A-Level The normal distribution Practice Questions

Free A-Level The normal distribution practice questions with full step-by-step worked solutions. Covers normal-distribution, probability, standardising, inverse-normal. Practise exam-style problems and check your method.

normal-distributionprobabilitystandardisinginverse-normalquantileproperties
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable X∼N(50, 82)X\sim N(50,\ 8^{2}). Find P(X<58)P(X<58), giving your answer to 4 decimal places.
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Worked solution

  1. Write down the distribution and required probability

    X∼N(50, 82),P(X<58)X\sim N(50,\ 8^{2}),\quad P(X<58)

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

  2. Standardise and use the standard normal

    Z=X−μσ=58−508=1; P(X<58)=P(Z<1)=0.8413Z=\frac{X-\mu}{\sigma}=\frac{58-50}{8}=1;\ P(X<58)=P(Z<1)=0.8413

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X<58)≈0.8413P(X<58)\approx 0.8413

    This is the required probability to 4 decimal places.

Answer
0.84130.8413
Question 2
2 markseasy
The random variable X∼N(40, 62)X\sim N(40,\ 6^{2}). Find P(X>31)P(X>31), giving your answer to 4 decimal places.
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Worked solution

  1. Write down the distribution and required probability

    X∼N(40, 62),P(X>31)X\sim N(40,\ 6^{2}),\quad P(X>31)

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

  2. Standardise and use the standard normal

    Z=X−μσ=31−406=−1.5; P(X>31)=P(Z>−1.5)=1−0.0668=0.9332Z=\frac{X-\mu}{\sigma}=\frac{31-40}{6}=-1.5;\ P(X>31)=P(Z>-1.5)=1-0.0668=0.9332

    Convert to Z, then read the probability from the standard normal.

  3. State the probability

    P(X>31)≈0.9332P(X>31)\approx 0.9332

    This is the required probability to 4 decimal places.

Answer
0.93320.9332
Question 3
3 marksintermediate
The random variable X∼N(15, 22)X\sim N(15,\ 2^{2}). Find P(X>12)P(X>12), giving your answer to 4 decimal places.
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Worked solution

  1. Write down the distribution and required probability

    X∼N(15, 22),P(X>12)X\sim N(15,\ 2^{2}),\quad P(X>12)

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

  2. Standardise the boundary value(s)

    Z=X−μσ=12−152=−1.5Z=\frac{X-\mu}{\sigma}=\frac{12-15}{2}=-1.5

    Convert X to the standard normal variable Z.

  3. Rewrite the probability in terms of Z

    P(X>12)=P(Z>−1.5)P(X>12)=P(Z>-1.5)

    Standardising expresses the probability using the standard normal.

  4. Evaluate using the standard normal distribution

    P(X>12)=P(Z>−1.5)=1−0.0668=0.9332P(X>12)=P(Z>-1.5)=1-0.0668=0.9332

    The cumulative distribution of Z gives the required area.

  5. Sketch the normal curve

    bell-shaped, symmetric about μ=15\text{bell-shaped, symmetric about }\mu=15

    A sketch helps identify the area required.

  6. State the probability

    P(X>12)≈0.9332P(X>12)\approx 0.9332

    This is the required probability to 4 decimal places.

Answer
0.93320.9332
Question 4
5 markshard
The random variable X∼N(1000, 502)X\sim N(1000,\ 50^{2}). Find the value of aa such that P(X<a)=0.2500P(X<a)=0.2500, giving your answer to 4 significant figures.
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Worked solution

  1. Write the condition on the required value

    X∼N(1000, 502),P(X<a)=0.2500X\sim N(1000,\ 50^{2}),\quad P(X<a)=0.2500

    We must find the value a with the given probability.

  2. Find the corresponding z-value

    P(Z<z)=0.2500⇒z=−0.6745P(Z<z)=0.2500\Rightarrow z=-0.6745

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

  3. Unstandardise to find a

    a=μ+σz=1000+50×−0.6745=966.3a=\mu+\sigma z=1000+50\times -0.6745=966.3

    Reversing the standardisation returns to the original scale.

  4. Sketch the normal curve

    bell-shaped, symmetric about μ=1000\text{bell-shaped, symmetric about }\mu=1000

    A sketch helps identify the area required.

  5. State the standard deviation

    σ=50\sigma=50

    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. State the value of a

    a≈966.3a\approx 966.3

    This is the required value to 4 significant figures.

Answer
966.3966.3
Question 5
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.
Show worked solution

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

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