Respuesta :
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, [tex]\mu=18[/tex]
and standard deviation, [tex]\sigma=2.6[/tex]
The z-score, for any arbitrary life span of x year, is given by
[tex]z=\frac{x-\mu}{\sigma}[/tex]
By using the given values, we have
[tex]z=\frac{x-18}{2.6} \cdots(i)[/tex]
The z-score for x=24, by using equation (i), is
[tex]z=\frac{24-18}{2.6}=\frac{6}{2.6}=2.31[/tex]
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.