Question

Based on data from a college, scores on a certain test are normally distributed with a mean of 1530 and a standard deviation of 318. Find the percentage of scores greater than 2007, Find the percentage of scores less than 1053, Find the percentage of scores between 894-2166.

Table

Full data set
Standard Scores and Percentiles for a Normal Distribution ​(cumulative values from the​ left)
Standard score	​%	Standard score	​%
minus−3.0	0.13	0.1	53.98
minus−2.5	0.62	0.5	69.15
minus−2	2.28	0.9	81.59
minus−1.5	6.68	1	84.13
minus−1	15.87	1.5	93.32
minus−0.9	18.41	2	97.72
minus−0.5	30.85	2.5	99.38
minus−0.1	46.02	3	99.87
0	50.00	3.5	99.98

Answer 1

Given:-

= 1530,  =318

Let X denote the score in the test

To find the P(X>2007) = 1 - P(X<2007)

Using Z = ( X - ) /

P(X>2007) = 1 - P( ( X - ) / < (2007 -  1530) / 318 )

A) P(X>2007) = 1 - P(Z < 1.5) = 1 - 0.93319 = 0.06681

to find the percentage of this probability we need to multiply the number by 100

0.06681 * 100 = 6.68%

B) P ( X < 1053) = P ( ( X - ) / < (1053 - 1530) / 318 )

  P ( X < 1053) = P ( Z < -1.5) = 0.06681

So,  0.06681* 100 = 6.68%

C) P ( 894 < X < 2166) =   P (  (894 -  1530) / 318 < ( X - ) / < (2166 - 1530) / 318 )

  P ( 894 < X < 2166) =   P (  -2.0 < Z < 2.0 )

P ( 894 < X < 2166) = P ( Z < 2.0) - P ( Z < -2.0 )

P ( 894 < X < 2166) = 0.97725 - 0.02275 = 0.9545

So, 0.9545 * 100 = 95.45%