# Based on data from a college,

Based on data from a college, scores on a certain test arenormally distributed with a mean of 1530 and a standard deviationof 318. Find the percentage of scores greater than 2007, Find thepercentage of scores less than 1053, Find the percentage of scoresbetween 894-2166.

Table

 Fulldata 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

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 multiplythe 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%

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