Suppose a normal distribution has a mean of 38 and a standard deviation of 2. What is the probability that a data value is between 37 and 41? Round your answer to the nearest tenth of a percent.

A. 56.5%
B. 62.5%
C. 83.5%
D. 77.5%

Question

Suppose a normal distribution has a mean of 38 and a standard deviation of 2. What is the probability that a data value is between 37 and 41? Round your answer to the nearest tenth of a percent.

 		A.	56.5%
 		B.	62.5%
 		C.	83.5%
 		D.	77.5%

giantrobot · Answer

distribution table is your god here, step one use the formula $$\frac{ value-mean }{ s.t.d }$$

So find the probability of something being less than 41, insert that into the formula you get a z value of 1.5 which according to the table is 0.9332

So the second bit is a tad tricky, you get a value of -.5, the problem is, this z value is for the probability of the value being less that 37. And remember, if you get a z score of negative, you will get treat is as a positive and 1-value obtained it. So you end up with
$$1-0.6915$$
now find the difference between these two, which is, 
$$0.9332-(1-0.6915)=0.6247$$

Round that off to get, 0.625 which is 62.5%