Question

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?

Answer 1

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

Explanation:

Problems of normally distributed samples are 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 = (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 = 189.3, \sigma = 23.5`

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

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

`Z = (180 - 189.3)/(23.5)`

`Z = -0.4`

`Z = -0.4` 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