Question

Assume that a set of test scores is normally distributed with a mean of 120120 and a standard deviation of 1010. Use the​ 68-95-99.7 rule to find the following quantities.

Suggest you make a drawing and label first.
a. Percentage of scores less than 100b. Relative frequency of scores less than 120c. Percentage of scores less than 140d. Percentage of scores less than 80e. Relative frequency of scores less than 60f. Percentage of scores greater than 120

Answer 1

a. 2.5%

b. 0.5

c. 97.5%

d. 0%

e. 0

f. 50%

Explanation:

• mean: 120

• standard deviation: 10

The​ 68-95-99.7 rule states, for this case:

• The probability of a score between 110 and 130 is 68%

• The probability of a score between 100 and 140 is 95%

• The probability of a score between 90 and 150 is 99.7%

See figure attached

a. Percentage of scores less than 100 = 0.15 + 2.35 = 2.5%

b. Relative frequency of scores less than 120 = 0.5

c. Percentage of scores less than 140 = 100 - (0.15 + 2.35) = 97.5%

d. Percentage of scores less than 80 = 0% (note: this value is out of the range -3 standard deviation, +3 standard deviation)

e. Relative frequency of scores less than 60 = 0 (note: this value is out of the range -3 standard deviation, +3 standard deviation)

f. Percentage of scores greater than 120 = 50%

Answer 2

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

`z=(x-100)/(10) \\x=10z+100`

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

b) P(X<120) = P(Z<2) =
`(95)/(2) =47.5%`

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) =
`(100-95)/(2) =25%`