Answer:
A) 34.13%
B) 15.87%
C) 95.44%
D) 97.72%
E) 49.87%
F) 0.13%
Step-by-step explanation:
To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:
[tex]z=\frac{x-m}{s}[/tex]
Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:
[tex]z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0[/tex]
So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:
P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)
= 0.5 - 0.1587 = 0.3413
It means that the PERCENT of scores that are between 90 and 100 is 34.13%
At the same way, we can calculated the percentages of B, C, D, E and F as:
B) Over 110
[tex]P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587[/tex]
C) Between 80 and 120
[tex]P( 80<x< 110 ) = P( \frac{80-100}{10} <z>\frac{120-100}{10})=P(-2<z<2)\\P(-2<z<2)=P(z<2) - P(z<-2) = 0.9772 - 0.0228 = 0.9544[/tex]
D) less than 80
[tex]P( x < 80 ) = P( z<\frac{80-100}{10})=P(z<-2) = 0.9772[/tex]
E) Between 70 and 100
[tex]P( 70<x< 100 ) = P( \frac{70-100}{10} <z>\frac{100-100}{10})=P(-3<z<0)\\P(-3<z<0)=P(z<0) - P(z<-3) = 0.5 - 0.0013 = 0.4987[/tex]
F) More than 130
[tex]P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013[/tex]