Question

The thumb length of fully grown females of a certain type of frog is normally distributed with a mean of 8.59 mm and a standard deviation of 0.63 mm. Calculate the probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.

Answer 1

Answer:

21.77% probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.

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:

$\mu = 8.59, \sigma = 0.63$

Calculate the probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.

This is 1 subtracted by the pvalue of Z when X = 9.08. So

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

$Z = \frac{9.08 - 8.59}{0.63}$

$Z = 0.78$

$Z = 0.78$ has a pvalue of 0.7823

1 - 0.7823 = 0.2177

21.77% probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.