Respuesta :
This question was not written properly
Complete Question
The lifespans of lions 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\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.
Answer:
Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585
Step-by-step explanation:
The empirical rule formula states that:
1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.
3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.
Mean is given in the question as: 12.5
Standard deviation : 2.4 years
We start by applying the first rule
1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.
μ - σ
12.5 -2.4
= 10.1
We apply the second rule
2) 95% of data falls 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
We apply the third rule
3)99.7% of data falls 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
From the above calculation , we can see that
5.3 years corresponds to one side of 99.7%
Hence,
100 - 99.7%/2 = 0.3%/2
= 0.15%
And 10.1 years corresponds to one side of 68%
Hence
100 - 68%/2 = 32%/2 = 16%
So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%
= 15.85%
Therefore, the probability of a lion living between 5.3 to 10. 1 years
is converted to decimal =
= 15.85/ 100
= 0.1585
Answer:
The probability of a particular lion living less than 10.1 years is 16%
