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8 July, 10:47

5. The length of similar components produced by a company are approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random a) what is the probability that the length of this component is between 4.98 and 5.02 cm

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  1. 8 July, 12:56
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    the probability that he length of this component is between 4.98 and 5.02 cm is 0.682 (68.2%)

    Step-by-step explanation:

    Since the random variable X = length of component chosen at random, is normally distributed, we can define the following standardized normal variable Z:

    Z = (X - μ) / σ

    where μ = mean of X, σ = standard deviation of X

    for a length between 4.98 cm and 5.02 cm, then

    Z₁ = (X₁ - μ) / σ = (4.98 cm - 5 cm) / 0.02 cm = - 1

    Z₂ = (X₂ - μ) / σ = (5.02 cm - 5 cm) / 0.02 cm = 1

    therefore the probability that the length is between 4.98 cm and 5.02 cm is

    P (4.98 cm ≤X≤5.02 cm) = P (-1 ≤Z≤ 1) = P (Z≤1) - P (Z≤-1)

    from standard normal distribution tables we find that

    P (4.98 cm ≤X≤5.02 cm) = P (Z≤1) - P (Z≤-1) = 0.841 - 0.159 = 0.682 (68.2%)

    therefore the probability that he length of this component is between 4.98 and 5.02 cm is 0.682 (68.2%)
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