X N(60, 16). Find P(X<66). [2 marks]

State the continuity correction you would use to approximate P(X 30) for a large binomial X. [1 mark]

SQA AH style: sum of normals-style practice: Independent variables X N(50, 16) and Y N(30, 9) are given. Find the distribution of T=X+Y and P(T>85).

A sum of independent normals is normal, with means and variances adding: E(T)=50+30=80, Var(T)=16+9=25, so T N(80, 25) (2 marks). Standardise: Z=85-80sqrt(25)=55=1 (1 mark). P(T>85)=P(Z>1)=1-0.8413=0.1587 (1 mark). Markers reward adding the means and variances, the standard deviation sqrt(25)=5, and the tail probability.

ScotlandStatisticsSyllabus dot point

How do you model a continuous measurement, and how do you find probabilities from a normal distribution?

Work with continuous random variables and the normal distribution, standardise to find probabilities, combine independent normal variables, and use the normal approximation to the binomial and Poisson distributions with a continuity correction.

A focused answer to the SQA Advanced Higher Statistics continuous random variables content: the normal distribution, standardising to the Z-distribution, finding probabilities and quantiles, combining independent normal variables, and the normal approximation to the binomial and Poisson with a continuity correction.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

For a continuous random variable, probability is the area under the density curve, so $P(X=x)=0$ and only intervals have non-zero probability. If $X\sim N(\mu,\sigma^2)$ , standardise with $Z=\dfrac{X-\mu}{\sigma}$ , where $Z\sim N(0,1)$ , and read the area from tables. A linear combination of independent normals is normal: for $T=X+Y$ , $E(T)=E(X)+E(Y)$ and $\text{Var}(T)=\text{Var}(X)+\text{Var}(Y)$ . When $n$ is large, $B(n,p)\approx N(np,\,np(1-p))$ and $\text{Po}(\lambda)\approx N(\lambda,\lambda)$ , each with a continuity correction of $\pm 0.5$ .

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What this dot point is asking
Continuous random variables
The normal distribution and standardising
Working backwards from a probability
Combining independent normal variables
Normal approximations with a continuity correction
Try this

What this dot point is asking

Most measurements (heights, masses, times) are continuous, and the normal distribution is the central model for them. The SQA wants you to find probabilities by standardising to the $Z$ -distribution, to work backwards from a probability to a value, to combine independent normal variables, and to approximate the binomial and Poisson by a normal when the counts are large, using a continuity correction.

Continuous random variables

A continuous random variable is described by a probability density function whose total area is one.

The normal distribution and standardising

The normal distribution $N(\mu,\sigma^2)$ is symmetric and bell-shaped, fixed by its mean $\mu$ and variance $\sigma^2$ . To find a probability you convert to the standard normal $Z\sim N(0,1)$ .

Find a probability between two values

To find the probability that a normal variable falls in an interval, we convert both endpoints to $z$ -scores, then express the interval probability as a difference of two cumulative probabilities that we can read from the standard normal table.

Step 1: Standardise both endpoints

For $X\sim N(100, 25)$ , the standard deviation is $\sigma=\sqrt{25}=5$ . Subtract the mean and divide by $\sigma$ to convert each boundary to a $z$ -score.

z_1=\dfrac{95-100}{5}=-1, \qquad z_2=\dfrac{108-100}{5}=1.6

Step 2: Express the probability as a difference of cumulative probabilities

The probability of the interval equals the cumulative probability up to the upper $z$ -value minus the cumulative probability up to the lower $z$ -value.

P(95<X<108)=P(Z<1.6)-P(Z<-1)

Step 3: Look up the table values and compute

$P(Z<1.6)=0.9452$ directly from the table. For the negative $z$ -value, use the symmetry rule $P(Z<-1)=1-P(Z<1)=1-0.8413=0.1587$ .

P(95<X<108)=0.9452-0.1587=0.7865

Final answer: $P(95<X<108)\approx 0.7865$ .

Working backwards from a probability

If you are given a probability and asked for a value (a quantile), find the $z$ -value first, then unstandardise with $X=\mu+z\sigma$ .

Find the value exceeded by 10% of a population

When given a probability and asked for the corresponding $x$ -value, we work backwards: find the $z$ -value from the table first, then convert back to the original scale using $X=\mu+z\sigma$ .

