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.

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

15.85%

Step-by-step explanation: Answer on Khan

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