Answer to Question #213191 in Statistics and Probability for Manasa Reddyrajula

Question

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?

Answer 1

$\mu = 80.00 \\ \sigma=10.00 \\ n=40$

a)

$P(\bar{x} ≤ 87.00) = P(Z≤ \frac{87.00-80.00}{10.00/ \sqrt{40}}) \\ = P(Z≤ 4.42) \\ = 0.9999$

b)

$P(\bar{x} ≥87.00) = 1 -P(Z<87.00) \\ = 1 - 0.9999 \\ = 0.0001$

c)

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