Question

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The distribution of the number of people in line at a grocery store has a mean...

The distribution of the number of people in line at a grocery store has a mean of 3 and a variance of 9. A sample of the numbers of people in line in 50 stores is taken.

(a) Calculate the probability that the sample mean is more than 4? Round values to four decimal places.

(b) Calculate the probability the sample mean is less than 2.5. Round answers to four decimal places.

(c) Calculate the probability that the the sample mean diﬀers from the population mean by less than 0.5. Round answers to four decimal places.

Please help, and show step by step. Thank you.

Solutions

Expert Solution

Population mean = 3

Population variance = 9

Population standard deviation = 3

Sample size, n = 50

For sample

Sample mean = Population mean = 3

Sample standard deviation, sd = Population standard deviation / sqrt(n) = 3/sqrt(50) = 0.4243

(a) Calculate the probability that the sample mean is more than 4?

Let, x_bar be the sample mean

P[ x_bar > 4 ] = P[ ( x_bar - mean )sd > ( 4 - mean )sd ]

P[ x_bar > 4 ] = P[ ( x_bar - 3 )0.4243 > ( 4 - 3 )0.4243 ]

P[ x_bar > 4 ] = P[ Z > 2.36 ]

P[ x_bar > 4 ] = 1 - P[ Z < 2.36 ]

P[ x_bar > 4 ] = 1 - 0.9909

P[ x_bar > 4 ] = 0.0091

(b) Calculate the probability the sample mean is less than 2.5.

Let, x_bar be the sample mean

P[ x_bar < 2.5 ] = P[ ( x_bar - mean )sd < ( 2.5 - mean )sd ]

P[ x_bar < 2.5 ] = P[ ( x_bar - 3 )0.4243 < ( 2.5 - 3 )0.4243 ]

P[ x_bar < 2.5 ] = P[ Z < -1.18 ]

P[ x_bar < 2.5 ] = 0.1190

Let, x_bar be the sample mean

P[ 3 - 0.5 < x_bar < 3 + 0.5 ] = P[ 2.5 < x_bar < 3.5 ]

P[ 2.5 < x_bar < 3.5 ] = P[ ( 2.5 - mean )sd < ( x_bar - mean )sd < ( 3.5 - mean )sd ]

P[ 2.5 < x_bar < 3.5 ] = P[ ( 2.5 - 3 )0.4243 < ( x_bar - 3 )0.4243 < ( 3.5 - 3 )0.4243 ]

P[ 2.5 < x_bar < 3.5 ] = P[ -1.181 < Z < 1.18 ]

P[ 2.5 < x_bar < 3.5 ] = P[ Z < 1.18 ] - P[ Z < -1.18 ]

P[ 2.5 < x_bar < 3.5 ] = 0.8810 - 0.1190

P[ 2.5 < x_bar < 3.5 ] = 0.7620

milcah answered 2 days ago

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