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The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kg/cm2 and a variance of 8100 (kg/cm2) 2. What is the probability that the compressive strength is between 5900 and 6000 kg/cm2?

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  1. 7 June, 00:37
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    Answer: the probability is 0.364

    Step-by-step explanation:

    Since the compressive strength of samples of cement can be modeled by a normal distribution, we would apply the formula for normal distribution which is expressed as

    z = (x - µ) / σ

    Where

    x = the compressive strength of samples of cement.

    µ = mean compressive strength.

    σ = standard deviation

    From the information given,

    µ = 6000 kg/cm²

    Variance = 8100

    σ = √variance = √8100 = 90

    We want to find the probability that the compressive strength is between 5900 and 6000 kg/cm2. It is expressed as

    P (5900 ≤ x ≤ 6000)

    For x = 5900,

    z = (5900 - 6000) / 90 = - 1.11

    Looking at the normal distribution table, the probability corresponding to the z score is 0.136

    For x = 6000,

    z = (6000 - 6000) / 90 = 0

    Looking at the normal distribution table, the probability corresponding to the z score is 0.5

    P (5900 ≤ x ≤ 6000) = 0.5 - 0.136

    P (5900 ≤ x ≤ 6000) = 0.364
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