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 < -0.5) - 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

8

step-by-step explanation:

c) yes, because of asa or aas