Question

Let x be a continuous random variable that has a normal distribution with μ = 60 and σ = 12. Assuming n ≤ 0.05N, where n = sample size and N = population size, find the probability that the sample mean, x¯, for a random sample of 24 taken from this population will be between 54.91 and 61.79.

Let x be a continuous random variable that has a normal distribution with μ = 60 and σ = 12. Assuming n ≤ 0.05N, where n = sample size and N = population size,


(a) Compute the mean, μx¯ of x¯ .

(b) Compute the standard deviation σx¯ of x¯ .

Answer 1

Solution :

Given that,

mean = = 60

standard deviation = = 12

n = 24

a) = = 60

b) = / n = 12 / 24 = 2.45

c) P(54.91 < < 61.79)  

= P[(54.91 - 60) / 2.45 < ( - ) / < (61.79 - 60) / 2.45)]

= P(-2.08 < Z < 0.73)

= P(Z < 0.73) - P(Z < -2.08)

Using z table,  

= 0.7673 - 0.0188

= 0.7485