Question

A farmer was interested in determining how many grasshoppers were in his field. He knows that the distribution of grasshoppers may not be normally distributed in his field due to growing conditions. As he drives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows he figures the mean number of flights to be 57 with a standard deviation of 12. What is the probability of the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor?

Answer 1

Answer:

There is a 5.59% probability that the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

After several rows he figures the mean number of flights to be 57 with a standard deviation of 12, so $\mu = 57, \sigma = 12$.

We have a sample of 40 rows, so we have to find the standard deviation of the sample to use in the place of $\sigma$ in the Z score formula.

$s = \frac{12}{\sqrt{40}} =1.897$

What is the probability of the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor?

This is 1 subtracted by the pvalue of Z when $X = 60$.

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

$Z = \frac{60 - 57}{1.89}$

$Z = 1.59$

$Z = 1.59$ has a pvalue of 0.9441.

This means that there is a 1-0.9441 = 0.0559 = 5.59% probability that the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor.