Let $X=$ weight of bread in grams, that is normally distributed. If $P(X<445)=0.31$, and $P(X>490)=0.04,$ what is $Pr(458 < X <486)$?

Question

$X=$weight of bread in grams, that is normally distributed. if $P(X<445)=0.31$, and $P(X>490)=0.04$. What what is $Pr(458 < X <486)$?

So we need to find $\mu$ and $\sigma$

We know:

$P\left(\cfrac{X-\mu}{\sigma}\lt \cfrac{445-\mu}{\sigma}\right)=(0.31)$

and $P\left(\cfrac{X-\mu}{\sigma}\gt\cfrac{490-\mu}{\sigma}\right)=(0.04)$

So we have two equations with two unknowns.

$$\cfrac{490-\mu}{\sigma}=\Phi^{-1}(0.04)$$

and

$$\cfrac{445-\mu}{\sigma}=\Phi^{-1}(0.31)$$

I'm getting confused over how to continue. My question is:

1. When using normal distribution, and we want to find $Pr(x \gt y)$, where $y$ $\in R$, is it $\Phi (y)$ or $1-\Phi(y)$ when looking for a probability greater than the $y$ we picked?

2. Since $0.31$ doesn't appear on my normal distribution table, I'd look for a value of $0.69$ on the table, correct?

Because $1-\Phi(x)=0.69$ $\rightarrow $$\Phi(x)=0.69\rightarrow x=\Phi^{-1}(0.69)$

Answer 1

1. $P(Z > y) = 1 - P(Z \le y) = 1 - \Phi(y)$ is $Z$ has a standard normal distribution. Thus, the equations for $\mu, \sigma$ are: $$ \frac{445-\mu}{\sigma}=\Phi^{-1}(0.31) \\ \frac{490-\mu}{\sigma}=\Phi^{-1}(1-0.04) \tag{*} $$

2. You are correct about where to look to get $\Phi^{-1}(.31)$, but you missed a sign. Since $.31 < .5$, $\Phi^{-1}(.31) < 0$. Use $\Phi(-x) = 1 - \Phi(x) = 1 - .31 = .69 $ to conclude $-x = \Phi^{-1}(.69) $

3. Once you have the right hand side of (*), multiply by $\sigma $ to get a syatem of two linear equations in two unknowns, which you can solve with your favourite method :)