Homework SourseUse the normal distribution of IQ scores which has a mean of
Use the normal distribution of IQ scores, which has a mean of 95 and a standard deviation of 17, and the following table with the standard scores and percentiles for a normal distribution to find the indicated quantity.
Percentage of scores greater than 52.5 is ____%.
(Round to two decimal places as needed.)
99.98
| Standard score | Percent |
| -3.0 | 0.13 |
| -2.5 | 0.62 |
| -2 | 2.28 |
| -1.5 | 6.68 |
| -1 | 15.87 |
| -0.9 | 18.41 |
| -0.5 | 30.85 |
| -0.1 | 46.02 |
| 0 | 50.00 |
| 0.10 | 53.98 |
| 0.5 | 69.15 |
| 0.9 | 81.59 |
| 1 | 84.13 |
| 1.5 | 93.32 |
| 2 | 97.72 |
| 2.5 | 99.38 |
| 3 | 99.87 |
| 3.5 | 99.98 |
Solution
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 52.5
u = mean = 95
s = standard deviation = 17
Thus,
z = (x - u) / s = -2.5
Thus, using the table, the right tailed area of this is
P(z > -2.5 ) = 100 - 0.62 = 99.38% [ANSWER]
