Hard A-Level The normal distribution Questions

Challenging, exam-style A-Level The normal distribution questions with worked solutions. Stretch yourself on the hardest normal-distribution, probability, inverse-normal, find-mean problems.

normal-distributionprobabilityinverse-normalfind-meanfind-sdfind-mean-sd
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
The random variable X∼N(75, 92)X\sim N(75,\ 9^{2}). Find P(X>60)P(X>60), giving your answer to 4 decimal places.
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Worked solution

  1. Write down the distribution and required probability

    X∼N(75, 92),P(X>60)X\sim N(75,\ 9^{2}),\quad P(X>60)

    We are given a normal distribution and asked for a probability.

  2. Standardise the boundary value(s)

    Z=X−μσ=60−759=−1.667Z=\frac{X-\mu}{\sigma}=\frac{60-75}{9}=-1.667

    Convert X to the standard normal variable Z.

  3. Rewrite the probability in terms of Z

    P(X>60)=P(Z>−1.667)P(X>60)=P(Z>-1.667)

    Standardising expresses the probability using the standard normal.

  4. Evaluate using the standard normal distribution

    P(X>60)=P(Z>−1.667)=1−0.0478=0.9522P(X>60)=P(Z>-1.667)=1-0.0478=0.9522

    The cumulative distribution of Z gives the required area.

  5. Sketch the normal curve

    bell-shaped, symmetric about μ=75\text{bell-shaped, symmetric about }\mu=75

    A sketch helps identify the area required.

  6. State the standard deviation

    σ=9\sigma=9

    The spread of the distribution is set by the standard deviation.

  7. Recall the standardisation formula

    Z=X−μσZ=\frac{X-\mu}{\sigma}

    Standardising converts X to the standard normal variable Z.

  8. State the standard normal distribution

    Z∼N(0,1)Z\sim N(0,1)

    The standardised variable has mean 0 and standard deviation 1.

  9. Recall the total probability

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x)\,dx=1

    The area under the whole curve is 1.

  10. Use the symmetry of the curve

    P(Z<0)=0.5P(Z<0)=0.5

    The standard normal is symmetric about 0.

  11. Recall the complement rule

    P(Z>z)=1−P(Z<z)P(Z>z)=1-P(Z<z)

    The two tails and the centre sum to 1.

  12. Interpret the probability as an area

    P=area under the density curveP=\text{area under the density curve}

    Probabilities correspond to areas under the density function.

  13. Note the mean, median and mode coincide

    μ=median=mode=75\mu=\text{median}=\text{mode}=75

    The symmetry of the curve makes these equal.

  14. Locate the points of inflection

    x=μ±σ=66, 84x=\mu\pm\sigma=66,\ 84

    The curve changes concavity one standard deviation from the mean.

  15. State the probability

    P(X>60)≈0.9522P(X>60)\approx 0.9522

    This is the required probability to 4 decimal places.

Answer
0.95220.9522
Question 2
8 markschallenging
A normally distributed variable XX has mean μ=0\mu=0. Given that P(X<2)=0.9000P(X<2)=0.9000, find the standard deviation σ\sigma, to 4 significant figures.
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Worked solution

  1. Write down what is known

    μ=0,P(X<2)=0.9000\mu=0,\quad P(X<2)=0.9000

    The mean is known but the standard deviation is not.

  2. Find the required z-value

    z=1.282z=1.282

    The inverse normal gives the z-value for this probability.

  3. Set up the standardisation equation

    2−0σ=1.282\frac{2-0}{\sigma}=1.282

    Standardising the boundary value gives an equation in sigma.

  4. Solve for the standard deviation

    σ=2−01.282=1.561\sigma=\frac{2-0}{1.282}=1.561

    Rearranging the equation gives the standard deviation.

  5. Sketch the normal curve

    bell-shaped, symmetric about μ=0\text{bell-shaped, symmetric about }\mu=0

    A sketch helps identify the area required.

  6. State the standard deviation

    σ=2\sigma=2

    The spread of the distribution is set by the standard deviation.

  7. Recall the standardisation formula

    Z=X−μσZ=\frac{X-\mu}{\sigma}

    Standardising converts X to the standard normal variable Z.

  8. State the standard normal distribution

    Z∼N(0,1)Z\sim N(0,1)

    The standardised variable has mean 0 and standard deviation 1.

