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.

P (37 < x < 41) = P (-0.5 < Z < 1.5) = 0.6247

Step-by-step explanation:

We know mean u = 38 standard dev. s = 2

We want P (37 < x < 41)

so

P ((37 - 38) / 2 < Z) = P (-0.5 < Z)

P (Z < (41 - 38) / 2) = P (Z < 1.5)

Find P (Z < - 0.5) = 0.3085

Find P (Z > 1.5) = 0.0668

so P (-0.5 < Z < 1.5) = 1 - P (Z 1.5)

P (-0.5 < Z < 1.5) = 1 - 0.3085 - 0.0668

P (-0.5 < Z < 1.5) = 0.6247

P (37 < x < 41) = P (-0.5 < Z < 1.5) = 0.6247