Question

Assume that a set of test scores is normally distributed with a mean of 80 and a standard deviation of 25. Use the​ 68-95-99.7 rule to find the following quantities.

c. The percentage of scores between 30 and 105 is ​ %.?

(Round to one decimal place as​ needed.)

Answer 1

68-95-99.7 rule :

z = (x-mean)/SD

= (x-80)/25

for 30 :

z = (30-80)/25 = -2

for 105

z = (105-80)/25 = 1

according to 68-95-99.7 rule

P(-2 < z < 2) = 95%

therefore P(-2 < z < 0) = P(-2 < z < 2)/2 {since normal distribution is symmetrical about z=0}

P(-2 < z < 0) = 95% / 2 = 47.5%

P(-1<z<1) = 68%

P(0<z<1) = 34%

P(30<x<105) = P(-2<z<1) = P(-2 < z < 0) + P(0<z<1)

= 47.5% + 34%

P(30<x<105) = 81.5%

P.S. (please upvote if you find the answer satisfactory)