5.5 The normal distribution

Learning objectives

5.5.1Normal distribution• understand the use of a normal distribution to model a continuous random variable, and use normal distribution tables Sketches of normal curves to illustrate distributions or probabilities may be required.5.5.2Problems concerning a variable X• solve problems concerning a variable X, where,XN 2+ nv_ i, including - finding the value of P 1Xx>_ i, or a related probability, given the values of x1, n, v. - finding a relationship between x1, n and v given the value of P 1Xx>_ i or a related probability For calculations involving standardisation, full details of the working should be shown, e.g. Z X v n = -_ i.5.5.3Binomial distribution• recall conditions under which the normal distribution can be used as an approximation to the binomial distribution, and use this approximation, with a continuity correction, in solving problems. n sufficiently large to ensure that both np > 5 and nq > 5.

$X\sim N(\mu,\sigma^2)$ models a continuous variable whose distribution is approximately bell-shaped, symmetric and unimodal. The centre is $\mu$ ; $\sigma$ controls spread; total area under the curve is 1.

Probability	Sketch/shading instruction
$P(X<a)$	shade left of $a$
$P(X>a)$	shade right of $a$
$P(a<X<b)$	shade between $a$ and $b$

$Standardise with$ Z=\frac{X-\mu}{\sigma},\qquad Z\sim N(0,1), $then use the standard normal table and symmetry or complements as needed.$

For $X\sim N(10,4)$ , the standard deviation is 2, so $P(X<12)=P(Z<(12-10)/2)=P(Z<1)=0.8413$ .

Curve height is not a probability; area is. Because the model is continuous, $P(X=a)=0$ , so $<$ and $\le$ give the same probability. In $N(\mu,\sigma^2)$ , the second parameter is variance.

Step	Direct probability	Inverse/relationship problem
1	sketch and identify lower, upper or interval area	convert the stated area to the correct signed $z$ -quantile
2	display $Z=(X-\mu)/\sigma$ with the numerical boundary	write $(x-\mu)/\sigma=z$
3	read the table; complement or subtract if needed	rearrange to $x=\mu+z\sigma$ or the required relationship

If $X\sim N(50,6^2)$ , then $P(X>62)=P\left(Z>\frac{62-50}{6}\right)=P(Z>2)=1-0.9772=0.0228.$ This displays the full standardisation required.

If $P(X<x)=0.90$ , tables give $z\approx1.282$ , so $(x-\mu)/\sigma=1.282$ and $x=\mu+1.282\sigma$ . If $x$ is known, the same equation gives a relationship between $\mu$ and $\sigma$ .

For a central probability, split the excluded area equally between two tails before finding $z$ . For example, central 90% leaves 0.05 per tail and uses bounds $\mu\pm1.645\sigma$ .

‘At least’ and ‘more than’ are upper-tail events. Do not feed an upper-tail probability directly into a lower-tail table, and do not omit the sign of $z$ for a boundary below $\mu$ .

$For $X\sim B(n,p)$, let $q=1-p$. Use the normal approximation only when$ np>5\quad\text{and}\quad nq>5, $then approximate with$ Y\sim N(np,npq).

Binomial event	Continuous normal event
$X\le k$	$Y<k+0.5$
$X<k$ (that is $X\le k-1$ )	$Y<k-0.5$
$X\ge k$	$Y>k-0.5$
$X>k$ (that is $X\ge k+1$ )	$Y>k+0.5$
$a\le X\le b$	$a-0.5<Y<b+0.5$

If $X\sim B(100,0.4)$ , then $np=40$ and $nq=60$ , so the conditions hold and $Y\sim N(40,24)$ . To approximate $P(X\le45)$ , use $P(Y<45.5)=P\left(Z<\frac{45.5-40}{\sqrt{24}}\right).$

The correction gives each integer value its full unit-wide bar: the bar for $X=45$ extends from 44.5 to 45.5. Sketching the included integer bars makes the correct boundary visible.

Check both approximation conditions before calculating. Use variance $npq$ but standard deviation $\sqrt{npq}$ when standardising, and apply continuity correction before converting to $z$ .

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—Cambridge International AS & A Level Mathematics (9709)Read

5.5 The normal distribution

Learning objectives

A normal probability is area under a symmetric continuous curve

Show the tail, standardise fully, then solve or invert

Move each binomial boundary by half a unit

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