Probability normal distribution

Question

A hospital is trying to improve quality in every area. One of the metrics they look closely at is the quality of physicians. Several individual measurements contribute to a physician's quality score, but they have observed physician quality rankings to be normally distributed, with an average quality score of 83, and a standard deviation of 4.8. A physician score of 94 or better is considered to be exceptional.

What portion of physicians would have a quality score between 73 and 93 (inclusive)  (round to the nearest 4 decimal places)?

Answer 1

Hi Anto,

In order to answer this question, we should apply the formula for "between" probabilities:

P(a < x < b) = P(x < b) - P(x < a)

(Don't worry about the fact that these are strict inequalities and the problem says "inclusive" - for a continuous distribution like the Normal, that really doesn't matter.)

So we have:

P(73 < x < 93) = P(x < 93) - P(x < 73)

Now, let's calculate the z-scores for each of the boundaries:

z₉₃ = (93 - 83)/4.8 = 2.08

z₇₃ = (73- 83)/4.8 = ^-2.08

Now, use your table of z-values to find the corresponding probabilities:

z = 2.08 gives a probability of 0.9812, and z = ^-2.08 gives a probability of 0.0188

Now, back to the "between" formula to get the final answer:

P(73 < x < 93) = P(x < 93) - P(x < 73)

= 0.9812 - 0.0188

=0.9624

There are other ways to do this problem since the boundaries are symmetric around the mean, but I think this is the way that is the easiest to understand (and the most general.) Does all that make sense?