A farmer was interested in determining how many grasshoppers were in his field. He knows that thedistribution of grasshoppers may not be normally distributed in his field due to growing conditions. As hedrives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows hefigures the mean number of flights to be 57 with a standard deviation of 12. What is the probability of thefarmer will count 52 or fewer flights or 60 or more flights on average in the next 40 rows down which he driveshis tractor?
A) 0.0612
B) 0.4959
C) 0.9388
D) 0.0530

Question

Quizplus · Accepted Answer

The probability of counting 52 or fewer flights or 60 or more flights can be found using the normal distribution formula and Z-scores. Given a mean ($\mu$) of 57 and a standard deviation ($\sigma$) of 12 for the distribution of grasshopper flights, and considering the sample size (n) of 40, the standard error (SE) of the mean is $\sigma / \sqrt{n} = 12 / \sqrt{40}$. To find the Z-scores for 52 and 60, we use $Z = (X - \mu) / SE$. Calculating these Z-scores and looking up their corresponding probabilities in the Z-table, then adding the probabilities for Z-scores below 52 and above 60 gives us the total probability. The correct answer, based on these calculations, matches the probability given in option A, which accounts for the tails of the distribution where the farmer would see 52 or fewer and 60 or more flights.

A Farmer Was Interested in Determining How Many Grasshoppers Were