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Statistics problem help - Printable Version +- Online Degrees and CLEP and DSST Exam Prep Discussion (https://degreeforum.net/mybb) +-- Forum: Main Category (https://degreeforum.net/mybb/Forum-Main-Category) +--- Forum: General Education-Related Discussion (https://degreeforum.net/mybb/Forum-General-Education-Related-Discussion) +--- Thread: Statistics problem help (/Thread-Statistics-problem-help) |
Statistics problem help - hunter2115 - 10-09-2007 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? Statistics problem help - blueyzz28 - 10-09-2007 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!! Statistics problem help - blueyzz28 - 10-09-2007 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. Statistics problem help - hunter2115 - 10-09-2007 blueyzz28 Wrote: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. Statistics problem help - blueyzz28 - 10-09-2007 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? Statistics problem help - hunter2115 - 10-09-2007 blueyzz28 Wrote: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. 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! Statistics problem help - blueyzz28 - 10-09-2007 You are correct. I used the estimates to get the 13.50% but your z-score is probably closer at 13.59%. Statistics problem help - jec959 - 10-09-2007 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? Statistics problem help - hunter2115 - 10-10-2007 jec959 Wrote: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! |