Question

In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 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 190 bushels of corn per acre?

Answer 1

Answer:

$P(X>190)= 1-0.5809=0.4191$

$r=1200*0.4191=502.92\approx 503 acres$

We would expect around 503 acres with more than 190 bushels of corn per acre

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Let X the random variable who represents the amount of corn yield. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =185.2,\sigma =23.5)$

A quality control inspector takes a sample of n=1200 acres. That represent the sample size.  

And we want to calculate the probability that random variable X would be more than 190 bushels of corn per acre. So we want to find:

$P(X>190)$

And we can use the z score given by this formula:

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

If we replace the values given we have:

$z=\frac{190-185.2}{23.5}=0.2042$

Sow e want this probability:

$P(z>0.2042)=1-P(Z<0.2042)= 1-0.5809=0.4191$

And in order to find how many would be expected to yield more than 190 bushels of corn per acre. We can multiply the faction obtained by the sample size

$r=1200*0.4191=502.92\approx 503 acres$

And we would expect around 503 acres with more than 190 bushels of corn per acre