Question

Ch 7 # 5: Please help answer the probability for a-e:

A small hair salon in Denver, Colorado, averages about 60 customers on weekdays with a standard deviation of 15. It is safe to assume that the underlying distribution is normal. In an attempt to increase the number of weekday customers, the manager offers a $4 discount on 7 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 7 weekday period jumps to 71. [You may find it useful to reference the z table.]

a. What is the probability to get a sample average of 71 or more customers if the manager had not offered the discount? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Probability:___________________

   b. Do you feel confident that the manager’s discount strategy has worked?

• No, there is good chance (more than 5%) of getting 71 or more customers without the discount.

• No, there is only a small chance (less than 5%) of getting 71 or more customers without the discount.

• Yes, there is good chance (more than 5%) of getting 71 or more customers without the discount.

• $\rightarrow$ (this is the correct answer) Yes, there is only a small chance (less than 5%) of getting 71 or more customers without the discount.

• Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 38.2 and 4.3 mpg, respectively. [You may find it useful to reference the z table.]

  c. What is the probability that a randomly selected passenger car gets more than 40 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

• Probability:___________________

  d. What is the probability that the average mpg of five randomly selected passenger cars is more than 40 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

  Probability:___________________

• e. If five passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 40 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

• Probability:___________________

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	60
std deviation   =σ=	15.0000
sample size       =n=	7
std error=σx̅=σ/√n=	5.6695

probability =

P(X>71)

=

P(Z>1.94)=

1-P(Z<1.94)=

1-0.9738=

0.0262

b)

Yes, there is only a small chance (less than 5%) of getting 71 or more customers without the discount.

c)

probability =

P(X>40)

=

P(Z>0.42)=

1-P(Z<0.42)=

1-0.6628=

0.3372

d)

probability that all of the passenger cars get more than 40 mpg =(0.3372)⁵ =0.0044