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S4.2

Understand and use the Normal distribution as a model; find probabilities using the Normal distribution; link to histograms, mean, standard deviation, points of inflection and the binomial distribution.

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Normal distribution

Worked answers and methods for S4.2 on Edexcel A-level Maths 9MA0.

Definition

Specification S4.2: understand and use the Normal distribution as a model; find probabilities using the Normal distribution; link to histograms, mean, standard deviation, points of inflection and the binomial distribution.

Explanation

  • Write the model as X∼N(μ,σ2)X\sim\mathrm N(\mu,\sigma^2), with the variance, not the standard deviation, in the bracket.
  • The curve is symmetric about μ\mu with points of inflection at μ±σ\mu\pm\sigma, so a bell-shaped histogram and a mean close to the median support a Normal model.
  • Probabilities come from the calculator, but the working must show the standardisation z=x−μσz=\dfrac{x-\mu}{\sigma} when μ\mu or σ\sigma is unknown: an inverse-Normal value such as 1.28161.2816 for the top 10%10\% gives one linear equation in μ\mu and σ\sigma, and two such equations are solved simultaneously.
  • Keep the same units throughout.
  • A Normal variable is continuous, so there is no continuity correction; the correction, P(X>95.5)P(X>95.5) for “more than 9595”, belongs only to approximating a binomial B(n,p)\mathrm B(n,p) by N(np, np(1−p))\mathrm N(np,\,np(1-p)) when nn is large and pp is near 0.50.5.
A Normal density is symmetric about its mean; its inflection points occur at μ−σ\mu-\sigma and μ+σ\mu+\sigma, and shaded area represents probability.

Worked example

1.

The lifetime, LL hours, of a component is modelled by L∼N(72,82)L\sim\mathrm N(72,8^2). Exactly 10%10\% of components last longer than ll hours. Find the value of ll, giving your answer to 1 decimal place.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.P(L>l)=0.10P(L>l)=0.10, so P(L<l)=0.90P(L<l)=0.90 and the inverse-Normal value is z=1.2816z=1.2816.
  2. 2.Standardise: l−728=1.2816\dfrac{l-72}{8}=1.2816.
  3. 3.So l=72+8×1.2816=82.25…=82.3l=72+8\times1.2816=82.25\ldots=82.3 hours.

Answer: l=82.3l=82.3

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1.P(L<l)=0.9P(L<l)=0.9 or z=1.2816z=1.2816 seenM1
Standardises: l−728=\dfrac{l-72}{8}= their zzM1
awrt 82.382.3A1
Notes
M1: awrt 1.28
M1: Must be a zz-value between 1 and 1.5

(3 marks)

Common mistakes

  • Don't use the variance as the denominator when standardising instead of the standard deviation.
  • Don't use the lower-tail quantile for a value exceeded by a stated percentage.

Mark-scheme wording to use

  • “1st M1 for attempting normal with mean = 100 or ft their answer to (c) … 1st A1 for correctly labelled standard deviation allow √(250/3) = √83.3... = 9.1(28….) or correctly labelled variance. Implied by N( μ, 250/3 ) or correct answer. … 2nd M1 for attempt at continuity correction i.e. sight of 95 ± 0.5 … 2nd A1 for awrt 0.689 [Answer only 4/4]” — Pearson 9MA0, June 2024 Paper 31, Q1(d)
  • “1st M1 for standardising 16.3 and setting equal to z value where 0.5 < |z| < 0.6 … 2nd M1 for standardising 29 and setting equal to z value where 1 < |z| < 1.5 … 3rd M1 dep on 1st or 2nd M1 for solving their two linear eq'ns – reach an eq'n in one variable … 1st A1 for σ = awrt 7.03 (but see 3rd case below)” — Pearson 9MA0, June 2024 Paper 31, Q5(d)
  • “B1 for using the correct model to obtain z awrt 0.842. M1 for standardising to an appropriate z-value between 0.5 and 1. A1 for sigma awrt 23.8 hours” — Pearson 9MA0, Specimen Paper 31, Q5(b)

What the examiner wants

A Normal used as a continuous model in the right units. The June 2024 Paper 31 report says errors in finding P(T<300)P(T<300) came from applying a continuity correction, P(T<299.5)P(T<299.5), or from not converting minutes to seconds, while the strongest answers to the unknown-parameter part wrote down a pair of linear equations in μ\mu and σ\sigma with suitable zz-values.