  9. Recall the total probability

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x)\,dx=1

    The area under the whole curve is 1.

  10. Use the symmetry of the curve

    P(Z<0)=0.5P(Z<0)=0.5

    The standard normal is symmetric about 0.

  11. Recall the complement rule

    P(Z>z)=1−P(Z<z)P(Z>z)=1-P(Z<z)

    The two tails and the centre sum to 1.

  12. Interpret the probability as an area

    P=area under the density curveP=\text{area under the density curve}

    Probabilities correspond to areas under the density function.

  13. Note the mean, median and mode coincide

    μ=median=mode=0\mu=\text{median}=\text{mode}=0

    The symmetry of the curve makes these equal.

  14. Locate the points of inflection

    x=μ±σ=−2, 2x=\mu\pm\sigma=-2,\ 2

    The curve changes concavity one standard deviation from the mean.

  15. State the standard deviation

    σ≈1.561\sigma\approx 1.561

    This is the required standard deviation to 4 significant figures.

Answer
1.5611.561
Question 3
8 markschallenging
The random variable X∼N(25, 42)X\sim N(25,\ 4^{2}). Find the value of aa such that P(X<a)=0.9950P(X<a)=0.9950, giving your answer to 4 significant figures.
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Worked solution

  1. Write the condition on the required value

    X∼N(25, 42),P(X<a)=0.9950X\sim N(25,\ 4^{2}),\quad P(X<a)=0.9950

    We must find the value a with the given probability.

  2. Find the corresponding z-value

    P(Z<z)=0.9950⇒z=2.576P(Z<z)=0.9950\Rightarrow z=2.576

    The inverse normal gives the z-value for the required probability.

  3. Unstandardise to find a

    a=μ+σz=25+4×2.576=35.3a=\mu+\sigma z=25+4\times 2.576=35.3

    Reversing the standardisation returns to the original scale.

  4. Sketch the normal curve

    bell-shaped, symmetric about μ=25\text{bell-shaped, symmetric about }\mu=25

    A sketch helps identify the area required.

  5. State the standard deviation

    σ=4\sigma=4

    The spread of the distribution is set by the standard deviation.

  6. Recall the standardisation formula

    Z=X−μσZ=\frac{X-\mu}{\sigma}

    Standardising converts X to the standard normal variable Z.

  7. State the standard normal distribution

    Z∼N(0,1)Z\sim N(0,1)

    The standardised variable has mean 0 and standard deviation 1.

  8. Recall the total probability

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x)\,dx=1

    The area under the whole curve is 1.

  9. Use the symmetry of the curve

    P(Z<0)=0.5P(Z<0)=0.5

    The standard normal is symmetric about 0.

  10. Recall the complement rule

    P(Z>z)=1−P(Z<z)P(Z>z)=1-P(Z<z)

    The two tails and the centre sum to 1.

  11. Interpret the probability as an area

    P=area under the density curveP=\text{area under the density curve}

    Probabilities correspond to areas under the density function.

  12. Note the mean, median and mode coincide

    μ=median=mode=25\mu=\text{median}=\text{mode}=25

    The symmetry of the curve makes these equal.

  13. Locate the points of inflection

    x=μ±σ=21, 29x=\mu\pm\sigma=21,\ 29

    The curve changes concavity one standard deviation from the mean.

  14. Keep full accuracy until the end

    round only at the final step\text{round only at the final step}

    Rounding early would introduce error into the answer.

  15. State the value of a

    a≈35.3a\approx 35.3

    This is the required value to 4 significant figures.

Answer
35.335.3
Question 4
8 markschallenging
The inverse normal function is used to:
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Worked solution

  1. Recall the properties of the normal distribution

    X∼N(μ,σ2)X\sim N(\mu,\sigma^{2})

    List the standard facts about the bell-shaped curve.

  2. Compare each option with these properties

    test each statement\text{test each statement}

    Only the statement consistent with the properties can be correct.

  3. Sketch the normal curve

    bell-shaped and symmetric\text{bell-shaped and symmetric}

    A sketch shows the key features of the distribution.

  4. Recall the standardisation

    Z=X−μσZ=\frac{X-\mu}{\sigma}

    Standardising uses the mean and standard deviation.

