Question #213191

Suppose the mean expenditure per customer at a tire store is $80.00, with a standard deviation


of $10.00. If a random sample of 40 customers is taken, what is the probability


a) that the sample average expenditure per customer for this sample will be $87.00 or


less?


b) that the sample average expenditure per customer for this sample will be $87.00 or


more?


c) that the sample average expenditure per customer for this sample will be between


$70.00 and $85.00?



1
Expert's answer
2021-07-06T12:13:00-0400

μ=80.00σ=10.00n=40\mu = 80.00 \\ \sigma=10.00 \\ n=40

a)

P(xˉ87.00)=P(Z87.0080.0010.00/40)=P(Z4.42)=0.9999P(\bar{x} ≤ 87.00) = P(Z≤ \frac{87.00-80.00}{10.00/ \sqrt{40}}) \\ = P(Z≤ 4.42) \\ = 0.9999

b)

P(xˉ87.00)=1P(Z<87.00)=10.9999=0.0001P(\bar{x} ≥87.00) = 1 -P(Z<87.00) \\ = 1 - 0.9999 \\ = 0.0001

c)

P(70.00<xˉ<85.00)=P(xˉ<85.00)P(xˉ<70.00)=P(Z<85.0080.0010.00/40)P(Z<70.0080.0010.00/40)=P(Z<3.16)P(Z<6.32)=0.999210.00003=0.99918P(70.00< \bar{x}<85.00) = P(\bar{x}<85.00) -P(\bar{x}<70.00) \\ =P(Z< \frac{85.00 -80.00}{10.00/ \sqrt{40}}) -P(Z< \frac{70.00 -80.00}{10.00/ \sqrt{40}}) \\ =P(Z<3.16) -P(Z< -6.32) \\ = 0.99921 -0.00003 \\ = 0.99918


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