Question

Suppose the population average amount of dog biscuits in a bag is 40 pounds with a standard deviation of 6 pounds. A simple random sample of 64 bags is selected. What is the probability the sample mean weight will be within 2.25 pounds of the population mean?

Question 17 options:

1)

.9973

2)

.0228

3)

.6827

4)

.9545

Answer 1

Solution:

We are given

µ = 40

σ = 6

n = 64

We have to find P(40 – 2.25 <Xbar<40+2.25) = P(Xbar < 42.25) – P(Xbar< 37.75)

Find P(Xbar<42.25)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (42.25 – 40)/[6/sqrt(64)]

Z = 3

P(Z< 3) = P(Xbar<42.25) = 0.99865

(by using z-table)

Now find P(Xbar<37.75)

Z = (37.75 – 40)/[6/sqrt(64)]

Z = -3

P(Z< -3) = P(Xbar<37.75) = 0.00135

(by using z-table)

Required probability = P(Xbar < 42.25) – P(Xbar< 37.75)

Required probability = 0.99865 - 0.00135

Required probability = 0.9973