Question

You are a farmer about to harvest the crop. To describe the uncertainty in the size of the harvest, you feel that it may be described as normal distribution with a mean of 80,000 bushels with a standard deviation  of 2500 bushels. Find the probability that your harvest will exceed 84,400 bushels.

Answer 1

Solution:

Given, mean, $\mu=80,000$

And standard deviation, $\sigma=2500$

Let X denotes the random variable such that $X\sim N(80000,\ 2500)$

We need to find $P(X>84400)$ .

$P(X>84400)=1-P(X\le84400)$

$=1-P(z\le \dfrac{84400-80000}{2500})$ [$\because z=\dfrac{X-\mu}{\sigma}$ ]

$=1-P(z\le \dfrac{4400}{2500})=1-P(z\le 1.76)$

$=1-0.96080=0.0392$

Thus, the required probability is 0.0392