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Normal distribution

The normal distribution: Finding probabilities

The normal distribution: Finding probabilities

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Explainer Video

Tutor: Bilal

Summary

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 μ\muμ and standard deviation σ\sigmaσ as well as the upper and lower limits for the area required.

​

Maths; Normal distribution; KS5 Year 13; The normal distribution: Finding probabilities

calculator tip

  Menu7:Distribution2:Normal CDLower:Upper:σ:μ:\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:\\Upper&:\\ \sigma&:\\ \mu&: \end{aligned}}72LowerUpperσμ​  Menu:Distribution:Normal CD::::​​​​

​

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≤a)P(X\le a)P(X≤a) or P(X>a)P(X \gt a)P(X>a)​), input an extreme value into the corresponding tail.


Example 1

The random variable XXX​ follows a normal distribution with mean 202020​ and variance 2.52.52.5. ​Find the following probabilities:

i) P(18≤X<24)P(18 \leq X \lt 24)P(18≤X<24)​

ii) P(X<21)P(X\lt 21)P(X<21)​


Part i):

​


​
Maths; Normal distribution; KS5 Year 13; The normal distribution: Finding probabilities


​


​
​  Menu7:Distribution2:Normal CDLower:18Upper:24σ:2.5μ:20\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:18\\Upper&:24\\ \sigma&:\sqrt{2.5}\\ \mu&:20 \end{aligned}}72LowerUpperσμ​  Menu:Distribution:Normal CD:18:24:2.5​:20​​​​
​p=                   0.8913\boxed{ p = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ 0.8913}p=                   0.8913​​​


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


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


Part ii):

​


​
Maths; Normal distribution; KS5 Year 13; The normal distribution: Finding probabilities


​


​
​  Menu7:Distribution2:Normal CDLower:−999999999Upper:21σ:2.5μ:20\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:-999999999\\Upper&:21\\ \sigma&:\sqrt{2.5}\\ \mu&:20 \end{aligned}}72LowerUpperσμ​  Menu:Distribution:Normal CD:−999999999:21:2.5​:20​​​​
​p=                   0.7365\boxed{ p = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ 0.7365}p=                   0.7365​​​


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


Example 2

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

i) Find the probability that a person selected at random is taller than 1.82m1.82m1.82m.

ii) A random person from the group is selected and their height is noted. This process happens 666 times. What is the probability that the person chosen is taller than 1.82m1.82m1.82m​ least twice?


Part i):

​


​
Maths; Normal distribution; KS5 Year 13; The normal distribution: Finding probabilities


​


​
​  Menu7:Distribution2:Normal CDLower:182Upper:99999999999σ:7μ:175\boxed{\begin{aligned} & \ \ Menu\\7&: Distribution\\2&: Normal \ CD\\Lower&:182\\Upper&:99999999999\\ \sigma&:7\\ \mu&:175 \end{aligned}}72LowerUpperσμ​  Menu:Distribution:Normal CD:182:99999999999:7:175​​​​
​p=                   0.1587\boxed{ p = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ 0.1587}p=                   0.1587​​​


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


Part ii):

Let XXX be the event that the person chosen is taller than 1.82m1.82m1.82m. There are 666 trials, and two outcomes (either taller than or shorter than 1.82m1.82m1.82m). XXX can therefore be modelled by a binomial distribution, as:

X∼B(6,0.1587)X \sim B(6,0.1587)X∼B(6,0.1587)​​


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

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


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


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Probability distributions

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The normal distribution

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The normal distribution

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The normal distribution: Finding probabilities

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FAQs - Frequently Asked Questions

How do you find probabilities of the form P(X < a) and P(X > a) for a normal distribution?

Set either the lower or upper end to be an extreme value (like 999999 or -9999999) and the other end to be a.

How do you find probabilities from a normal distribution?

You can find probabilities using a calculator.

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