Question

Assume that a set of test scores is normally distributed with a mean of 80 and a standard deviation of 15

Use the​ 68-95-99.7 rule to find the following quantities.

a. The percentage of scores less than

80 is ___%.

​(Round to one decimal place as​ needed.)

b. The percentage of scores greater than 95 is ___​%

​(Round to one decimal place as​ needed.)

c. The percentage of scores between 50 and 95 is ___​%.

​(Round to one decimal place as​ needed.)

Answer 1

Solution:

Scores follows normal distribution.

Mean = 80

SD = 15

Part a

We have to find P(X<80)

Total area under the curve = 1.00

So, half area under the curve = 0.500

That is, P(X<80) = 0.500

Answer: 50.0%

Part b

Here, we have to find P(X>95)

Mean + 1*SD = 80 + 1*15 = 95

So, according to 68-95-99.7 rule, area within 1 standard deviation from mean is 68%.

So, remaining area at both sides = 100 – 68 = 32%

Remaining area at right side = 32/2 = 16%

Answer: 16.0%

Part c

We have to find P(50<X<95)

P(50<X<95) = P(X<95) – P(X<50)

We have P(X>95) = 0.16 (from part b)

So, P(X<95) = 1 – 0.16 = 0.84

P(X<95) = 0.84

Now, we have following

Mean - 2*SD = 80 - 2*15 =80 – 30 = 50

So, according to 68-95-99.7 rule, area within 2 standard deviation from mean is 95%.

So, remaining area at both sides = 1 - .95 = 0.05

So, area remaining at left = 0.05/2 = 0.025

So, P(X<50) = 0.025

P(50<X<95) = P(X<95) – P(X<50)

P(50<X<95) = 0.84 - 0.025

P(50<X<95) = 0.815

Answer: 81.5%