Statistics problem help

[hunter2115]

I'm having trouble with a particular problem in my Penn Foster Business Stats course. I thought maybe someone could help me out. The problem is as follows:
"The Burger Bin fast-food restaurant sells a mean of 24 burgers an hour and its burger sales are normally distributed. If hourly sales fall between 24 and 42 burgers 49.85% of the time, the standard deviation is"?

The second part goes on to ask:
"Using the information from above, what is the probability that the Burger Bin will sell 12 to 18 burgers in an hour"?

I think my main problem lies in figuring the standard deviation of this normal distribution (part a). My thinking is: 42 - 24 = 18 x .4985 = 8.973 (the variance). The stadard deviation would be the square root of the variance, or 2.995

Anyone?

[blueyzz28]

If the mean is 24 and distribution is normally distributed, then your standard deviation would be 3. I get that because 49.85% is above the mean, if you double that number you arrive at at 99.7%. Remember that 99.7% is 3 standard deviations when dealing with the normal distribution.

As for the probability of serviing 12-18 hamburgers or hotdogs, I do not remember which one we are serving, it would be 13.50%, I believe. You get that by plugging in the standard deviations on both sides of the distribution and the percentages associated with them. 12-24 hamburgers would be two standard deviations below the mean which would equal 47.5%. Between 18 and 24 would be 34%. So, the difference between the two is 13.5%.

I think that is correct but I could be very wrong!!

[blueyzz28]

I forgot to add that the standard deviation number is 6. The number of standard deviations that contains the 24-42 hamburgers is 3. Again, I could be totally wrong, but I believe that is correct.

[hunter2115]

I forgot to add that the standard deviation number is 6. The number of standard deviations that contains the 24-42 hamburgers is 3. Again, I could be totally wrong, but I believe that is correct.

Is there a formula for calculating the standard deviation in the 1st part of the question? I think if I understand the standard deviation, I will be able to figure the 2ns part.

[blueyzz28]

Assuming I am correct, you cannot figure the standard deviation the normal way because you have no data values. What I did reading your problem was the following. I know that the normal distribution has a mean of 24 and 49.85% of the values lie between 24 and 42. So, I know that this equates to 3 standard deviations above the mean. So, I took 42-24 which equals 18 and divided by the number of standard deviations above the mean, which is 3. This gives me 18/3 which equals 6.

Does that make any sense?

[hunter2115]

Assuming I am correct, you cannot figure the standard deviation the normal way because you have no data values. What I did reading your problem was the following. I know that the normal distribution has a mean of 24 and 49.85% of the values lie between 24 and 42. So, I know that this equates to 3 standard deviations above the mean. So, I took 42-24 which equals 18 and divided by the number of standard deviations above the mean, which is 3. This gives me 18/3 which equals 6.

Does that make any sense?

Got it! I think you are absolutely correct on this. The second part of the question can then be figured using z-scores, when:
z= x - the mean, divided by the standard deviation

12-24 divided by 6 = -2, or P= .4772 (from the z score table)
18-24 divided by 6 = -1, or P= .3414

.4772 - .3414 = .1359 (the probability of selling between 12 and 18). Does that seem correct?
Thanks for your help!

[blueyzz28]

You are correct. I used the estimates to get the 13.50% but your z-score is probably closer at 13.59%.

[jec959]

Since my proctor hasn't received the exam packages for my math and finance course I haven't started this yet. I was wondering how the course was?

[hunter2115]

Since my proctor hasn't received the exam packages for my math and finance course I haven't started this yet. I was wondering how the course was?

It's a little tricky considering that you really dont have alot of support. But, as you can see from this thread... we have friends out there!! (thanks Bluezyzz28)
The course is graded 2/3 on 3 unproctored exams and 1/3 on 1 proctored final (1 hour)exam. I scored 100% on exam #1, but it's getting tougher fast.
I am definately one of those "math-phobic" types and I'm not really sure if I could pass the DSST Stats exam or a course in Stats, so I figure this is my best option. Time will tell!