It is possible to use the standard normal distribution together with the inverse normal distribution function to find the values of $\mu$ and $\sigma$ for another random variable that follows a normal distribution.

Finding one unknown parameter

Use the standard normal distribution $Z \sim (0,1^2)$ with the inverse normal function on a calculator to find either the mean or the standard deviatiation, given a probability.

Example 1

Given that $X \sim N(\mu, 0.2^2)$ and $P(X \leq 3) = 0.24$ , find the value of $\mu$ to two decimal places.

Transform the probability to the standard normal distribution:

$Z = \dfrac{X - \mu}{\sigma} = \dfrac{X - \mu}{0.2} \\X = 0.2Z + \mu$ 

$\begin{aligned} P(X \leq 3) &= 0.24\\P(0.2Z+\mu \leq 3) &=0.24\\P\left(Z \leq \dfrac{3 - \mu}{0.2}\right) &=0.24 \end{aligned}$ 

Use the inverse normal distribution to find a value that corresponds to a standard normal probability of $0.24$ :

$P(Z \leq a) = 0.24 \Rightarrow a = -0.7063$ 

Equate this with the expression to solve for $\mu$ :

$P\left(Z \leq \dfrac{3 - \mu}{0.2}\right) = 0.24 \Rightarrow \dfrac{3-\mu}{0.2}=-0.7063\\ \begin{aligned} 3 - \mu &= -0.14126\\\mu &= 3.14126 \end{aligned}$

$\mu = \underline{3.14 \ (2 \ d.p.)}$

Finding two unknown parameters

When the mean and standard deviation are both unknown, a similar process can be used, however two different probabilities must be given. This will give enough information to form two equations that can be solved simultaneously to find the mean and standard deviation.

Example 2

Given that $X \sim N(\mu, \sigma^2)$ , $P(X \gt 5) = 0.0478$ and $P(X \lt 4.2)=0.1587$ , find the values of $\mu$ and $\sigma$ to one decimal place.

Transform both probabilities into the standard normal distribution:

$X = \sigma Z+\mu$ 

$\begin{aligned}P(X \gt 5) &= 0.0478\\P(\sigma Z+\mu \gt 5)&= 0.0478\\P\left(Z \gt \dfrac{5-\mu}{\sigma}\right) &=0.0478\\P\left( Z \lt \dfrac{5-\mu}{\sigma}\right) &=1- 0.0478\\P\left( Z \lt \dfrac{5-\mu}{\sigma}\right) &=0.9522 \end{aligned}$

$\begin{aligned}P(X \lt 4.2)&=0.1587\\P\left(Z \lt \dfrac{4.2 - \mu}{\sigma} \right) &=0.1587 \end{aligned}$

Use the inverse normal distribution on both probabilities to create two simultaneous equations in $\mu$ and $\sigma$ :

$P(Z \lt a) = 0.9522 \Rightarrow a = 1.6666\\P\left( Z \lt \dfrac{5-\mu}{\sigma}\right) =0.9522 \Rightarrow \dfrac{5-\mu}{\sigma} = 1.6666$

$1.6666\sigma + \mu = 5$

$P(Z \lt a)= 0.1587 \Rightarrow a = -0.9998\\ P\left(Z \lt \dfrac{4.2 - \mu}{\sigma} \right) =0.1587 \Rightarrow \dfrac{4.2 - \mu}{\sigma} = -0.9998$

$-0.9998\sigma + \mu = 4.2$

Solve the two simultaneous equations:

 $1.6666\sigma + \mu = 5\\-0.9998\sigma + \mu = 4.2 \\ \rule{85pt}{0.9pt}\\ 2.6664 \sigma = 0.8\\ \sigma = 0.3 \ (1 \ d.p.)$ 

$1.6666 (0.3) + \mu = 5\\\mu = 5 - 0.49998\\ \mu = 4.5 \ (1 \ d.p.)$ 

$\mu = \underline{4.5}, \sigma = \underline{0.3}$ 

Note: It is acceptable to solve the simultaneous equations directly using a calculator.

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