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 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

Answer 1

Answer: B. 461 acres

Step-by-step explanation:

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. $\mu=185.2\ \ , \ \sigma=23.5$

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.

$P(x>190)=P(\dfrac{x-\mu}{\sigma}>\dfrac{190-185.2}{23.5})$

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

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

$1-0.5809405$   [using z-value calculator or table]

$=0.4190595$

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

$0.4190595\times1100=460.96545\approx461\text{ acres}$

hence, the correct answer is B. 461 acres .