Question

the number of ants per acre in the forest is normally distributed with mean 42000 and standard deviation 12275. let x = number of ants in a randomly selected acre of the forest. Round all answers to 4 decimal places where possible. Find the probability that a randomly selectd acre has between 32647 and 43559 ants.

Answer 1

Answer:

0.3182 = 32.81% probability that a randomly selected acre has between 32647 and 43559 ants.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 42000, \sigma = 12275$

Find the probability that a randomly selectd acre has between 32647 and 43559 ants.

This is the pvalue of Z when X = 43559 subtracted by the pvalue of Z when X = 32647. So

X = 43559:

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

$Z = \frac{43559 - 42000}{12275}$

$Z = 0.13$

$Z = 0.13$ has a pvalue of 0.5517.

X = 32647:

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

$Z = \frac{32647 - 42000}{12275}$

$Z = -0.76$

$Z = -0.76$ has a pvalue of 0.2236

0.5517 - 0.2236 = 0.3281

0.3182 = 32.81% probability that a randomly selected acre has between 32647 and 43559 ants.