Hi there...

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

I believe I've solved the first two parts of the question but am confused by the probability distribution of x (I hope I'm correct that it is .30, .45, .15, .09, which add up to a probability of 1). I am to calculate the expected value of x - the mean - and although I arrived at 45 as the mean, I'm not confident I am calculating this correctly as I think I may be confusing the formulas for discrete vs continuous variables.

Any advice would be greatly appreciated...Shannon

The expected value formulas for discrete and continuous variables follow the same basic principle:

Summation of [ x * p(x) ]

Your answer is essentially correct, but you may want to:

1.make the midpoint of each category as 34.5, 44.5, 54.5, 64.5
- for example, the midpoint of 30,31,32,33,34,35,36,37,38,39 is 34.5

2.and extend the probabilities by one decimal place.

You'll get an answer of 44.85, which is a bit more precise.

That's great, John...thanks so much! I actually had calculated 44.85 but for some reason, decided to round it off (long day, I guess!).

This forum has been a terrific help. While I'm a loooong way from being proficient, I'm feeling a lot less frustration as I continue to learn from this site.

Have a great day...Shannon