Answer:
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
