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------------------------------------------------ Assume that a set of test scores is normally distributed with a mean of [tex]120[/tex] and a standard deviation of [tex]10[/tex]. Use the 68-95-99.7 rule to find the following quantities:

a. Percentage of scores less than [tex]100[/tex]
b. Relative frequency of scores less than [tex]120[/tex]
c. Percentage of scores less than [tex]140[/tex]
d. Percentage of scores less than [tex]80[/tex]
e. Relative frequency of scores less than [tex]60[/tex]
f. Percentage of scores greater than [tex]120[/tex]

Answer :

Answer:

Step-by-step explanation:

Given that it is assumed that a set of test scores is normally distributed with a mean of 120 and a standard deviation of 10.

If X is set of test scores, then X is N(120,10)

We can convert any x score to z score and vice versa as

[tex]z=\frac{x-100}{10} \\x=10z+100[/tex]

a) P(X<100) =P(Z<0) =0.50=50%

b) P(X<120) = P(Z<2) = [tex]\frac{95}{2} =47.5%[/tex]

c) P(X<140) = P(Z<4) = 1

d) P(X<60) = P(Z<-4)=0

e) P(X<60) = 0

f) P(X>120)=P(Z>2) = [tex]\frac{100-95}{2} =25%[/tex]

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