Question

The mean number of words per minute (WPM) read by sixth graders is 99 with a variance of 441. If 82 sixth graders are randomly selected, what is the probability that the sample mean would differ from the population mean by less than 3.69 WPM? Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = = 99

Variance = 2 = 441

standard deviation = = 2 = 441 = 21

n = 82

=   = 99

= / n = 21 / 82 = 2.32

99 ± 3.69 = 95.31, 102.69

P(95.31 < < 102.69)  

= P[(95.31 - 99) / 21 < ( - ) / < (102.69 - 99) / 21)]

= P( - 0.18 < Z < 0.18 )

= P(Z < 0.18) - P(Z < - 0.18)

Using z table,  

= 0.5714 - 0.4286  

= 0.1428