  5. State the standard normal

    Z∼N(0,1)Z\sim N(0,1)

    The standardised variable has this distribution.

  6. Total area under the curve

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x)\,dx=1

    All the probability sums to 1.

  7. Symmetry about the mean

    P(X<μ)=P(X>μ)=0.5P(X<\mu)=P(X>\mu)=0.5

    Each half of the curve carries half of the area.

  8. Mean, median and mode

    μ=median=mode\mu=\text{median}=\text{mode}

    Symmetry makes the three averages coincide.

  9. Points of inflection

    x=μ±σx=\mu\pm\sigma

    The curve changes concavity at these points.

  10. Empirical rule within one s.d.

    P(μ−σ<X<μ+σ)≈0.68P(\mu-\sigma<X<\mu+\sigma)\approx0.68

    A useful benchmark for the spread.

  11. Empirical rule within two s.d.

    P(μ−2σ<X<μ+2σ)≈0.95P(\mu-2\sigma<X<\mu+2\sigma)\approx0.95

    A useful benchmark for the spread.

  12. Empirical rule within three s.d.

    P(μ−3σ<X<μ+3σ)≈0.997P(\mu-3\sigma<X<\mu+3\sigma)\approx0.997

    Almost all of the data lie within three standard deviations.

  13. Eliminate the impossible options

    discard invalid claims\text{discard invalid claims}

    Rule out statements that break the standard facts.

  14. Confirm with the definition

    N(μ,σ2) notationN(\mu,\sigma^{2})\text{ notation}

    The two parameters are the mean and the variance.

  15. Select the correct statement

    x=μ+σ Φ−1(p)x=\mu+\sigma\,\Phi^{-1}(p)

    This matches a known property of the normal distribution.

Answer
find a value of x given a cumulative probability.
Question 5
8 markschallenging
For a continuous variable XX, P(a<X<b)P(a<X<b) can be written as:
Show worked solution

Worked solution

  1. Recall the properties of the normal distribution

    X∼N(μ,σ2)X\sim N(\mu,\sigma^{2})

    List the standard facts about the bell-shaped curve.

  2. Compare each option with these properties

    test each statement\text{test each statement}

    Only the statement consistent with the properties can be correct.

  3. Sketch the normal curve

    bell-shaped and symmetric\text{bell-shaped and symmetric}

    A sketch shows the key features of the distribution.

  4. Recall the standardisation

    Z=X−μσZ=\frac{X-\mu}{\sigma}

    Standardising uses the mean and standard deviation.

  5. State the standard normal

    Z∼N(0,1)Z\sim N(0,1)

    The standardised variable has this distribution.

  6. Total area under the curve

    ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x)\,dx=1

    All the probability sums to 1.

  7. Symmetry about the mean

    P(X<μ)=P(X>μ)=0.5P(X<\mu)=P(X>\mu)=0.5

    Each half of the curve carries half of the area.

  8. Mean, median and mode

    μ=median=mode\mu=\text{median}=\text{mode}

    Symmetry makes the three averages coincide.

  9. Points of inflection

    x=μ±σx=\mu\pm\sigma

    The curve changes concavity at these points.

  10. Empirical rule within one s.d.

    P(μ−σ<X<μ+σ)≈0.68P(\mu-\sigma<X<\mu+\sigma)\approx0.68

    A useful benchmark for the spread.

  11. Empirical rule within two s.d.

    P(μ−2σ<X<μ+2σ)≈0.95P(\mu-2\sigma<X<\mu+2\sigma)\approx0.95

    A useful benchmark for the spread.

  12. Empirical rule within three s.d.

    P(μ−3σ<X<μ+3σ)≈0.997P(\mu-3\sigma<X<\mu+3\sigma)\approx0.997

    Almost all of the data lie within three standard deviations.

  13. Eliminate the impossible options

    discard invalid claims\text{discard invalid claims}

    Rule out statements that break the standard facts.

  14. Confirm with the definition

    N(μ,σ2) notationN(\mu,\sigma^{2})\text{ notation}

    The two parameters are the mean and the variance.

  15. Select the correct statement

    P(a<X<b)=P(X<b)−P(X<a)P(a<X<b)=P(X<b)-P(X<a)

    This matches a known property of the normal distribution.

Answer
P(X<b)−P(X<a)P(X<b)-P(X<a)

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