Question

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mean 1,200 pounds and standard deviation 100 pounds.
a. what is the probability that demand will exceed 1,000 pounds?
b. what is the probability that demand will be between 1,100 and 1,300 pounds?
c. the probability is 0.10 that demand will be more than how many pounds?

Answer 1

a.
$\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)$

Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.

$\mathbb P(Z>-2)=\mathbb P(|Z|<2)+\mathbb P(Z>2)=0.95+0.025=0.975$

b.
$\mathbb P(1100<X<1300)=\mathbb P\left(\dfrac{1100-1200}{100}<\dfrac{X-1200}{100}<\dfrac{1300-1200}{100}\right)=\mathbb P(-1<Z<1)$

About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.

c. You're looking for $k$ such that

$\mathbb P(X>k)=0.10$

Since

$\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10$

occurs for $k^*\approx1.2816$, it follows that

$\dfrac{k-1200}{100}=1.2816\implies k\approx1328$

So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.