Step 1: Convert the upper-tail probability to a cumulative probability and find $z$

"Exceeded by 10%" means $P(X>x)=0.10$ , so $P(X<x)=0.90$ . We look up $0.90$ in the body of the standard normal table to find the corresponding $z$ -value.

P(Z<z)=0.90 \implies z\approx 1.282

Step 2: Unstandardise to recover $X$

Rearranging $Z=\dfrac{X-\mu}{\sigma}$ gives $X=\mu+z\sigma$ . Substitute $\mu=100$ , $\sigma=5$ and $z=1.282$ .

X=100+1.282\times 5=100+6.41=106.41

Step 3: Interpret in context

Values above $106.4$ account for the top 10% of the population.

Final answer: the value exceeded by 10% of the population is approximately $106.4$ .

Combining independent normal variables

A linear combination of independent normal variables is itself normal, so you only need the new mean and variance.

Normal approximations with a continuity correction

When a binomial or Poisson count is large, a normal curve approximates it well; because you are using a continuous curve for a discrete count, apply a continuity correction of $\pm 0.5$ .

Approximate a binomial probability

When $n$ is large, the binomial is well approximated by a normal with the same mean and variance. Because we are using a continuous distribution to approximate a discrete one, we apply a continuity correction of $\pm 0.5$ to each boundary before standardising.

Step 1: Set up the normal approximation

For $X\sim B(100, 0.5)$ , compute $np=100\times 0.5=50$ (mean) and $np(1-p)=100\times 0.5\times 0.5=25$ (variance). Approximate by $Y\sim N(50, 25)$ , so $\sigma=\sqrt{25}=5$ .

Step 2: Apply the continuity correction

The event $P(X\le 45)$ in the discrete binomial corresponds to $P(Y<45.5)$ in the continuous normal, because the integers up to $45$ are represented by the interval up to $45.5$ . Standardise the corrected boundary:

z=\dfrac{45.5-50}{5}=\dfrac{-4.5}{5}=-0.9

Step 3: Look up the probability

Use the symmetry rule $P(Z<-0.9)=1-P(Z<0.9)=1-0.8159$ .

P(Y<45.5)=P(Z<-0.9)=1-0.8159=0.1841

Final answer: $P(X\le 45)\approx 0.1841$ .

Try this

Q1. $X\sim N(60, 16)$ . Find $P(X<66)$ . [2 marks]

Cue. $z=\dfrac{66-60}{4}=1.5$ , so $P(X<66)=P(Z<1.5)=0.9332$ .

Q2. State the continuity correction you would use to approximate $P(X\ge 30)$ for a large binomial $X$ . [1 mark]

Cue. Use $P(Y>29.5)$ : for $P(X\ge 30)$ the boundary shifts down by $0.5$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: normal prob3 marksThe mass of an apple is normally distributed with mean

120

g and standard deviation

15

g. Find the probability that an apple has mass greater than

145

Show worked answer →

Standardise: $Z=\dfrac{X-\mu}{\sigma}=\dfrac{145-120}{15}=\dfrac{25}{15}\approx 1.667$ (1 mark).

Required probability is $P(Z>1.667)=1-P(Z<1.667)=1-0.9522=0.0478$ (1 mark).

So about $0.048$ , roughly a $4.8\%$ chance an apple exceeds $145$ g (1 mark). Markers reward the standardisation, the use of the upper tail and the final probability.

AH style: sum of normals4 marksIndependent variables

X\sim N(50, 16)

and

Y\sim N(30, 9)

are given. Find the distribution of

T=X+Y

and

P(T>85)

Show worked answer →

A sum of independent normals is normal, with means and variances adding: $E(T)=50+30=80$ , $\text{Var}(T)=16+9=25$ , so $T\sim N(80, 25)$ (2 marks).

Standardise: $Z=\dfrac{85-80}{\sqrt{25}}=\dfrac{5}{5}=1$ (1 mark).

$P(T>85)=P(Z>1)=1-0.8413=0.1587$ (1 mark). Markers reward adding the means and variances, the standard deviation $\sqrt{25}=5$ , and the tail probability.

Related dot points

Sources & how we know this

SQA Advanced Higher Statistics Course Specification (C803 77) — SQA (2023)
SQA Advanced Higher Statistics Data Booklet — SQA (2019)