Exam tip

Convert an exceedance percentage to the corresponding lower-tail probability before standardising or using an inverse Normal calculation.

Worked practice

Q1
Tier 1 · Easy

1.

A random variable has distribution X∼N⁡(50,62)X\sim\operatorname{N}(50,6^2). Find P(X<56)P(X<56) to 44 decimal places.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionSchemeMarks
1
  • 0.84130.8413
2
Notes
Standardise: z=(56−50)/6=1z=(56-50)/6=1. Therefore P(X<56)=P(Z<1)=0.841344…P(X<56)=P(Z<1)=0.841344\ldots, which is 0.84130.8413.

(2 marks)

Q2
Tier 2 · Standard

2.

A random variable XX is modelled by X∼N⁡(64,42)X\sim\operatorname{N}(64,4^2). Find P(58<X<70)P(58<X<70) to 44 decimal places.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
QuestionSchemeMarks
2
  • 0.86640.8664
4
Notes
Standardising gives z=(58−64)/4=−1.5z=(58-64)/4=-1.5 and z=(70−64)/4=1.5z=(70-64)/4=1.5. Hence P(58<X<70)=P(−1.5<Z<1.5)=Φ(1.5)−Φ(−1.5)=0.93319…−0.06681…=0.8664P(58<X<70)=P(-1.5<Z<1.5)=\Phi(1.5)-\Phi(-1.5)=0.93319\ldots-0.06681\ldots=0.8664.

(4 marks)

Q3
Tier 3 · Hard

3.

Let X∼B⁡(200,0.35)X\sim\operatorname{B}(200,0.35). Use a Normal approximation with a continuity correction to estimate P(X≥82)P(X\geq82), giving your answer to 44 decimal places.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionSchemeMarks
3
  • 0.04410.0441
5
Notes
Use Y∼N⁡(np,np(1−p))=N⁡(70,45.5)Y\sim\operatorname{N}(np,np(1-p))=\operatorname{N}(70,45.5). The continuity correction gives P(X≥82)≈P(Y>81.5)P(X\geq82)\approx P(Y>81.5). Thus z=(81.5−70)/45.5=1.7049…z=(81.5-70)/\sqrt{45.5}=1.7049\ldots, so the upper-tail probability is 1−Φ(1.7049…)=0.04411-\Phi(1.7049\ldots)=0.0441.

(5 marks)

Q4
Tier 1 · Easy

4.

The random variable TT has distribution T∼N⁡(72,92)T\sim\operatorname{N}(72,9^2). State the two values of TT at which the Normal density curve has points of inflection.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionSchemeMarks
4
  • 6363 and 8181
2
Notes
A Normal density has points of inflection at μ−σ\mu-\sigma and μ+σ\mu+\sigma. Here these are 72−9=6372-9=63 and 72+9=8172+9=81.

(2 marks)

Q5
Tier 2 · Standard

5.

A variable XX is modelled by X∼N⁡(μ,52)X\sim\operatorname{N}(\mu,5^2). Given that P(X<42)=0.10P(X<42)=0.10 and that the 0.100.10 standard Normal quantile is −1.2816-1.2816, find μ\mu to 33 significant figures.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
QuestionSchemeMarks
5
  • μ=48.4\mu=48.4 to 33 significant figures
4
Notes
Standardising the tenth percentile gives (42−μ)/5=−1.2816(42-\mu)/5=-1.2816. Hence 42−μ=−6.40842-\mu=-6.408, so μ=48.408\mu=48.408, which is 48.448.4 to 33 significant figures.

