [tex]\mu=93[/tex]
Standard deviation : [tex]\sigma = 22[/tex]
Sample size : [tex]n=30[/tex]
The formula to calculate z-score is given by :_
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 97.95, we have
[tex]z=\dfrac{97.95-93}{\dfrac{22}{\sqrt{30}}}\approx1.23[/tex]
The P-value = [tex]P(z>1.23)=1-P(z<1.23)=1-0.8906514=0.1093486\approx0.1093[/tex]
Hence, the probability that the sample mean would be greater than 97.95 WPM =0.1093
