Answer:
0.0104
Step-by-step explanation:
Given that, the lifespans of crocodiles have an approximately Normal distribution, with a mean of 18 years and a standard deviation of 2.6 years.
So, mean, ![\mu=18](https://tex.z-dn.net/?f=%5Cmu%3D18)
and standard deviation, ![\sigma=2.6](https://tex.z-dn.net/?f=%5Csigma%3D2.6)
The z-score, for any arbitrary life span of x year, is given by
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
By using the given values, we have
![z=\frac{x-18}{2.6} \cdots(i)](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-18%7D%7B2.6%7D%20%5Ccdots%28i%29)
The z-score for x=24, by using equation (i), is
![z=\frac{24-18}{2.6}=\frac{6}{2.6}=2.31](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B24-18%7D%7B2.6%7D%3D%5Cfrac%7B6%7D%7B2.6%7D%3D2.31)
From the table, the area to the left side of z=2.31 =0.98956
But, the proportion of crocodiles have lifespans of at least 24 years
= Area to the right side of the z=2.31
=1- (Area to the left side of the z=2.31)
=1-0.98956
=0.01044
So, the proportion of crocodiles that have lifespans of at least 24 years is 0.0104.
Hence, option (a) is correct.