Question

Nike’s annual report says that the average American buys 6.5 pairs of sport shoes per year. Suppose the
population standard deviation is 2.1 and that a sample of 25 customers will be examined next year.
Use this information to determine the following questions based on sampled means:
a. What is the standard error of the mean in this experiment?
b. What is the likelihood that the sample mean is between 6 and 7 pairs of sports shoes?
c. What is the likelihood that the sample mean is greater than 7 pairs?
d. What is the likelihood that the sample mean is less than 5 pairs?

Answer 1

Solution :

Given that,

mean = = 6.5

standard deviation = = 2.1

a) n = 25

= = 6.5

= / n = 2.1 / 25 = 0.42

b) P(6 < < 7)  

= P[(6 - 6.5) /0.42 < ( - ) / < (7 - 6.5) / 0.42)]

= P(-1.19 < Z < 1.19)

= P(Z < 1.19) - P(Z < -1.19)

Using z table,  

= 0.8830 - 0.1170   

= 0.7660

c) P( > 7) = 1 - P( < 7)

= 1 - P[( - ) / < (7 - 6.5) / 0.42 ]

= 1 - P(z < 1.19)

Using z table,    

= 1 - 0.8830

= 0.1170

d) P( < 5) = P(( - ) / < (5 - 6.5) / 0.42 )

= P(z < -3.57)

Using z table

= 0.0002