Question

Suppose a normal distribution has a mean of 49 and a standard deviation of

5. What is the probability that a data value is between 48 and 52? Round your


answer to the nearest tenth of a percent.


A. 26. 2%


B. 30. 5%


C. 31. 1%


D. 28. 8%

Answer 1

30.5%

Explanation:

use the z-score formula to convert the values.

z = (X - υ) / σ

where X is the test statistic, υ is the mean, σ is the standard deviation.

52 is a test statistic:

z = (52 - 49) / 5

= 0.6.

find this area in a z-score table (find +0.6 on the left column and .00 on top row. where these two meet is the area. that is 0.72575).

48 is a test statistic:

z = (48 - 49) / 5

= -0.2

find this area in a z-score table (find -0.2 on the left column and .00 on top row. where these two meet is the area. that is 0.42074).

these are both areas to the left of 52 and 48. we want the area (probability) between them.

so we subtract 0.42074 from 0.72575.

0.72575 - 0.42074 = 0.30501. that is 30.501% = 30.5%