(4 marks)

Q6
Tier 3 · Hard

6.

A variable XX is modelled by X∼N⁡(30,σ2)X\sim\operatorname{N}(30,\sigma^2). Given that P(26<X<34)=0.80P(26<X<34)=0.80, find σ\sigma to 33 significant figures. Hence find P(X>36∣X>32)P(X>36\mid X>32) to 44 decimal places. Use the unrounded value of σ\sigma in your working. You may use the 0.900.90 standard Normal quantile 1.28161.2816.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
QuestionSchemeMarks
6
  • σ=3.12\sigma=3.12
  • P(X>36∣X>32)=0.1046P(X>36\mid X>32)=0.1046
6
Notes
By symmetry, P(X<34)=0.90P(X<34)=0.90, so 4/σ=1.28164/\sigma=1.2816 and σ=3.1211…\sigma=3.1211\ldots. Since X>36X>36 implies X>32X>32, the conditional probability is P(X>36)/P(X>32)P(X>36)/P(X>32). The corresponding standardised values are 1.92241.9224 and 0.64080.6408, giving tail probabilities 0.0272777…0.0272777\ldots and 0.260826…0.260826\ldots. Their ratio is 0.104582…0.104582\ldots, giving 0.10460.1046.

(6 marks)

Q7
Tier 2 · Standard

7.

A variable XX is modelled by X∼N⁡(μ,σ2)X\sim\operatorname{N}(\mu,\sigma^2). Given P(X<42)=0.20P(X<42)=0.20 and P(X<58)=0.80P(X<58)=0.80, find μ\mu and σ\sigma. Give σ\sigma to 33 significant figures. You may use the standard Normal quantiles −0.8416-0.8416 and 0.84160.8416.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
QuestionSchemeMarks
7
  • μ=50\mu=50
  • σ=9.51\sigma=9.51 to 33 significant figures
5
Notes
Standardising the two percentiles gives (42−μ)/σ=−0.8416(42-\mu)/\sigma=-0.8416 and (58−μ)/σ=0.8416(58-\mu)/\sigma=0.8416. Adding the equations, or using symmetry, gives μ=50\mu=50. Then 8/σ=0.84168/\sigma=0.8416, so σ=8/0.8416=9.50570…\sigma=8/0.8416=9.50570\ldots, giving 9.519.51 to 33 significant figures.

(5 marks)

Q8
Tier 3 · Hard

8.

A process measurement XX is modelled by a Normal distribution whose density curve has points of inflection at 492492 and 508508. Measurements outside the interval 484≤X≤516484\leq X\leq516 are rejected. Find the probability that a measurement is accepted and the expected number rejected from 12001200 measurements. Given that a measurement is accepted, find the probability that it exceeds 505505. Give probabilities to 44 decimal places and the expected number to the nearest integer. Use unrounded values in your working.

(7)

(Total for Question 8 is 7 marks)

Mark scheme

Mark scheme for question 8
QuestionSchemeMarks
8
  • μ=500\mu=500 and σ=8\sigma=8
  • P(484≤X≤516)=0.9545P(484\leq X\leq516)=0.9545
  • Expected number rejected =55=55
  • P(X>505∣484≤X≤516)=0.2548P(X>505\mid 484\leq X\leq516)=0.2548
7
Notes
The inflection points are μ−σ\mu-\sigma and μ+σ\mu+\sigma, so their midpoint gives μ=500\mu=500 and half their separation gives σ=8\sigma=8. The acceptance bounds standardise to −2-2 and 22, hence the acceptance probability is Φ(2)−Φ(−2)=0.954499…\Phi(2)-\Phi(-2)=0.954499\ldots. The expected rejected count is 1200[1−0.954499…]=54.6003…1200[1-0.954499\ldots]=54.6003\ldots, giving 5555. Within the acceptance interval, exceeding 505505 means 505<X≤516505<X\leq516; the lower standardised value is 0.6250.625. Thus the conditional probability is [Φ(2)−Φ(0.625)]/[Φ(2)−Φ(−2)]=0.254830…=0.2548[\Phi(2)-\Phi(0.625)]/[\Phi(2)-\Phi(-2)]=0.254830\ldots=0.2548.

