Question

The lifespan of lions in a particular zoo are normally distributed. The average lion lives 10 years; the standard deviation is 1.4 years. Use the empirical rule (68-95-99.7) to estimate the probability of a lion living less that 7.2 years. I need an answer that provides detailed how each number of percentage was accumulated. The answer is suppose to be 2.5% but there is not explanation of how the answer is 2.5%

Answer 1

From the given information, The average lifespan of lion is 10 years. The standard deviation for the lifespan of lion is 1.4 years. Empirical rule (68-95-99.7) states that, Percentage of observations that are within one standard deviation's is 68% U-0 =10 -1.4 = 8.6 1+0 = 10 +1.4 =11.4 That is 68% of observations are within (8.6,11.4). Percentage of observations that are within two standard deviation's is 95% 1-20 = 10-2(1.4) = 10 -2.8 = 7.2 + 20 = 10+2(1.4) = 10 +2.8 = 12.8 That is 95% of observations are within (7.2,12.8). Percentage of observations that are within three standard deviation's is 99.7% 11-30 = 10-3(1.4) =10 -4.2 = 5.8

U1+30 =10+3(1.4) = 10 +4.2 = 14.2 That is 99.7% of observations are within (5.8,14.2). Since, from the above calculations, 95% of observations lie within 2 standard deviations from the mean. Therefore, (1-0.95) Required probability = 0.05 2 = 0.025 = 2.5% Alternative method: 3496 3496 0.1596 20 The probability of a lion living less than 7.2 years is calculated as follows: P(X<7.2)= less than (4-20) = 0.15% +2.35% = 2.5%