How should I approach this question?

Question

At a particular university, students' grades in introductory statistic classes are generally unimodal and skewed to the left with a mean of u = 63 and a standard deviation of o = 18.0. (Round to 2 decimal places for all z-values and round all other answers to 4 decimal places, if needed.)

(a) The distribution of students' grades is ______(b2).

(b) If n = 40 students are selected at random, the distribution of the sample mean grade is _______(b1) with a mean of ______ and a standard deviation of ____.

(c) The probability that the sample mean grade for these 40 students is less than 67.0 is ____.

(d) If n = 40 students are selected at random, the distribution of the sample total grade is ______(b1) with a mean of _______ and a standard deviation of ____.

(e) The probability that the total grade for these 40 students is less than 2680.0 is _____.

(b2) options= (left skewed) or (right skewed) or (may/may not be normal) or (exactly normal) or (approximately normal)

(b1) options= (left skewed) or (right skewed) or (not normal) or (approximately normal)

Answer 1

Mia, this is more an addendum or clarification of Mr d's answer. (check out the surprise ending however)

To make things clear. Assume you have a population with measurement which has an undefined distribution. You can take a sample of the population to try and estimate the mean of that measure. If you take a sample of 20 measurements and get a mean, that is one estimate. If you take another sample of 20, that is another estimate. None of the estimates will be exactly equal to the population mean but they will be close. It is these multiple estimates which will have (as the number of estimates get larger) which will have a distribution which approaches a normal distribution. The population is still an undefined distribution. This is why we call the measure of the variability of these population mean estimates the standard deviation of the MEAN, to differentiate it from the standard deviation of the population itself. Some texts use the term standard error to further emphasize the difference.

This works not just for mean but any measurement which is a measure of central tendency (as in the central limit theory). However for question d) which is talking NOT about the mean but the total of the 40 samples, the distribution of each one of those samples of 40 (assuming others would be made) would not be normal , because we are talking about the total, not the mean.

Sorry to take so long because here is the major change. Mr D didn't mention this part at all.

While the mean of the sum of 2 or more independent variables is equal to the sum of their expected values, the standard deviation is NOT. The Variance is the sum of the variances, which is not the same.

So, the standard deviation of the sum of a set of independent random variables is equal to the square root of the sum of their variances.

So for question d) we have

E(Sum of 40 samples) = 40* 63 = 2520. V(Sum of 40 samples) = 40 * 18² = 12960.

The standard deviation of the total of multiple samples of 40 students is

sd = √(12960) or 113.8. (not the mean so not the sd of the mean - get it?)

To answer d) we calculate the Z score as (2680 - 2520)/113.8. which is 1.406 P(z<1.406) ≈ .920

Note, because the sum of multiple variables retains the original distribution, this answer is approximate as the distribution of the total values is left skewed.

Answer 2

Hi Mia L.

Here's the crucial piece of information which you might easily miss (and end up in the left-skewed portion of student grades....?):

Individual data have mean, sample deviation (usually termed standard deviation, but (n-1)ot quite the same!), and a distribution shape (normal, skewed variously, kurtosis, etc.). But whenever n>=2 of those data, randomly selected, are pooled together (i.e. by averaging = taking the mean), the new set of pooled-data points are ALWAYS normally distributed. The mean will not change on going to pooled data (such as you are considering in parts (b) onwards above), but the sample deviation becomes smaller, by the factor of (n)^(0.5), where n points of original data are pooled together.

This fact has implications. If you ever want to consider rejecting "erroneous" data, or consider if data are skewed, you MUST do it on the original data set, not on any averaged or replicated measurements.

I reckon you can cope with the probability tables on standard distributions?

--Cheers, -- Mr. d.