The Normal Distribution

What you'll learn

How to recognise and interpret normal notation such as $\sim N(\mu,\sigma^2)$ .
How to standardise values and find probabilities from a normal distribution.
How to work backwards from percentages to find cut-off values, means or standard deviations.
How the normal distribution is used in hypothesis tests for a population mean.

1. The normal model

A random variable is a quantity whose value depends on chance. For example, the weight of coffee in a jar, a journey time, or a battery lifetime.

A continuous random variable can take any value in an interval, not just whole-number values. For continuous variables, probabilities are represented by areas under a curve.

Definition

Normal distribution

A normal distribution is a continuous probability model with a symmetric bell-shaped curve. If $\sim N(\mu,\sigma^2)$ , then $μ\mu$ is the mean, $σ\sigma$ is the standard deviation, and $σ2\sigma^2$ is the variance.

The mean is the centre of the curve. The standard deviation measures the spread: larger $σ\sigma$ means a wider, flatter curve.

A normal curve is centred at the mean, with larger standard deviation giving a wider and flatter distribution.

Tip

Strict or non-strict inequalities

For a continuous normal variable, $P(X<a)=P(X≤a)P(X<a)=P(X\le a)$ because the probability of getting exactly one value is zero.

Example

Reading normal notation

Suppose the fill weight of a packet is modelled by $\sim N(120, 16)$ .

Identify the mean from the first number:
$μ=120\mu = 120$
The second number is the variance, so take the square root to get the standard deviation:
$σ=16=4\sigma = \sqrt{16}=4$
By symmetry, half the values lie below the mean:
$P (W < 120) = 0.5$

Common Mistake

Do not use the variance as sigma

In $\sim N(20,25)$ , the standard deviation is 5, not 25. The second number is $σ2\sigma^2$ .

2. Standardising to the standard normal

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1. We write this as $\sim N(0,1)$ .

To convert a value of $X$ into a standard normal value, use:

Z=X−μσZ=\frac{X-\mu}{\sigma}

This tells you how many standard deviations above or below the mean the value is.

Standardising converts an original value x into a z-value measuring its distance from the mean in standard deviations.

Key Idea

Standardising

Most normal probability questions become much easier once you convert the boundary value into a $z$ -value.

Example

Finding normal probabilities

The mass of cereal in a box is normally distributed with mean 250 g and standard deviation 5 g. Find the probability that a box contains: more than 257 g, less than 243 g, and between 246 g and 255 g.

The three probability regions in the cereal example correspond to an upper tail, a lower tail, and an interval under the normal curve.

Write the distribution carefully:
$\sim N(250,5^2)$
For more than 257 g, standardise 257:
$z=257−2505=1.4z=\frac{257-250}{5}=1.4$
Use the upper tail of the standard normal distribution:
$P(X>257)=P(Z>1.4)≈0.0808P(X>257)=P(Z>1.4)\approx 0.0808$
For less than 243 g, standardise 243:
$z=243−2505=−1.4z=\frac{243-250}{5}=-1.4$
Use the lower tail:
$P(X<243)=P(Z<−1.4)≈0.0808P(X<243)=P(Z<-1.4)\approx 0.0808$
For the interval, standardise both ends:
$P(246<X<255)=P(−0.8<Z<1)≈0.629P(246<X<255)=P(-0.8<Z<1)\approx 0.629$

3. Working backwards with inverse normal

Sometimes you are given a probability and asked for the value of the variable. This is an inverse normal question.

A percentile is a cut-off value. For example, the 90th percentile is the value below which 90% of observations lie.

The 90th percentile is the cut-off with 90% of the normal distribution shaded to its left.

If $z_p$ is the standard normal value with $P(Z<z_p)=p$ , then:

x=μ+σzpx=\mu+\sigma z_p

Example

Finding a cut-off time

The time taken to complete an online task is normally distributed with mean 62 minutes and standard deviation 10 minutes. Find the time by which 90% of people complete the task.

