The mean amount purchased by a typical customer at Churchill\'s Grocery Store is

Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is $29.50 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 76 customers, answer the following questions.

What is the likelihood the sample mean is at least $31.50? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

What is the likelihood the sample mean is greater than $28.50 but less than $31.50? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

Within what limits will 95 percent of the sample means occur? (Round your answers to 2 decimal places.)

Explanation / Answer

a)

z(31.50) = (31.50-29.50)/(6/sqrt(6)) = 2.906

P(x-bar >=31.50)
= P(z > 2.906)
= 1P ( Z<2.906 )
=10.9982
=0.0018

b)

z(28.50) = (28.50-29.50)/(6/sqrt(76)) = -1.4529

P(28.50< x-bar < 31.50) = P(-1.4529 < z < 2.906) = normalcdf(-1.4529,2.906) = 0.9247

c)

Corresponding sample mean interval ::
x-bar = -1.96*(6/sqrt(76)) + 29.50 = $28.15
x-bar = +1.96*(6/sqrt(76))+ 29.50 = $30.84
Ans: (28.15 ,30.84)

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