Question

Suppose the average checkout bill at a large supermarket is 73.25 dollars, with a standard deviation of 24.75 dollars.

22 percent of the time when a random sample of 41 customer bills is examined, the sample average will exceed what value?

Note that this is a question about the sample mean.

Enter a numeric value correct to two digits after the decimal point.

In a random sample of 41 customer bills, what is the probability that the sample average will be less than 70.16?

Give the answer as a percent (no percent symbol), correct to two decimal digits:

Answer 1

Solution :

Given that ,

mean = = 73.25

standard deviation = = 24.75

n = 41

= = 73.25 and

= / n = 24.75 / 41 = 3.8653

1)

The z - distribution of the 22 % is,

P( Z > z ) = 22%

1 - P( Z < z ) = 0.22

P( Z <  ) = 1 - 0.22

P( Z < z ) = 0.78

P( Z < 0.77) = 0.78

z = 0.77

Using z - score formula,

= z * +

= 0.77 * 3.8653 + 73.25

= 76.23

2)

P( < 70.16) =   P(( - ) / < (70.16 - 73.25) / 3.8653)

= P(z < -0.80)   Using standard normal table.   

= 0.2119

= 21.19