The phrase “by which 90% complete” means a lower-tail probability of 0.90:
$P (T < t) = 0.90$
Use inverse normal to find the standard normal value:
$\Rightarrow z\approx 1.282$
Convert back to the original scale:
$t=62+10(1.282)≈74.8t=62+10(1.282)\approx 74.8$
So the required time is about 75 minutes.

4. Finding a missing mean or standard deviation

If a question gives you a percentage and asks for $μ\mu$ or $σ\sigma$ , turn the percentage into a $z$ -value first.

The key equation is:

x=μ+σzx=\mu+\sigma z

If you have one unknown, you need one equation. If both $μ\mu$ and $σ\sigma$ are unknown, you need two probability statements and therefore two equations.

Example

Finding a standard deviation

The heights of a group are normally distributed with mean 168 cm. It is known that 12% of the group are shorter than 154 cm. Find the standard deviation.

The known lower-tail percentage fixes the position of 154 cm below the mean, allowing the standard deviation to be found.

Translate the sentence into probability notation:
$P (H < 154) = 0.12$
Use inverse normal on the standard normal distribution:
$\Rightarrow z\approx -1.175$
Substitute into $x=μ+σzx=\mu+\sigma z$ :
$154=168+σ(−1.175)154=168+\sigma(-1.175)$
Solve for the positive standard deviation:
$σ=141.175≈11.9\sigma=\frac{14}{1.175}\approx 11.9$

Example

Finding both the mean and standard deviation

A sprinter’s 100 m times are normally distributed. The sprinter runs faster than 9.8 seconds on 12% of attempts, and slower than 10.9 seconds on 25% of attempts. Find the mean and standard deviation.

The two given probability statements locate two cut-offs on the same normal curve, giving two equations for the unknown mean and standard deviation.

“Faster than 9.8 seconds” means the time is less than 9.8:
$P (T < 9.8) = 0.12$
“Slower than 10.9 seconds” means the time is greater than 10.9, so:
$P (T < 10.9) = 0.75$
Convert both probabilities to $z$ -values:
$z1≈−1.175,z2≈0.674z_1\approx -1.175,\qquad z_2\approx 0.674$
Form two equations:
$9.8−μσ=−1.175,10.9−μσ=0.674\frac{9.8-\mu}{\sigma}=-1.175,\qquad \frac{10.9-\mu}{\sigma}=0.674$
Rearrange:
$9.8=μ−1.175σ,10.9=μ+0.674σ9.8=\mu-1.175\sigma,\qquad 10.9=\mu+0.674\sigma$
Subtract the first equation from the second:
$1.1=1.849σ⇒σ≈0.5951.1=1.849\sigma \Rightarrow \sigma\approx 0.595$
Substitute back to find the mean:
$μ=10.9−0.674(0.595)≈10.5\mu=10.9-0.674(0.595)\approx 10.5$

Common Mistake

Faster means smaller

For race times, “faster than 10 seconds” means $T < 10$ , not $T > 10$ .

5. Conditional probability with normal variables

A conditional probability is the probability of an event happening given that another event has already happened.

P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}

For lifetimes, “lasts another 10 hours after already lasting 45 hours” is not just $P (X > 55)$ . You must condition on the fact it has already lasted 45 hours.

Conditional lifetime probabilities compare the smaller tail beyond the later time with the larger tail beyond the time already survived.

Example

Battery lasting longer after use

A battery lifetime $B$ is normally distributed with mean 60 hours and standard deviation 8 hours. Given that a battery has already lasted 55 hours, find the probability that it lasts at least another 12 hours.

