1. ## Normal Prob Distribution

Anyone help with this problem?

A company is trying to anticipate demand for a product. After historical sales history is analyzed for similar products, a sales forecaster predicts expected demand of 40,000 units with a .90 probability that demand would be between 30,000 units and 50,000 units?

1)What is the mean and std deviation? Can we use the normal approach to the binomial distribution to figure mean and std dev? Is this a continuous pd with std normal probability distribution required?
Should I use...

z= x- m
s

Thanks!

2. Originally Posted by Stormy-1
Anyone help with this problem?

A company is trying to anticipate demand for a product. After historical sales history is analyzed for similar products, a sales forecaster predicts expected demand of 40,000 units with a .90 probability that demand would be between 30,000 units and 50,000 units?

1)What is the mean and std deviation? Can we use the normal approach to the binomial distribution to figure mean and std dev? Is this a continuous pd with std normal probability distribution required?
Should I use...

z= x- m
s

Thanks!
Let X be the random variable number of units demanded.

It's reasonable I think to assume X ~ Normal($\displaystyle \mu = 40,000, ~ \sigma$). The task then is to get $\displaystyle \sigma$.

Note that Pr(30,000 < X < 50,000) = 0.9 => Pr(X > 50,000) = Pr(X < 30,000) = 0.05 by symmetry.

Consider Pr(X > 50,000):

$\displaystyle z_{50,000} = \frac{x - \mu}{\sigma} \Rightarrow z = \frac{50,000 - 40,000}{\sigma} = \frac{10,000}{\sigma}$.

But $\displaystyle \Pr(X > 50,000) = \Pr(z > z_{50,000}) = 0.05 \Rightarrow z_{50,000} = 1.64485$ (corect to five decimal places).

Therefore $\displaystyle 1.64485 = \frac{10,000}{\sigma} \Rightarrow \sigma = .......$