The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. (a) What is the likelihood the sample mean is at least $25.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
(c) Within what limits will 90 percent of the sample means occur? (Round your answers to 2 decimal places.)

a.     ( to four decimal places)

b.   ( to four decimal places )

c. The limits will be between the interval of   ( 22.33,24.67 )

Step-by-step explanation:

Given that :

mean = 23.50

standard deviation = 5.00

sample size = 50

The objective is to calculate the following:

(a)  What is the likelihood the sample mean is at least $25.00?

Let X be the random variable, the probability that the sample mean is at least 25.00 is:

  to two decimal places

From the normal tables :

    ( to four decimal places)

(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00?

 to four decimal places

(c) Within what limits will 90 percent of the sample means occur?

At 90 % confidence interval, level of significance = 1 - 0.90 = 0.10

The critical value for the = 1.65

Standard Error =

Standard Error =  

Standard Error = 0.7071

Therefore, at 90 percent of the sample means, the limits will be between the intervals of :

Lower limit =  ( 23.5 - (1.65×0.707) )

Lower limit =  ( 23.5 - 1.16655 )

Lower limit = 22.33345

Lower limit = 22.33    (to two decimal places).

Upper Limit = ( 23.5 + (1.65*0.707) )

Upper Limit = ( 23.5 + 1.16655 )

Upper Limit = 24.66655

Upper Limit = 24.67

The limits will be between the interval of   ( 22.33,24.67 )