Question

Given a normal distribution with μ = 103 σ= 25 you select a sample of n = 25. What is the probability that X overbar is greater than 104? P(Xoverbar>104) =

Answer 1

Solution :

Given that ,

mean = = 103

standard deviation = = 25

n = 25

= 103 and

= / n = 25 / 25 = 25 / 5 = 5

P( > 104) = 1 - P( < 104)

= 1 - P(( - ) / < (104 - 103) / 5)

= 1 - P(z < 0.20)

= 1 - 0.5793

= 0.4207

Probability = 0.4207