FIND PROBABILITIES USING THE NORMAL DISTRIBUTION

Question

X

1)X is a normally distribute random variable with mean 31, SD 6, Prob of x between 13 and 43=

(13-31)/6= -3 .997/2 = .4985

(43-13)/6= 2 .95/2 = .475

.4985+.475= .974

2) X is a normally distribute random variable with mean 1, SD 16, Prob of x between 33 and 49=

(33-1)/16=2 .4985

(49-33)/16= 1 .475

.4985-.475= .024

My question is, why do I add the totals in #1, yet subtract them in #2?

I have a normal distribution chart, but am confused about what whether to add or subtract in the final step.

Answer 1

In #1, you got the correct z-scores although the second formula you wrote down should be (43-31)/6 = 2. From the z-score table, the percentile for z = 2 is 0.9772. The percentile for z = -3 is 0.0013. All you have to do now is subtract the lower percentile from the higher one:

P(between 13 and 43) = 0.9772 - 0.0013 = 0.9759

I'm not sure what you did. It looks as if you got the wrong percentile for z = -3 from the table: it should be 0.0013. There is no need to divide by 2. When finding the probability of a score between two other scores, you always subtract the smaller from the larger percentile.

In #2, your z-scores are wrong. The formula is:

z = (x-μ)/σ

which yields:

z = (49-1)/16 = 48/16 = 3

z = (33-1)/16 = 32/16 = 2

Look up the percentiles for z = 3 and z = 2 in a z-score table. Subtract the percentile for z = 2 from the percentile for z = 3 to get the probability of a score between 33 and 43.