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 1100 acres, about how many would be expected to yield more than 190 bushels of corn per acre?
A. 639 acresB. 461 acresC. 419 acresD. 503 acres

Given : 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.

i.e. [tex]\mu=185.2\ \ , \ \sigma=23.5[/tex]

Let x denotes the amount of corn yield.

Now, the probability that the amount of corn yield is more than 190 bushels of corn per acre.

[tex]P(x>190)=P(\dfrac{x-\mu}{\sigma}>\dfrac{190-185.2}{23.5})[/tex]

[Formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]]

[tex]=P(z>0.2043)=1-P(z<0.2043)[/tex] [∵ P(Z>z)=1-P(Z<z)]

[tex]1-0.5809405[/tex] [using z-value calculator or table]

[tex]=0.4190595[/tex]

Now, If a study included 1100 acres then the expected number to yield more than 190 bushels of corn per acre :-

[tex]0.4190595\times1100=460.96545\approx461\text{ acres}[/tex]

hence, the correct answer is B. 461 acres .