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

Explanation:

Given: Mean :
`\mu=93`

Standard deviation :
`\sigma = 22`

Sample size :
`n=30`

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

`z=(x-\mu)/((\sigma)/(√(n)))`

For x= 97.95, we have

`z=(97.95-93)/((22)/(√(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