(7 marks)

Q9
Tier 3 · Hard

9.

A variable XX is modelled by X∼N⁡(μ,σ2)X\sim\operatorname{N}(\mu,\sigma^2). Given that P(X<44)=P(X>68)P(X<44)=P(X>68) and P(50<X<62)=0.60P(50<X<62)=0.60, find μ\mu and σ\sigma. You may use the 0.800.80 standard Normal quantile 0.84160.8416. Hence find the two values of XX at which the density curve has points of inflection, giving them to 33 significant figures. A histogram represents 900900 such measurements, with bar area equal to frequency. Estimate the total frequency between the points of inflection, and explain why agreement with this frequency alone would not establish that a Normal model is suitable.

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
QuestionSchemeMarks
9
  • μ=56\mu=56 and σ=7.13\sigma=7.13 to 33 significant figures
  • The points of inflection are at X=48.9X=48.9 and X=63.1X=63.1 to 33 significant figures.
  • Estimated frequency between the points of inflection =614=614 to the nearest whole number.
  • Matching one central histogram area does not check the model's symmetry, unimodality, tails or possible outliers, so the full histogram and context must also be considered.
7
Notes
Equal opposite-tail probabilities in a Normal distribution place the mean halfway between 4444 and 6868, so μ=56\mu=56. The interval 50<X<6250<X<62 is therefore μ−6<X<μ+6\mu-6<X<\mu+6. Its central probability is 0.600.60, leaving 0.200.20 in each tail, so 6/σ=0.84166/\sigma=0.8416 and σ=6/0.8416=7.12928…\sigma=6/0.8416=7.12928\ldots. The points of inflection are μ±σ=48.8707…\mu\pm\sigma=48.8707\ldots and 63.1293…63.1293\ldots. The probability between them is P(−1<Z<1)=0.682689…P(-1<Z<1)=0.682689\ldots, so the histogram area represents an estimated frequency 900(0.682689…)=614.420…900(0.682689\ldots)=614.420\ldots, giving 614614 to the nearest whole number. Agreement over only this interval is not a complete model check.

(7 marks)

Q10
Tier 3 · Hard

10.

A variable XX is modelled by X∼N⁡(μ,σ2)X\sim\operatorname{N}(\mu,\sigma^2). Given that P(X<40)=0.10P(X<40)=0.10 and P(X<55)=0.70P(X<55)=0.70, find μ\mu and σ\sigma to 33 significant figures. Hence find P(45<X<60)P(45<X<60) to 44 decimal places. Use unrounded values in your working. You may use the standard Normal quantiles −1.2816-1.2816 and 0.52440.5244.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
QuestionSchemeMarks
10
  • μ=50.6\mu=50.6 and σ=8.31\sigma=8.31 to 33 significant figures
  • P(45<X<60)=0.6216P(45<X<60)=0.6216
7
Notes
Standardising gives (40−μ)/σ=−1.2816(40-\mu)/\sigma=-1.2816 and (55−μ)/σ=0.5244(55-\mu)/\sigma=0.5244. Subtraction gives 15/σ=1.806015/\sigma=1.8060, so σ=8.305647…\sigma=8.305647\ldots and μ=40+1.2816σ=50.644518…\mu=40+1.2816\sigma=50.644518\ldots. Therefore P(45<X<60)=Φ((60−μ)/σ)−Φ((45−μ)/σ)=0.621622…=0.6216P(45<X<60)=\Phi((60-\mu)/\sigma)-\Phi((45-\mu)/\sigma)=0.621622\ldots=0.6216.

(7 marks)

Verified exam appearances

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