Question 6 2 pts

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?

346 acres
654 acres
785 acres
415 acres

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 represent the random variable amount of corn yield

Assuming the X follows a normal distribution

[tex]X \sim N(\mu, \sigma)[/tex]

And we know that the average [tex]\bar X [/tex] is dsitributed:

[tex]\bar X \sim N(\mu=189.3, \sigma_{\bar X}=23.5)[/tex]

And we are interested on this probability:

[tex]P(\bar X>180)[/tex]

For this case we can use the z score formula given by:

[tex]z=\frac{\bar X -\mu}{\sigma_{\bar X}}[/tex]

If we apply this we got:

[tex]P(\bar X>180)=P(Z>\frac{180-189.3}{23.5})=P(Z>-0.396)=1-P(Z<-0.396)=1-0.346=0.654[/tex]

And since we have a proportion estimated and we hava a total of 1200 acres the expected to yield more than 180 bushels of corn per acre would be:

[tex]r=1200*0.654=784.8[/tex]

And if we round up this amount we got [tex]\approx 785[/tex] and that would be the best option for this case.