Question 1071705: The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 62 customers, answer the following questions.
a.
What is the likelihood the sample mean is at least $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
b.
What is the likelihood the sample mean is greater than $25.00 but less than $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
c.
Within what limits will 90 percent of the sample means occur? (Round your answers to 2 decimal places.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 62 customers, answer the following questions.
a.
What is the likelihood the sample mean is at least $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
z(27) = (27-26)/(6/sqrt(62)) = 1.31
P(x-bar >=27) = P(z > 1.31) = normalcdf(1.31,100) = 0.0951
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b.
What is the likelihood the sample mean is greater than $25.00 but less than $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
z(25) = (25-26)/(6/sqrt(62)) = -1.31
P(25< x-bar < 27) = P(-1.31< z < 1.31) = normalcdf(-1.31,1.31) = 0.8098
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c.
Within what limits will 90 percent of the sample means occur? (Round your answers to 2 decimal places.)
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Left tail of that interval has 0.05 of the area under the curb
Left boundary of the interval:: z = invNorm(0.05) = -1.6449
Right boundary of the interval:: z = invNorm(0.95) = 1.6449
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Corresponding sample mean interval ::
x-bar = -1.6449*(6/sqrt(62)) + 26 = $24.75
x-bar = +1.6449*(6/sqrt(62))+26 = $27.25
Ans: (24.75,27,25)
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Cheers,
Stan H.
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