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