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?
Answers
Given : mean expenditure per customer at a tire store is $80.00, with a standard deviation of $10.00
a random sample of 40 customers is taken
To find : 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?
Solution:
Mean = 80
SD = σ = 10
SE = standard error of Mean
n = sample size = 40
SE = σ/ √n
=> SE =10/√40
=> SE = 1.58
sample average expenditure per customer for this sample will be $87.00
Z Score = ( 87 - 80 )/ 1.58 = 4.43
≈ 100 % sample average expenditure per customer for this sample will be $87.00 or less
Hence probability 1
≈ 0 % sample average expenditure per customer for this sample will be $87.00 or or more
probability 0
between $70.00 and $85.00
=> Between z scores = ( - 10/ 1.58) and ( 5/1.58)
=> -6.33 and 3.16
Probability = 0.9992 - 0 = 0.9992
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