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

Question

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.

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Eden 1 month 2021-10-17T03:33:05+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-17T03:34:06+00:00

    Answer:

    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

    0
    2021-10-17T03:34:54+00:00

    Answer:

    62.5%

    Step-by-step explanation:

    just took the quiz

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45:7+7-4:2-5:5*4+35:2 =? ( )