The service time of the first service of a BMW car is found to be normally distributed with a mean of 70 minnutes and a variance of 81 minutes.

Mr F says: For questions 1 - 3, let X be the random variable 'service time of the first service of a BMW car (minutes)'.

1. If a customer brings her BMW car for its first service, what is the probability that the car will be ready

within one hour?

Mr F says: Calculate Pr(X < 60). I assume you know how to convert X = 60 into a z-value and use standard normal tables to do the calculation .....

2. What is the probability that the job will take more than 90 min?

Mr F says: Calculate Pr(X > 90). I assume you know how to convert X = 90 into a z-value and use standard normal tables to do the calculation .....

3. What percentage of the first service will be completed between 50 min and 60 min?

Mr F says: Calculate Pr(50 < X < 60) = Pr(X < 60) - Pr(X < 50). I assume you know how to convert X = 50 and X = 60 into z-values and use standard normal tables to do the calculations ..... Multiply the final answer yb 10 to convert probability into a percentage.

4. The BMW dealer has policy to give its customers a 15% discount on the cost if the first service is not completed within 80 min. From a sample of 80 customers who brought their cars in for the first service, how many are likely to receive the 15% discount?

Mr F says: Calculate p = Pr(X > 80). I assume you know how to convert X = 80 into a z-value and use standard normal tables to do the calculation .....

Let Y be the random variable 'number of customers who receive the 15% discount'.

Y ~ **Binomial**(p = value calculated above, n = 80).

Calculate E(Y).

5. If the dealer wants to ensure that no more than 5% of all its first service will take longer than 80 min, what should the mean service time be? Mr F says: I assume the variance is still 81 ....? Use your standard normal tables (or otherwise) to find the value z* such that Pr(Z > z*) = 0.05. Now solve for :