Question #339830

1.) In a National Achievement Test, the mean was found to be 75 and the standard deviation was 15. The scores also approximate the normal distribution.


a. What is the minimum score that belongs to the upper 15% of the group?


b. What is the two extreme scores outside of which 15% of the group are expected to fall?


c. What is the score that divide the distribution into two such that 75% of the cases below it?


d. Estimate the range of scores that will include the middle 45% of the distribution.




1
Expert's answer
2022-05-12T12:42:51-0400

a.

P(X>x)=1P(Zx7515)=0.15P(X>x)=1-P(Z\le \dfrac{x-75}{15})=0.15

x7515=1.036433\dfrac{x-75}{15}=1.036433

x=90.5465x=90.5465

b.


P(X<x1)=0.152P(X<x_1)=\dfrac{0.15}{2}

P(Zx17515)=0.075P(Z\le \dfrac{x_1-75}{15})=0.075

x17515=1.4395\dfrac{x_1-75}{15}=-1.4395

x1=53.4075x_1=53.4075


P(X>x2)=0.152P(X>x_2)=\dfrac{0.15}{2}

1P(Zx27515)=0.0751-P(Z\le \dfrac{x_2-75}{15})=0.075

x27515=1.4395\dfrac{x_2-75}{15}=1.4395

x2=96.5925x_2=96.5925

c.


P(X<x)=P(Z<x7515)=0.75P(X<x)=P(Z< \dfrac{x-75}{15})=0.75

x7515=0.6749\dfrac{x-75}{15}=0.6749

x=85.1235x=85.1235


d.

Middle 45% is contained in the range of 27.5%-72.5%


P(X<x1)=P(Z<x17515)=0.275P(X<x_1)=P(Z< \dfrac{x_1-75}{15})=0.275

x17515=0.59776\dfrac{x_1-75}{15}=-0.59776

x1=66.0336x_1=66.0336


P(X<x2)=P(Z<x27515)=0.725P(X<x_2)=P(Z< \dfrac{x_2-75}{15})=0.725

x27515=0.59776\dfrac{x_2-75}{15}=0.59776

x2=83.9664x_2=83.9664


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