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14 April, 17:24

The lifespan of a lion in a particular zoo are normally distributed. The average lion lives 12.5 years the. Standard deviation is 2.4 years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between 5.3 and 10.1 years.

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  1. 14 April, 17:34
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    0.1585

    Step-by-step explanation:

    Solution:-

    The lifespan of a lion in a particular zoo is normally distributed with average lion lives:

    Mean u = 12.5 years

    Standard deviation s = 2.4 years

    We are to use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between 5.3 and 10.1.

    - The empirical rule (68-95-99.7%) states:

    P (u - s < X < u + s) = 68%

    P (u - 2s < X < u + 2s) = 95%

    P (u - 3s < X < u + 3s) = 99.7%

    - The test have the following number of standard deviations (s):

    u - s < X < u + s = 12.5 - 2.4 < X < 12.5 + 2.4 = 10.1 < X < 14.9

    u - 2s < X < u + 2s = 12.5 - 4.8 < X < 12.5 + 4.8 = 7.7 < X < 17.3

    u - 3s < X < u + 3s = 12.5 - 7.2 < X < 12.5 + 7.2 = 5.3 < X < 19.7

    Hence,

    P (10.1 < X < 14.9) = 0.68

    P (7.7 < X < 17.3) = 0.95

    P (5.3 < X < 19.7) = 0.997

    - We need P (X < 10.1) and P (X < 5.3):

    P (X < 10.1) = [ 1 - P (10.1 < X < 14.9) ] / 2

    = [ 1 - 0.68 ] / 2

    = 0.16

    P (X < 5.3) = [ 1 - P (5.3 < X < 19.7) ] / 2

    = [ 1 - 0.997 ] / 2

    = 0.0015

    Hence,

    P (5.3 < X < 10.1) = P (X < 10.1) - P (X < 5.3)

    = 0.16 - 0.0015

    = 0.1585
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