Question

The amount of money spent by Superstore customers is normally distributed with mean 150 and a standard deviation of 12. Suppose that a sample of 64 customers are selected. Answer the following questions.

What is the mean of the sampling distribution?

150

140

170

240

What is the standard error of the sampling distribution?

1.23

1.50

1.55

1.20

What is the probability that the average spending by the customers is greater than 154?

0.9962

0.9966

0.0038

0.0048

What is the probability that the average spending by the customers is less than 154?

0.9962

0.9966

0.0038

0.0048

If the distribution of data is skewed to the left:

Median < Median < Mode

Mean, mode and median are equal

Only mean and mode are equal

None of the above

Answer 1

Solution :

Given that,

mean = = 150

standard deviation = = 12

n = 64

=   = 150

= / n = 12 / 64 = 1.50

a) P( > 154) = 1 - P( < 154)

= 1 - P[( - ) / < (154 - 150) / 1.50 ]

= 1 - P(z <2.67)

Using z table,    

= 1 - 0.9962

= 0.0038

b) P( < 154) = P(( - ) / < (154 - 150) / 1.50)

= P(z < 2.67)

Using z table

= 0.9962

c) Mean < Median < Mode