Mean of Test scores

Question

the mean of normal distributed test score is 79 and the standard deviation is 5 if there are 242 test scores in the data sample how many of them were in the 84 to 94 range

Answer 1

Question: The mean of normal distributed test score is 79 and the standard deviation is 5 if there are 242 test scores in the data sample how many of them were in the 84 to 94 range?

Answer: Test scores are usually a phenomenon that follows a normal distribution. You also have large data. So your strategy becomes this: Let X be a variable representing test scores (i.e., X = 85 means that a certain test score is 85). The questions asks for this: P(84 < X < 94). If you show this on an illustrated bell curve, you'll see an "in-between" kind of shading going on.

There are two ways to do this problem:

TI-83/84: 2nd VARS (to get to the DISTR menu) --> 2: normalcdf(

You need to enter the arguments for normalcdf in this order: normalcdf(lower_bound, upper_bound, mean, SD). Calculate normalcdf(84, 94, 79, 5). This is approximately 0.1573. This means that 15.73% of scores are between 84 and 94.

You could also get the same answer by doing normalcdf(1, 3), but those two numbers come only from finding the z-scores corresponding to the 84 and 94 scores (see the manual method below to find these z-scores). Doing normalcdf(1, 3) assumes a mean of 0 and SD of 1 (the standard normal distribution).

15.73% of 242 test scores is .1573(242) = approximately 38 of the total scores.

Manually (requires a z-table): Find two z-scores: one for the 84 raw score of interest, and another for the 94 raw score of interest. We need to see where these scores would fall on a z-distribution shaped by mean of 79 and SD of 5.

For 84: z = (84 - 79)/5 = 1

For 94: z = (94 - 79)/5 = 3

So now our question of P(84 < X < 94) can be thought of as P(1 < z < 3). The z-scores help to standardize raw scores on a level playing field, the normal distribution. Go to a z-table and find the area beneath z = 3 and the area beneath z = 1. Then find: (area below z = 3) - (area below z = 1) = 0.9987 - 0.8413 = 0.1574.

This result means that 15.74% of scores are between 84 and 94. This is off from the calculator answer by 0.0001 because values in a printed z-table are usually rounded to three or four places after the decimal point.

15.74% of 242 test scores is .1574(242) = approximately 38 of the total scores.