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How do you model continuous data with the normal distribution, and how do you find probabilities and unknown parameters?

The normal distribution as a model for continuous data, finding probabilities, the standard normal distribution, using the inverse normal, and approximating the binomial.

A focused answer to the Edexcel A-Level Mathematics normal distribution content, covering the normal distribution as a model for continuous data, finding probabilities, standardising, the inverse normal for unknown parameters, and the normal approximation to the binomial.

Generated by Claude Opus 4.810 min answer

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What this dot point is asking

Edexcel wants you to use the normal distribution N(μ,σ2)N(\mu, \sigma^2) as a model for continuous data, find probabilities with a calculator, standardise to the standard normal ZN(0,1)Z \sim N(0, 1), use the inverse normal to find values or unknown μ\mu or σ\sigma, recognise when a normal model is suitable, and use the normal approximation to the binomial.

The answer

The normal model

Standardising

Inverse normal and approximation

The inverse normal finds the value of XX for a given cumulative probability, which lets you solve for an unknown μ\mu or σ\sigma from a known proportion. For a binomial with large nn, you can approximate B(n,p)B(n, p) by N(np,np(1p))N(np, np(1 - p)), applying a continuity correction.

When a normal model is suitable

A normal model fits data that are roughly symmetric and bell-shaped about a central value, with most values near the mean and few in the tails. Heights, masses and measurement errors often qualify. It is a poor model for strongly skewed data, for counts that cannot be negative but have a long upper tail, or for data with two separate peaks. The normal approximation to the binomial works best when nn is large and pp is close to 0.50.5, so that npnp and n(1p)n(1 - p) are both reasonably large.

Examples in context

Try this

Q1. For XN(50,16)X \sim N(50, 16), find P(X>54)P(X > 54). [3 marks]

  • Cue. Z=54504=1Z = \dfrac{54 - 50}{4} = 1, so P(Z>1)0.159P(Z > 1) \approx 0.159.

Q2. State the approximating normal distribution for XB(100,0.5)X \sim B(100, 0.5). [2 marks]

  • Cue. np=50np = 50, np(1p)=25np(1 - p) = 25, so N(50,25)N(50, 25).

Exam-style practice questions

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

Edexcel 20206 marksThe heights of a population are modelled by XN(170,64)X \sim N(170, 64) cm. Calculate P(X>180)P(X > 180) and the height exceeded by only 10%10\% of the population.
Show worked answer →

Here μ=170\mu = 170 and σ=64=8\sigma = \sqrt{64} = 8 (B1).

Standardise for P(X>180)P(X > 180) (M1): Z=1801708=1.25Z = \dfrac{180 - 170}{8} = 1.25, so P(Z>1.25)0.1056P(Z > 1.25) \approx 0.1056 (A1).

For the top 10%10\%, find zz with P(Z>z)=0.10P(Z > z) = 0.10, so z1.2816z \approx 1.2816 (M1).

Convert back: X=μ+zσ=170+1.2816×8180.3X = \mu + z\sigma = 170 + 1.2816 \times 8 \approx 180.3 cm (M1, A1).

Markers reward the standard deviation, standardising, the upper-tail probability, the inverse zz, and converting back to XX.

Edexcel 20234 marksThe random variable XN(μ,σ2)X \sim N(\mu, \sigma^2) satisfies P(X<50)=0.20P(X < 50) = 0.20 and P(X<70)=0.90P(X < 70) = 0.90. Determine the values of μ\mu and σ\sigma.
Show worked answer →

Convert each probability to a zz-value (M1): P(Z<z)=0.20P(Z < z) = 0.20 gives z=0.8416z = -0.8416; P(Z<z)=0.90P(Z < z) = 0.90 gives z=1.2816z = 1.2816.

Form two equations using X=μ+zσX = \mu + z\sigma (M1): 50=μ0.8416σ50 = \mu - 0.8416\sigma and 70=μ+1.2816σ70 = \mu + 1.2816\sigma.

Subtract: 20=2.1232σ20 = 2.1232\sigma, so σ9.42\sigma \approx 9.42 (A1).

Substitute back: μ=50+0.8416×9.4257.9\mu = 50 + 0.8416 \times 9.42 \approx 57.9 (A1).

Markers reward the two zz-values, the simultaneous equations, and solving for σ\sigma then μ\mu.

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