Question

Use the normal distribution of IQ​ scores, which has a mean of 90 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.

The percentage of scores between 73 and 124 is ____%

Standard score	Percent
-3	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
0.1	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

Answer 1

Answer is:

81.85 %

Explanation:

= 90

= 17

To find P(73 < X < 124):

Case 1: For X from 73 to mid value:
Z = (73 - 90)/17 = - 1

From Table, Percentage of scores from 73 to mid value = 50 - 15.87= 34.13

Case 2: For X from mid value to 124:
Z = (124 - 90)/17 = - 2

From Table, Percentage of scores from mid value to 124 = 97.72 - 50 = 47.72

So,

Percentage of scores between 73 to 124 = 34.13 + 47.72 = 81.85

So

Answer is:

81.85 %