Lasting another 12 hours means the total lifetime must exceed 67 hours:
$P(B>67∣B>55)P(B>67\mid B>55)$
Since $B > 67$ is inside the event $B > 55$ , use:
$P(B>67∣B>55)=P(B>67)P(B>55)P(B>67\mid B>55)=\frac{P(B>67)}{P(B>55)}$
Calculate the numerator:
$P(B>67)=P(Z>67−608)=P(Z>0.875)≈0.1908P(B>67)=P\left(Z>\frac{67-60}{8}\right)=P(Z>0.875)\approx 0.1908$
Calculate the denominator:
$P(B>55)=P(Z>55−608)=P(Z>−0.625)≈0.7340P(B>55)=P\left(Z>\frac{55-60}{8}\right)=P(Z>-0.625)\approx 0.7340$
Divide:
$P(B>67∣B>55)≈0.19080.7340≈0.260P(B>67\mid B>55)\approx \frac{0.1908}{0.7340}\approx 0.260$

6. Hypothesis tests using the normal distribution

A hypothesis test checks whether sample evidence is strong enough to challenge a claim about a population.

The null hypothesis, $H_0$ , is the starting assumption.
The alternative hypothesis, $H_1$ , is what you are testing for.
The significance level is the cut-off probability for deciding that the evidence is unusually extreme.
A p-value is the probability, assuming $H_0$ is true, of getting a result at least as extreme as the sample result.

If the population standard deviation is $σ\sigma$ and the sample size is $n$ , then the sample mean has distribution:

Xˉ∼N(μ,σ2n)\bar X \sim N\left(\mu,\frac{\sigma^2}{n}\right)

The standard deviation of $Xˉ\bar X$ is called the standard error:

σn\frac{\sigma}{\sqrt n}

Example

Testing whether a mean has decreased

A machine is set to fill bottles with mean 500 ml. The standard deviation is known to be 5.5 ml. A manager suspects the mean fill has decreased. A random sample of 25 bottles has mean 497.8 ml. Test at the 5% significance level.

A lower-tailed hypothesis test rejects the null hypothesis when the sample mean lies unusually far below 500 ml.

State the hypotheses:
$H0:μ=500,H1:μ<500H_0:\mu=500,\qquad H_1:\mu<500$
Under $H_0$ , write the sampling distribution:
$Xˉ∼N(500,5.5225)\bar X \sim N\left(500,\frac{5.5^2}{25}\right)$
Calculate the test statistic:
$z=497.8−5005.5/25=−2.00z=\frac{497.8-500}{5.5/\sqrt{25}}=-2.00$
Find the lower-tail p-value:
$P(Z<−2.00)≈0.0228P(Z<-2.00)\approx 0.0228$
Compare with 5%:
$0.0228 < 0.05$
Reject $H_0$ . There is evidence at the 5% level that the mean fill has decreased.

If you are given summary statistics, use:

xˉ=∑xn\bar x=\frac{\sum x}{n}

For a large sample where the population standard deviation is not known, you may estimate it using the sample standard deviation $s$ :

s=∑x2−(∑x)2nn−1s=\sqrt{\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1}}

Tip

One-tailed or two-tailed

Words like “less than” or “more than” give a one-tailed test. Words like “changed” or “different” usually mean a two-tailed test.

Exam technique

In the exam

Read $N(μ,σ2)N(\mu,\sigma^2)$ carefully and take the square root of the second number to get $σ\sigma$ .
Draw a quick bell curve and shade the required region before using your calculator.
Convert “more than” probabilities into lower-tail probabilities when using inverse normal.
In hypothesis tests, always state $H_0$ , $H_1$ , the sampling distribution, your p-value or critical comparison, and a conclusion in context.

Self review

Check yourself

If $\sim N(40,16)$ , what are the mean and standard deviation?
How would you rewrite $P (X > 70) = 0.2$ as a lower-tail probability?
If a manager believes the mean has decreased, what should the alternative hypothesis be?

What you'll learn

How to recognise and interpret normal notation such as $\sim N(\mu,\sigma^2)$ .
How to standardise values and find probabilities from a normal distribution.
How to work backwards from percentages to find cut-off values, means or standard deviations.
How the normal distribution is used in hypothesis tests for a population mean.

