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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.

User Diver Dan
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1 Answer

6 votes

Answer:

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

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


Z = (43559 - 42000)/(12275)


Z = 0.13


Z = 0.13 has a pvalue of 0.5517.

X = 32647:


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


Z = (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.

User Yamile
by
8.8k points
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