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, 
and standard deviation, 
The z-score, for any arbitrary life span of x year, is given by

By using the given values, we have

The z-score for x=24, by using equation (i), is

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.