1. The normal model

A random variable is a quantity whose value depends on chance. For example, the weight of coffee in a jar, a journey time, or a battery lifetime.

A continuous random variable can take any value in an interval, not just whole-number values. For continuous variables, probabilities are represented by areas under a curve.

Definition

Normal distribution

The mean is the centre of the curve. The standard deviation measures the spread: larger $σ\sigma$ means a wider, flatter curve.

A normal curve is centred at the mean, with larger standard deviation giving a wider and flatter distribution.

Tip

Strict or non-strict inequalities

For a continuous normal variable, $P(X<a)=P(X≤a)P(X<a)=P(X\le a)$ because the probability of getting exactly one value is zero.

Example

Reading normal notation

Suppose the fill weight of a packet is modelled by $\sim N(120, 16)$ .

Identify the mean from the first number:
$μ=120\mu = 120$
The second number is the variance, so take the square root to get the standard deviation:
$σ=16=4\sigma = \sqrt{16}=4$
By symmetry, half the values lie below the mean:
$P (W < 120) = 0.5$

Common Mistake

Do not use the variance as sigma

In $\sim N(20,25)$ , the standard deviation is 5, not 25. The second number is $σ2\sigma^2$ .

2. Standardising to the standard normal

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1. We write this as $\sim N(0,1)$ .

To convert a value of $X$ into a standard normal value, use:

Z=X−μσZ=\frac{X-\mu}{\sigma}

This tells you how many standard deviations above or below the mean the value is.

Standardising converts an original value x into a z-value measuring its distance from the mean in standard deviations.

Key Idea

Standardising

Most normal probability questions become much easier once you convert the boundary value into a $z$ -value.

Example

Finding normal probabilities

The three probability regions in the cereal example correspond to an upper tail, a lower tail, and an interval under the normal curve.

Write the distribution carefully:
$\sim N(250,5^2)$
For more than 257 g, standardise 257:
$z=257−2505=1.4z=\frac{257-250}{5}=1.4$
Use the upper tail of the standard normal distribution:
$P(X>257)=P(Z>1.4)≈0.0808P(X>257)=P(Z>1.4)\approx 0.0808$
For less than 243 g, standardise 243:
$z=243−2505=−1.4z=\frac{243-250}{5}=-1.4$
Use the lower tail:
$P(X<243)=P(Z<−1.4)≈0.0808P(X<243)=P(Z<-1.4)\approx 0.0808$
For the interval, standardise both ends:
$P(246<X<255)=P(−0.8<Z<1)≈0.629P(246<X<255)=P(-0.8<Z<1)\approx 0.629$

3. Working backwards with inverse normal

Sometimes you are given a probability and asked for the value of the variable. This is an inverse normal question.

A percentile is a cut-off value. For example, the 90th percentile is the value below which 90% of observations lie.

The 90th percentile is the cut-off with 90% of the normal distribution shaded to its left.

If $z_p$ is the standard normal value with $P(Z<z_p)=p$ , then:

x=μ+σzpx=\mu+\sigma z_p

Example

Finding a cut-off time

The time taken to complete an online task is normally distributed with mean 62 minutes and standard deviation 10 minutes. Find the time by which 90% of people complete the task.

The phrase “by which 90% complete” means a lower-tail probability of 0.90:
$P (T < t) = 0.90$
Use inverse normal to find the standard normal value:
$\Rightarrow z\approx 1.282$
Convert back to the original scale:
$t=62+10(1.282)≈74.8t=62+10(1.282)\approx 74.8$
So the required time is about 75 minutes.

4. Finding a missing mean or standard deviation

If a question gives you a percentage and asks for $μ\mu$ or $σ\sigma$ , turn the percentage into a $z$ -value first.

The key equation is:

x=μ+σzx=\mu+\sigma z

If you have one unknown, you need one equation. If both $μ\mu$ and $σ\sigma$ are unknown, you need two probability statements and therefore two equations.

