Question

Question 6 2 pts

Answer 1

Answer:

785 acres

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 represent the random variable amount of corn yield

Assuming the X follows a normal distribution  

$X \sim N(\mu, \sigma)$  

And we know that the average $\bar X$ is dsitributed:

$\bar X \sim N(\mu=189.3, \sigma_{\bar X}=23.5)$

And we are interested on this probability:

$P(\bar X>180)$

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

$z=\frac{\bar X -\mu}{\sigma_{\bar X}}$

If we apply this we got:

$P(\bar X>180)=P(Z>\frac{180-189.3}{23.5})=P(Z>-0.396)=1-P(Z$

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:

$r=1200*0.654=784.8$

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