Question

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 3408 miles, with a variance of 249,001. If he is correct, what is the probability that the mean of a sample of 36 cars would be less than 3245 miles

Answer 1

Answer:

The probability that the mean of a sample of 36 cars would be less than 3245 miles

P(X⁻≤3245) = 0.975

Step-by-step explanation:

Step(i):-

The mean number of miles between services

                   μ= 3408 miles  

The Variance of miles between services

                σ² = 249,001

                σ = √249,001 = 499

Let 'X' be a random variable in a normal distribution

Given sample size 'n' =36

Step(ii):-

           $Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

          Z = -163/83.16 = 1.96

The probability that the mean of a sample of 36 cars would be less than 3245 miles

P(X⁻≤3245)  =   P(Z≤1.96)

                   =   0.5 + A(1.96)

                   =  0.5 + 0.4750

                  =   0.975

Final answer:-

The probability that the mean of a sample of 36 cars would be less than 3245 miles

P(X⁻≤3245) =0.975