Example

Finding a standard deviation

The heights of a group are normally distributed with mean 168 cm. It is known that 12% of the group are shorter than 154 cm. Find the standard deviation.

The known lower-tail percentage fixes the position of 154 cm below the mean, allowing the standard deviation to be found.

Translate the sentence into probability notation:
$P (H < 154) = 0.12$
Use inverse normal on the standard normal distribution:
$\Rightarrow z\approx -1.175$
Substitute into $x=μ+σzx=\mu+\sigma z$ :
$154=168+σ(−1.175)154=168+\sigma(-1.175)$
Solve for the positive standard deviation:
$σ=141.175≈11.9\sigma=\frac{14}{1.175}\approx 11.9$

Example

Finding both the mean and standard deviation

The two given probability statements locate two cut-offs on the same normal curve, giving two equations for the unknown mean and standard deviation.

“Faster than 9.8 seconds” means the time is less than 9.8:
$P (T < 9.8) = 0.12$
“Slower than 10.9 seconds” means the time is greater than 10.9, so:
$P (T < 10.9) = 0.75$
Convert both probabilities to $z$ -values:
$z1≈−1.175,z2≈0.674z_1\approx -1.175,\qquad z_2\approx 0.674$
Form two equations:
$9.8−μσ=−1.175,10.9−μσ=0.674\frac{9.8-\mu}{\sigma}=-1.175,\qquad \frac{10.9-\mu}{\sigma}=0.674$
Rearrange:
$9.8=μ−1.175σ,10.9=μ+0.674σ9.8=\mu-1.175\sigma,\qquad 10.9=\mu+0.674\sigma$
Subtract the first equation from the second:
$1.1=1.849σ⇒σ≈0.5951.1=1.849\sigma \Rightarrow \sigma\approx 0.595$
Substitute back to find the mean:
$μ=10.9−0.674(0.595)≈10.5\mu=10.9-0.674(0.595)\approx 10.5$

Common Mistake

Faster means smaller

For race times, “faster than 10 seconds” means $T < 10$ , not $T > 10$ .

5. Conditional probability with normal variables

A conditional probability is the probability of an event happening given that another event has already happened.

P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)}

For lifetimes, “lasts another 10 hours after already lasting 45 hours” is not just $P (X > 55)$ . You must condition on the fact it has already lasted 45 hours.

Conditional lifetime probabilities compare the smaller tail beyond the later time with the larger tail beyond the time already survived.

Example

Battery lasting longer after use

Lasting another 12 hours means the total lifetime must exceed 67 hours:
$P(B>67∣B>55)P(B>67\mid B>55)$
Since $B > 67$ is inside the event $B > 55$ , use:
$P(B>67∣B>55)=P(B>67)P(B>55)P(B>67\mid B>55)=\frac{P(B>67)}{P(B>55)}$
Calculate the numerator:
$P(B>67)=P(Z>67−608)=P(Z>0.875)≈0.1908P(B>67)=P\left(Z>\frac{67-60}{8}\right)=P(Z>0.875)\approx 0.1908$
Calculate the denominator:
$P(B>55)=P(Z>55−608)=P(Z>−0.625)≈0.7340P(B>55)=P\left(Z>\frac{55-60}{8}\right)=P(Z>-0.625)\approx 0.7340$
Divide:
$P(B>67∣B>55)≈0.19080.7340≈0.260P(B>67\mid B>55)\approx \frac{0.1908}{0.7340}\approx 0.260$

6. Hypothesis tests using the normal distribution

A hypothesis test checks whether sample evidence is strong enough to challenge a claim about a population.

The null hypothesis, $H_0$ , is the starting assumption.
The alternative hypothesis, $H_1$ , is what you are testing for.
The significance level is the cut-off probability for deciding that the evidence is unusually extreme.
A p-value is the probability, assuming $H_0$ is true, of getting a result at least as extreme as the sample result.

