Suppose a normal distribution has a mean of 48 and a standard deviation of 2. What is the probability that a data value is between 43 and 46? Round your answer to the nearest tenth of a percent.

A. 17.2%
B. 16.3%
C. 14.2%
D. 15.2%

Choice D.  15.2%

Step-by-step explanation:

We have a normal...

mean u = 48

standard deviation s = 2

We want P(43 < X < 46)

We standardize.

Consider  P(43 < X)  = P( (43 - 48)/2 < Z) = P(-2.5 < Z)

P( X < 46) = P( Z < (46 - 48)/2 ) = P(Z < -1)

We want  P( -2.5 < Z < -1)

Look at Z-scores.

P( Z < -2.5)  = 0.0062

P(Z  < -1) = 0.1587

so  P(-2.5 < Z < -1) =  P(Z < -1) - P(Z < -2.5) = 0.1587 - 0.0062 = 0.1525 = 15.2%

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