A-Level Normal approximation to binomial Practice Questions

Free A-Level Normal approximation to binomial practice questions with full step-by-step worked solutions. Covers normal-approximation, mean-variance, conditions, model-choice. Practise exam-style problems and check your method.

normal-approximationmean-varianceconditionsmodel-choicecontinuity-correctionprobability
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable X∼B(40, 0.5)X\sim B(40,\ 0.5) is approximated by a normal distribution. Find the mean of that normal distribution.
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Worked solution

  1. Identify the binomial parameters

    n=40, p=0.5n=40,\ p=0.5

    Read off the number of trials and the success probability.

  2. Apply the binomial mean and variance

    μ=np=20, σ2=np(1−p)=10\mu=np=20,\ \sigma^2=np(1-p)=10

    The normal approximation matches these two moments.

  3. State the mean

    μ=20\mu=20

    This is the mean of the approximating normal distribution.

Answer
2020
Question 2
2 markseasy
The random variable X∼B(60, 0.5)X\sim B(60,\ 0.5) is approximated by a normal variable YY. Which continuity-corrected statement correctly represents P(X≤35)P(X\le 35)?
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Worked solution

  1. State the discrete probability required

    P(X≤35)P(X\le 35)

    Identify the exact binomial probability to be approximated.

  2. Recall the continuity correction

    replace each integer boundary by a value ±0.5\text{replace each integer boundary by a value }\pm 0.5

    A discrete integer maps to a continuous interval of width 1.

  3. Select the corrected statement

    P(Y<35.5)P(Y< 35.5)

    This applies the continuity correction in the right direction.

Answer
P(Y<35.5)P(Y< 35.5)
Question 3
3 marksintermediate
The random variable X∼B(80, 0.45)X\sim B(80,\ 0.45). Which statement best explains why a normal approximation is appropriate here?
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Worked solution

  1. Identify the parameters

    n=80, p=0.45n=80,\ p=0.45

    Read off n and p from the binomial model.

  2. Check the validity conditions

    np=36, n(1−p)=44np=36,\ n(1-p)=44

    Both should comfortably exceed 5 for a good approximation.

  3. Recall the binomial mean and variance

    E[X]=np, Var(X)=np(1−p)E[X]=np,\ \mathrm{Var}(X)=np(1-p)

    These standard results supply the normal parameters.

  4. Substitute the given values into the mean

    np=80×0.45=36np=80\times 0.45=36

    The mean of the approximating normal is np.

  5. Compute the variance

    np(1−p)=80×0.45×0.55=19.8np(1-p)=80\times 0.45\times 0.55=19.8

    The variance of the approximating normal is np(1-p).

  6. Select the correct explanation

    nn is large with np=36np=36 and n(1−p)=44n(1-p)=44 both exceeding 55, and pp close to 0.50.5

    The approximation needs a large n and p not too extreme.

Answer
nn is large with np=36np=36 and n(1−p)=44n(1-p)=44 both exceeding 55, and pp close to 0.50.5
Question 4
5 markshard
The random variable X∼B(90, 0.5)X\sim B(90,\ 0.5) is approximated by a normal variable YY. Which continuity-corrected statement correctly represents P(X<40)P(X< 40)?
Show worked solution

Worked solution

  1. State the discrete probability required

    P(X<40)P(X< 40)

    Identify the exact binomial probability to be approximated.

  2. Recall the continuity correction

    replace each integer boundary by a value ±0.5\text{replace each integer boundary by a value }\pm 0.5

    A discrete integer maps to a continuous interval of width 1.

  3. Recall the binomial mean and variance

    E[X]=np, Var(X)=np(1−p)E[X]=np,\ \mathrm{Var}(X)=np(1-p)

    These standard results supply the normal parameters.

  4. Substitute the given values into the mean

    np=90×0.5=45np=90\times 0.5=45

    The mean of the approximating normal is np.

  5. Compute the variance

    np(1−p)=90×0.5×0.5=22.5np(1-p)=90\times 0.5\times 0.5=22.5

    The variance of the approximating normal is np(1-p).

  6. Find the standard deviation

    σ=22.5=4.7434\sigma=\sqrt{22.5}=4.7434

    The standard deviation is the square root of the variance.

  7. Check the first validity condition

    np=45>5np=45>5

    The expected number of successes should exceed 5.

  8. Check the second validity condition

    n(1−p)=45>5n(1-p)=45>5

    The expected number of failures should also exceed 5.

  9. Write the approximating distribution

    X≈N(45, 22.5)X\approx N(45,\ 22.5)

    Match the normal mean and variance to the binomial.

  10. Select the corrected statement

    P(Y<39.5)P(Y< 39.5)

    This applies the continuity correction in the right direction.

Answer
P(Y<39.5)P(Y< 39.5)
Question 5
8 markschallenging
The random variable X∼B(250, 0.6)X\sim B(250,\ 0.6) is approximated by a normal variable YY. Which continuity-corrected statement correctly represents P(X>160)P(X> 160)?
Show worked solution

Worked solution

  1. State the discrete probability required

    P(X>160)P(X> 160)

    Identify the exact binomial probability to be approximated.

  2. Recall the continuity correction

    replace each integer boundary by a value ±0.5\text{replace each integer boundary by a value }\pm 0.5

    A discrete integer maps to a continuous interval of width 1.

  3. Recall the binomial mean and variance

    E[X]=np, Var(X)=np(1−p)E[X]=np,\ \mathrm{Var}(X)=np(1-p)

    These standard results supply the normal parameters.

  4. Substitute the given values into the mean

    np=250×0.6=150np=250\times 0.6=150

    The mean of the approximating normal is np.

  5. Compute the variance

    np(1−p)=250×0.6×0.4=60np(1-p)=250\times 0.6\times 0.4=60

    The variance of the approximating normal is np(1-p).

  6. Find the standard deviation

    σ=60=7.746\sigma=\sqrt{60}=7.746

    The standard deviation is the square root of the variance.

  7. Check the first validity condition

    np=150>5np=150>5

    The expected number of successes should exceed 5.

  8. Check the second validity condition

    n(1−p)=100>5n(1-p)=100>5

    The expected number of failures should also exceed 5.

  9. Write the approximating distribution

    X≈N(150, 60)X\approx N(150,\ 60)

    Match the normal mean and variance to the binomial.

  10. State the continuity-correction principle

    shift each discrete boundary by ±0.5\text{shift each discrete boundary by }\pm 0.5

    A discrete value corresponds to a continuous interval of width 1.

  11. Introduce the standardising transform

    Z=X−μσZ=\frac{X-\mu}{\sigma}

    Standardising converts to the standard normal variable.

  12. Recall the standard normal distribution

    Z∼N(0,1)Z\sim N(0,1)

    Probabilities are read from the standard normal CDF.

  13. Distinguish variance from standard deviation

    σ2=60, σ=7.746\sigma^2=60,\ \sigma=7.746

    The variance is np(1-p); its square root is the standard deviation.

  14. Confirm p is close to one half

    p=0.6p=0.6

    The closer p is to 0.5, the more symmetric the distribution.

  15. Select the corrected statement

    P(Y>160.5)P(Y> 160.5)

    This applies the continuity correction in the right direction.

Answer
P(Y>160.5)P(Y> 160.5)

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