Question

Laboratory experiment shows that the life of the average butterfly is normally distributed with a mean of 18.8 days and a standard deviation of 2 days. Find the probability that a butterfly will live between 12.04 and 18.38 days.

Answer 1

41.64% probability that a butterfly will live between 12.04 and 18.38 days.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 18.8, \sigma = 2`

Find the probability that a butterfly will live between 12.04 and 18.38 days.

This is the pvalue of Z when X = 18.38 subtracted by the pvalue of Z when X = 12.04. So

X = 18.38

`Z = (X - \mu)/(\sigma)`

`Z = (18.38 - 18.8)/(2)`

`Z = -0.21`

`Z = -0.21` has a pvalue of 0.4168

X = 12.04

`Z = (X - \mu)/(\sigma)`

`Z = (12.04 - 18.8)/(2)`

`Z = -3.38`

`Z = -3.38` has a pvalue of 0.0004

0.4168 - 0.0004 = 0.4164

41.64% probability that a butterfly will live between 12.04 and 18.38 days.