Question

PLEASE NEED HELP

Answer 1

Using the normal distribution, it is found that 0.4068 of crocodiles live between 17.5 and 20.5 years.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given as follows:

$\mu = 18, \sigma = 2.6$

The proportion of crocodiles live between 17.5 and 20.5 years is the p-value of Z when X = 20.5 subtracted by the p-value of Z when X = 17.5, hence:

X = 20.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.5 - 18}{2.6}$

Z = 0.96

Z = 0.96 has a p-value of 0.8315.

X = 17.5:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17.5 - 18}{2.6}$

Z = -0.19

Z = -0.19 has a p-value of 0.4247.

0.8315 - 0.4247 = 0.4068

0.4068 of crocodiles live between 17.5 and 20.5 years.

More can be learned about the normal distribution at  brainly.com/question/4079902

#SPJ1