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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