If the population standard deviation is $σ\sigma$ and the sample size is $n$ , then the sample mean has distribution:

Xˉ∼N(μ,σ2n)\bar X \sim N\left(\mu,\frac{\sigma^2}{n}\right)

The standard deviation of $Xˉ\bar X$ is called the standard error:

σn\frac{\sigma}{\sqrt n}

Example

Testing whether a mean has decreased

A lower-tailed hypothesis test rejects the null hypothesis when the sample mean lies unusually far below 500 ml.

State the hypotheses:
$H0:μ=500,H1:μ<500H_0:\mu=500,\qquad H_1:\mu<500$
Under $H_0$ , write the sampling distribution:
$Xˉ∼N(500,5.5225)\bar X \sim N\left(500,\frac{5.5^2}{25}\right)$
Calculate the test statistic:
$z=497.8−5005.5/25=−2.00z=\frac{497.8-500}{5.5/\sqrt{25}}=-2.00$
Find the lower-tail p-value:
$P(Z<−2.00)≈0.0228P(Z<-2.00)\approx 0.0228$
Compare with 5%:
$0.0228 < 0.05$
Reject $H_0$ . There is evidence at the 5% level that the mean fill has decreased.

If you are given summary statistics, use:

xˉ=∑xn\bar x=\frac{\sum x}{n}

For a large sample where the population standard deviation is not known, you may estimate it using the sample standard deviation $s$ :

s=∑x2−(∑x)2nn−1s=\sqrt{\frac{\sum x^2-\frac{(\sum x)^2}{n}}{n-1}}

Tip

One-tailed or two-tailed

Words like “less than” or “more than” give a one-tailed test. Words like “changed” or “different” usually mean a two-tailed test.

Exam technique

In the exam

Read $N(μ,σ2)N(\mu,\sigma^2)$ carefully and take the square root of the second number to get $σ\sigma$ .
Draw a quick bell curve and shade the required region before using your calculator.
Convert “more than” probabilities into lower-tail probabilities when using inverse normal.
In hypothesis tests, always state $H_0$ , $H_1$ , the sampling distribution, your p-value or critical comparison, and a conclusion in context.

Self review

Check yourself

If $\sim N(40,16)$ , what are the mean and standard deviation?
How would you rewrite $P (X > 70) = 0.2$ as a lower-tail probability?
If a manager believes the mean has decreased, what should the alternative hypothesis be?

What you'll learn

1. The normal model

2. Standardising to the standard normal

3. Working backwards with inverse normal

4. Finding a missing mean or standard deviation

5. Conditional probability with normal variables

6. Hypothesis tests using the normal distribution

Recap questions

Review The Normal Distribution by teaching Genie

Remember key concepts with flashcards

The Normal Distribution

What you'll learn

1. The normal model

2. Standardising to the standard normal

3. Working backwards with inverse normal

4. Finding a missing mean or standard deviation

5. Conditional probability with normal variables

6. Hypothesis tests using the normal distribution

Recap questions

Review The Normal Distribution by teaching Genie

Remember key concepts with flashcards

The Normal Distribution

What you'll learn

1. The normal model

2. Standardising to the standard normal

3. Working backwards with inverse normal

4. Finding a missing mean or standard deviation

5. Conditional probability with normal variables

6. Hypothesis tests using the normal distribution

Recap questions

Review The Normal Distribution by teaching Genie

Remember key concepts with flashcards

The Normal Distribution Revision Guide

The Normal Distribution

What you'll learn

1. The normal model

2. Standardising to the standard normal

3. Working backwards with inverse normal

4. Finding a missing mean or standard deviation

5. Conditional probability with normal variables

6. Hypothesis tests using the normal distribution

Recap questions

Review The Normal Distribution by teaching Genie

Remember key concepts with flashcards

The Normal Distribution Revision Guide