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klewis
02-10-2006, 03:43 PM
Two Question first one I have answered just checking to see if process is correct...........Second question I just cant get my head around

First Question

One of Canada's well know distance learning institutions published the age profile of its first-time students. A random sample of 33 has ten students in thirties, fifteen students in forties, five students in fifties and three students in sixties. A student is randomly selected from the sample. Let x be the age of the selected student. Calculate

a. probability that the student is of age 30?

ans 10/33 .303030330 or round to 30%

b. probability that the student is of age 50 or 60?

ans P(x=50 or x=60) = P(x=50)+P(x=60)
= 5/33 + 3/33 = 8/33 or 24.24 or 24%

c. probability distribution of x

x P(x) x*P(x) x2 x2*P(x)

30 10/33 or .30 9.09 or 9 900 272.72 or 270
40 15/33 or .45 18.18 or 18 1600 727.27 or 720
50 5/33 or .15 7.57 or 7.5 2500 378.78 or 375
60 3/33 or .09 5.45 or 5.4 3600 327.27 or 324

Total 33/33 or .99 ???

because of repeating numbers after decimal do I use fraction for calculations or do I use decimals rounded accordingly?? I know you get the picture if I have over stated

d. expected value of x

mean = the sum of [x*p(x)] = 39.9 or 40.29 ?? depending on fraction or decimal calculation

Second Question

In grading apples into "A","B" and "C" a large orchard uses weights to distinguish apples. Any apple weighing more than 2 ounces is classified as Grade "A" while apple weighing less than 0.75 ounces is classified as Grade "C". If the days pick shows 16.6% are Grade "A" and 6.68% are Grade C, determine the mean and standard deviation. Assume weights are normally distributed.

Here is a HINT that I dont get - First calculate the appropriate z values for the probabilities given, then set up your equations to solve for the mean and standard deviation

I gather we know percentage of Grade A being 16.6% but we don't know how many? same with Grade C 6.68%.....I am guessing once we find out these two Grades we can solve for Grade B and then mean and standard deviation

JohnM
02-10-2006, 07:45 PM
Problem 1 looks OK to me. For part c, 33/33 = 1.0. In part d, I get 40.3 for the mean, which is close to yours.

The second problem has actually been worked out in the forum a while ago. Here's a link:

http://www.talkstats.com/probability/166-calculating-z-value-mean-standard-deviation.html?highlight=orchard