Question

Can you please explain me this Question? Thanks!

An investment club has 500 members. Their annual income is normally distributed with a mean of μ= $125,400 and a standard deviation σ= $27,642

a) If a random sample of 100 members is taken, what is the probability that their average is less than $125,00?

b) What is the probability that a randomly selected salary will be more than $125,000 hours?

Answer 1

Part a

Solution:

We are given

µ = 125400

σ = 27642

n = 100

We have to find P(X̄<125000)

Z = (X̄ - µ)/[σ/sqrt(n)]

Z = (125000 - 125400)/(27642/sqrt(100))

Z = -0.14471

P(Z<-0.14471) = 0.442471

(We can find this probability by using calculator/excel/z-table)

P(X̄<125000) = 0.442471

Required probability = 0.442471

Part b

Solution:

Solution:

We are given

µ = 125400

σ = 27642

We have to find P(X>125000)

P(X>125000) = 1 - P(X<125000)

Z = (X̄ - µ)/σ

Z = (125000 - 125400)/27642

Z = -0.01447

P(Z<-0.01447) = 0.494227

(We can find this probability by using calculator/excel/z-table)

P(X<125000) = 0.494227

P(X>125000) = 1 - P(X<125000)

P(X>125000) = 1 - 0.494227

P(X>125000) = 0.505773

Required probability =0.505773