Help with a question

Question

Consider a normal distribution with a mean of 12 and variance of 4. Approximately 82% of the area lies between which values?

a. 6 and 13

b. 10 and 16

c. 9 and 15

d. 10 and 18

e. Not enough information provided to solve

 

I am not sure exactly if there is a particular formula to use for a question like this. I know that since the variance is 4 then the standard deviation would then be 2. At first I thought I would find the answer from ± + σ =14      ±- σ =10

But that is not wrong so I am not sure what I am missing. Any help would be great.

Answer 1

Using the calculator [TI-83 Plus & its DISTR menu's NORMALDCDF (a,b,μ,σ) command with each of the four proffered "a,b" choices], I find that none of the probabilities from a to b equals 0.82.

 

I conclude that E, none of the above, is the best answer.

Answer 2

Obtain a z-score chart:

http://www.utdallas.edu/dept/abp/zscoretable.pdf

 

a.

for 6: 

z = (X - μ)/σ

z = (6 - 12)/4

z = -6/4

z = -1.5

0.0668

for 13

z = (13 - 12)/4

z = 1/4

z = 0.25

0.5793

0.5793 - 0.0668 = 0.5101

 

The others are done the same way.