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 |
Answer:
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%