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:[tex]P(X>190)= 1-0.5809=0.4191[/tex][tex]r=1200*0.4191=502.92\approx 503 acres[/tex]We would expect around 503 acres with more than 190 bushels of corn per acreStep-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: [tex]X\sim N(\mu =185.2,\sigma =23.5)[/tex] 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: [tex]P(X>190)[/tex] And we can use the z score given by this formula:[tex]z=\frac{X-\mu}{\sigma}[/tex]If we replace the values given we have:[tex]z=\frac{190-185.2}{23.5}=0.2042[/tex]Sow e want this probability:[tex]P(z>0.2042)=1-P(Z<0.2042)= 1-0.5809=0.4191[/tex]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[tex]r=1200*0.4191=502.92\approx 503 acres[/tex]And we would expect around 503 acres with more than 190 bushels of corn per acre