Finding probabilities for normal distributions

​​In a nutshell

Probabilities for a normal distribution are given by the area under a normal distribution curve. It is not easy to integrate to find the area under the curve, so you can usually find the areas (or probabilities) using the normal distribution function on a calculator.

Find probabilities using a calculator

Go to the normal distribution function in your calculator to enter the required information to find probabilities. Enter the mean $$\mu$$ and standard deviation $$\sigma$$ as well as the upper and lower limits for the area required.

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calculator tip

$$\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:\\Upper&:\\ \sigma&:\\ \mu&: \end{aligned}}$$​​

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The "upper" and "lower" corresponds to the upper and lower bounds of the cumulative probability. When asked to find a "one-tailed" probability (i.e. $$P(X\le a)$$ or $$P(X \gt a)$$​), input an extreme value into the corresponding tail.

Example 1

The random variable $$X$$​ follows a normal distribution with mean $$20$$​ and variance $$2.5$$. ​Find the following probabilities:

i) $$P(18 \leq X \lt 24)$$​

ii) $$P(X\lt 21)$$​

Part i):

​ ​ ​ ​	​$$\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:18\\Upper&:24\\ \sigma&:\sqrt{2.5}\\ \mu&:20 \end{aligned}}$$​​
	​$$\boxed{ p = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ 0.8913}$$​​

​$$P(18 \leq X \leq 24) = \underline{0.8913}$$

Note: Remember to square root the variance to find the standard deviation.

Part ii):

​ ​ ​ ​	​$$\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:-999999999\\Upper&:21\\ \sigma&:\sqrt{2.5}\\ \mu&:20 \end{aligned}}$$​​
	​$$\boxed{ p = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ 0.7365}$$​​

​$$P(X \lt 21) = \underline{0.7365}$$​​

Example 2

The heights of people in a group ($$H$$) in centimetres has distribution $$H \sim N(175,49)$$.

i) Find the probability that a person selected at random is taller than $$1.82m$$.

ii) A random person from the group is selected and their height is noted. This process happens $$6$$ times. What is the probability that the person chosen is taller than $$1.82m$$​ least twice?

Part i):

​ ​ ​ ​	​$$\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:182\\Upper&:99999999999\\ \sigma&:7\\ \mu&:175 \end{aligned}}$$​​
	​$$\boxed{ p = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ 0.1587}$$​​

​$$P(H \gt 1.82m) = P(H \gt 182cm) = \underline{0.1587}$$​​

Part ii):

Let $$X$$ be the event that the person chosen is taller than $$1.82m$$. There are $$6$$ trials, and two outcomes (either taller than or shorter than $$1.82m$$). $$X$$ can therefore be modelled by a binomial distribution, as:

$$X \sim B(6,0.1587)$$​​

Using the binomial distribution function on a calculator, the desired probability is:

$$\begin{aligned}P(X \ge 2 )&= 1 - P(X \le1)\\&= 1-0.75588... \\&= 0.2441 \ (4 \ d.p.) \end{aligned}$$​​

$$P(X \geq 2 ) = \underline{0.2441}$$​​