# Normal distribution involving $\Phi(z)$ and standard deviation

The random variable X has normal distribution with mean $\mu$ and standard deviation $\sigma$. $\mathbb{P}(X>31)=0.2743$ and $\mathbb{P}(X<39)=0.9192$. Find $\mu$ and $\sigma$.

Hint:

Write,

$$\tag{1}\textstyle P[\,X>31\,] =P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr]=.2743\Rightarrow {31-\mu\over\sigma} = z_1$$ $$\tag{2}\textstyle P[\,X<39\,] =P\bigl[\,Z<{39-\mu\over\sigma}\,\bigr]=.9192\Rightarrow {39-\mu\over\sigma} =z_2 ,$$

where $Z$ is the standard normal random variable.

You can find the two values $z_1$ and $z_2$ from a cdf table for the standard normal distribution. Then you'll have two equations in two unknowns. Solve those for $\mu$ and $\sigma$.

For example, to find $z_1$ and $z_2$, you can use the calculator here. It gives the value $z$ such that $P[Z<z]=a$, where you input $a$.

To use the calculator for the first equation first write $$\textstyle P\bigl[\,Z<\underbrace{31-\mu\over\sigma}_{z_1}\,\bigr]=1-P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr] =1-.2743=.7257.$$ You input $a=.7257$, and it returns $z_1\approx.59986$.

To use the calculator for the second equation, $$\textstyle P\bigl[\,Z<\underbrace{39-\mu\over\sigma}_{z_2}\,\bigr]= .9192,$$ input $a=.9192$, the calculator returns $z_2\approx1.3997$.

So, you have to solve the system of equations: \eqalign{ {31-\mu\over\sigma}&=.59986\cr {39-\mu\over\sigma}&=1.3997\cr } (The solution is $\sigma\approx 10$, $\mu\approx 25$.)

• mean=25 and sigma=10. Thanks! Commented Feb 26, 2012 at 13:29