Question

As of this time, University of Michigan-Dearborn has registered 8,000 students; their average age is 24 with a standard deviation of 9 years. We have decided to select 36 students for an experiment involving sleep deprivation. What would be the probability that the sample mean would be larger than 23 years old (to four decimal places)?

Answer 1

Solution :

Given that,

mean = = 24

standard deviation = = 9

n = 36

= = 24

= / n = 9 / 36 = 1.5

P( > 23) = 1 - P( < 23)

= 1 - P[( - ) / < (23 - 24) / 1.5]

= 1 - P(z < -0.67)   

= 1 - 0.2514

= 0.7486