Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 50% discount on the charges. The company wants to limit this discount to at most 5% of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.
I worked out the problem and found out that the z-value that corresponds to the required x-value. I found the z value from the normal distribution table for .0500. I used the standard normal distribution table to find out that the z-value is - 1.645. Substituting the values of ì, standard deviation(ó), and z in the formula x= ì+zó, we obtain
x = ì+zó = 15 +(-1.64)(2.4)=
15-3.936 = 11.064
I looked in the back of the book and the answer said approximately 19. I understand how they got to approximately 19. They figured out that the z-value was 1.64 and then they plugged it into the formula and that is how they got approx. 19. But I'm having trouble understanding why its 1.64 and not -1.64. If anyone could help me figure out where I'm going wrong I'd appreciate it. Thank you!
They want to give a discount to 5% of the people that are inconvenienced by a longer waiting time.
That means that 95% or .95 of the people should be satisfied.
So you want to find the prob that the area of region BELOW a certain waiting time is .95
The z-score of -1.94 would mean that the people get their service in a shorter time than the mean.
That would be good!
So they would not be inconvenienced, why would they expect a discount ?
I find this little applet very useful.
http://davidmlane.com/hyperstat/z_table.html
Whether you look up stats values in a chart, on a graph, a calculator, or a divise such as this, really makes no difference to me. Nobody actually does the real calculations in any of these methods.
The neat thing about this applet is that you don't even have to find the z-scores, you can enter the mean and standard deviation directly and you get a good visual of the situation.
Use the second of the applets, enter the mean and standard deviation and enter .95 in the "shaded area", click on the "below" and you should get 18.9476 which they rounded to 19
The weights (in pounds) of a sample of five boxes being sent by UPS are: 12, 6, 7, 3, and 10.
a. Compute the mean for this sample. Show your work!!
b. Compute the range for this sample. Show your work!!
c. Compute the variance for this sample. Show your work!!
d. Compute the standard deviation for this sample. Show your work!!
Please show all of your work when answering the following:
I. Use Chebyshev’s theorem to find what percent of the values will fall between 236 and 338 for a data set with a mean of 287 and standard deviation of 17.
II. Use the Empirical Rule to find what two values 95% of the data will fall between for a data set with a mean of 294 and standard deviation of 13
Well, it seems like you're having a bit of trouble with the sign of the z-value. Let me try to clarify it for you with a touch of humor!
Imagine you're standing in line at a comedy show, waiting for your turn to enter. The mean waiting time for everyone is 15 minutes, but you really don't want to spend more than that. So, you set a maximum waiting time for yourself, let's say 19 minutes.
Now, let's find out how far away from the mean (15 minutes) your desired waiting time is. If we calculate that, we get 19 - 15 = 4 minutes. This means our desired waiting time is 4 minutes longer than the mean.
In the language of statistics, we call that difference a positive value. And when we look up the z-value in the standard normal distribution table, we're actually looking for the area under the curve to the right of our desired waiting time.
Since we want to find the area to the right of our desired waiting time (19 minutes), we need the z-value for that positive area! That's why we need to use the z-value of approximately 1.64, rather than -1.64.
So, in summary, when determining the z-value and the corresponding waiting time, we need to consider the direction we're interested in. And in this case, it's the positive direction, like wanting to be at the front of the line at a comedy show!
I hope that clears things up for you. If you have any more questions or if you need another dose of humor, feel free to ask!
To understand why the z-value used in this problem is positive (+1.645), let's first consider the problem statement. The management of the Fast Auto Service wants to limit the discount to at most 5% of the customers.
In order to determine the maximum waiting time that guarantees at most a 5% discount, we need to find the corresponding z-value for a cumulative probability of 0.05 (5%). This is because the z-value represents the number of standard deviations away from the mean.
When using a standard normal distribution table, the table provides the cumulative probability to the left of the z-value. In this case, the table provides the cumulative probability to the left of the z-value, which represents the probability of a waiting time being less than a certain value.
In other words, using the standard normal distribution table, we need to find the z-value that corresponds to a cumulative probability of 0.95 (100% - 5%).
Since the table only provides the cumulative probability to the left of the z-value, we look for the closest value to 0.95 in the table, which is 0.950. This value corresponds to a z-value of approximately +1.645.
This positive z-value indicates that the waiting time of the customer should be greater than the mean to limit the discount to at most 5% of the customers.
Therefore, by substituting the positive z-value (+1.645) into the formula x = ì + zó, we obtain:
x = 15 + (1.645)(2.4) ≈ 18.9
Hence, the maximum guaranteed waiting time should be approximately 19 minutes to ensure that at most 5% of customers receive a discount.