# Statistics

• Sep 25th 2008, 08:00 AM
gibonwa33
Statistics
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.

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?

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

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

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?

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? (Rofl)
• Sep 25th 2008, 03:05 PM
mr fantastic
Quote:

Originally Posted by gibonwa33
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? (Rofl)

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 $\displaystyle {\color{red}\mu}$: $\displaystyle {\color{red}Z = \frac{X - \mu}{\sigma} \Rightarrow z* = \frac{80 - \mu}{9}}$

..