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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%) to estimate the probability of a lion living longer than 10.1 years.

User Maak
by
4.8k points

2 Answers

2 votes

Answer:

16

Explanation:

User Rahul Raut
by
5.1k points
7 votes

We have been given that 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. We are asked to find the probability of a lion living longer than 10.1 years using empirical rule.

First of all, we will find the z-score corresponding to sample score 10.1.


z=(x-\mu)/(\sigma), where,

z = z-score,

x = Random sample score,


\mu = Mean


\sigma = Standard deviation.


z=(10.1-12.5)/(2.4)


z=(-2.4)/(2.4)


z=-1

Since z-score of 10.1 is
-1. Now we need to find area under curve that is below one standard deviation from mean.

We know that approximately 68% of data points lie between one standard deviation from mean.

We also know that 50% of data points are above mean and 50% of data points are below mean.

To find the probability of a data point with z-score
-1, we will subtract half of 68% from 50%.


(68\%)/(2)=34\%


50\%-34\%=16\%

Therefore, the probability of a lion living longer than 10.1 years is approximately 16%.

User Atsushi
by
6.4k points
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