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

Answer:

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

Step-by-step 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:

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

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

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

$Z = \frac{18.38 - 18.8}{2}$

$Z = -0.21$

$Z = -0.21$ has a pvalue of 0.4168

X = 12.04

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

$Z = \frac{12.04 - 18.8}{2}$

$Z = -3.38$

$Z = -3.38$ 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.