Question

Use the normal distribution of IQ​ scores, which has a mean of 130 and a standard deviation of 12​,

and the following table with the standard scores and percentiles for a normal distribution to find the indicated quantity.

The percentage of scores between 100 and 160 is

​(Round to two decimal places as​ needed.)

Standard score            %                     standard score     %

-3.0	0.13	0.1	53.98
-2.5	0.62	0.5	69.15
-2	2.28	0.9	81.59
-1.5	6.68	1	84.13
-1	15.87	1.5	93.32
-0.9	18.41	2	97.72
-0.5	30.85	2.5	99.38
-0.1	46.02	3	99.87
0	50.00	3.5	99.98

Answer 1

Given that,

mean = = 130

standard deviation = = 12

P (100 < x < 160 )

P (100 - 130 /12) < ( x -  / ) < ( 160 - 130 / 12)

P ( - 30 / 12 < z < 30 / 12 )

P (-2.5< z < 2.5)

P ( z < 2.5) - P ( z < -2.5)

ng z table

=0.9938 - 0.0062

=0.9876

Probability = 0.9876

The percentage of scores between 99.38% and 0.62%