This question is poorly phrased
Complete Question
The lifespans of lions at a specific zoo follow a normal distribution. The average lifespan is 12.5 years with a standard deviation of 2.4 years. Apply the empirical rule (68-95-99.7%) to estimate the likelihood of a lion living between 5.3 and 10.1 years.
Answer:
The likelihood of a lion living from 5.3 to 10.1 years is 0.1585
Step-by-step explanation:
According to the empirical rule:
1) 68% of the data falls within 1 standard deviation of the mean, meaning between μ - σ and μ + σ.
2) 95% of the data is contained within 2 standard deviations around the mean - between μ - 2σ and μ + 2σ.
3) 99.7% of the data lies within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
The mean provided is: 12.5
Standard deviation: 2.4 years
Starting with the first rule:
1) 68% falls within 1 standard deviation from the mean, implying between μ - σ and μ + σ.
μ - σ
12.5 - 2.4
= 10.1
We now apply the second rule:
2) 95% of the data lies within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.
μ – 2σ
12.5 - 2 × 2.4
12.5 - 4.8
= 7.7
Now applying the last rule:
3)99.7% of the data resides within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
μ - 3σ
= 12.5 - 3(2.4)
= 12.5 - 7.2
= 5.3
The calculations indicate that
5.3 years is at one side of 99.7%
Therefore,
100 - 99.7%/2 = 0.3%/2
= 0.15%
Moreover, 10.1 years corresponds to one side of 68%
Thus
100 - 68%/2 = 32%/2 = 16%
Consequently, the percentage of a lion living between 5.3 to 10.1 years is evaluated as 16% - 0.15%
= 15.85%
Thus, the estimated probability of a lion surviving between 5.3 and 10.1 years
is represented as a decimal =
= 15.85/ 100
= 0.1585
