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5 March, 19:23

Miguel buys a large bottle and a small bottle of juice. The amount of juice that the manufacturer puts in the large bottle is a random variable with a mean of 1016 ml and a standard deviation of 8 ml. The amount of juice that the manufacturer puts in the small bottle is a random variable with a mean of 510 ml and a standard deviation of 5ml. If the total amount of juice in the two bottles can be described by a normal model, what's the probability that the total amount of juice in the two bottles is more than 1540.2 ml?

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  1. 5 March, 19:29
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    Pr (X>1540.2) = 0.0655

    Step-by-step explanation:

    Expected value of large bottle,

    E (Large) = 1016

    Expected value of small bottle,

    E (small) = 510

    Expected value of total

    E (total) = 1016 + 510 = 1526

    So the new mean is 1526

    Find standard deviation of new amount by variance

    Variance of large bottle,

    v (large) = 8^2 = 64

    Variance of small bottle,

    v (small) = 5^2 = 25

    Variance of total

    v (total) = 64+25 = 89

    So the new standard deviation

    sd (new) = sqrt (89) = 9.434

    Find probability using the new mean and s. d.

    Pr (X>1540.2)

    Z score, z = (x-mean) / sd

    = (1540.2 - 1526) / 9.434

    = 1.505

    value in z score

    P (z<1.51) = 0.9345

    For probability of x > 1540.2

    P (z > 1.51) = 1 - 0.9345 = 0.0655
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