In an agricultural study, the average amount of corn yield is normally distributed with a mean of 189.3 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn. If a study included 1200 acres, about how many would be expected to yield more than 180 bushels of corn per acre?

About 786 would be expected to yield more than 180 bushels of corn per acre

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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:

Proportion of acres with more than 180 bushels of corn per acre:

This is 1 subtracted by the pvalue of Z when X = 180. So

has a pvalue of 0.3446.

1 - 0.3446 = 0.6554

Out of 1200:

0.6554*1200 = 786.48

About 786 would be expected to yield more than 180 bushels of corn per acre