standard deviation

Question

The scores on a particular exam are normally distributed with a mean of 57 and a standard deviation of 8.

1. What is the probability that a randomly selected student scored below a 50%?
2. What proportion of students passed? (received a score of 70% or above)
3. What score represents the 90th percentile for test-takers?
4. Round each answer to four decimal places.

Answer 1

To find the percent of a samples in a certain range of a Normal distribution, you will need to use Z-scores (found in a table of your Stats book, or online).

The Z score is the value for how many standard deviations a number is from the mean of the distribution. It is calculated as (value - mean) / (standard deviation).

In this example with a mean of 57 and a standard deviation of 8, the value 65 has a Z-score of +1, and the value 49 has a Z-score of -1. But Z-scores do not need to be integers.

The Z-score you need for Part 1 is calculated, as above:

z = (value - mean) / (standard deviation)

which here is:

z = (50 - 57) / 8

When you look up this Z-score on a table, it will tell the proportion of the distribution that lies below 50 (which is the answer!).

If you needed to know the probability of a score above a value (as in Part 2), then you would take 1 minus (the probability found from the Z-score table).

For Part 3, the 90th percentile is the score at which 90% of the population scored at or below. You'll do this in the reverse way as Parts 1 and 2. First you must find the Z-score that corresponds to 90% (or 0.9). Then, using the same equation you used above, calculate the value of an exam score that corresponds to this Z-score.