In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the mean and S.D. of the distribution.
I need some help on the above number. Thanx.
$\displaystyle \Pr(X < 45) = 0.31$.
Find the value $\displaystyle a$ of $\displaystyle Z$ such that $\displaystyle \Pr(Z < a) = 0.31$.
From $\displaystyle Z = \frac{X - \mu}{\sigma}$ it follows that $\displaystyle a = \frac{45 - \mu}{\sigma}$.
Substitute your value of $\displaystyle a$ to get equation (1) with the unknowns $\displaystyle \mu$ and $\displaystyle \sigma$.
$\displaystyle \Pr(X > 64) = 0.08$.
Find the value $\displaystyle b$ of $\displaystyle Z$ such that $\displaystyle \Pr(Z > b) = 0.08$.
From $\displaystyle Z = \frac{X - \mu}{\sigma}$ it follows that $\displaystyle b = \frac{64 - \mu}{\sigma}$.
Substitute your value of $\displaystyle b$ to get equation (2) with the unknowns $\displaystyle \mu$ and $\displaystyle \sigma$.
Solve equations (1) and (2) simultaneously.
I doubt $\displaystyle \sigma$ is correct - for starters you have it negative!! Even ignoring the -ve as a typo it's size is way too big.
If you post your calculations I will review them for the mistake(s) (but be patient because I'm very busy and might take some time to post).
If $\displaystyle \Pr(Z < a) = 0.31$ then you're expected to be able to get (using tables or technology) that a = -0.4959 since $\displaystyle \Pr(Z < -0.4959) = 0.31$
Similar with b.
This is the inverse normal problem - you must have met questions like this in your earlier work ....