Hard A-Level Normal approximation to binomial Questions

Challenging, exam-style A-Level Normal approximation to binomial questions with worked solutions. Stretch yourself on the hardest normal-approximation, continuity-correction, probability, mean-variance problems.

normal-approximationcontinuity-correctionprobabilitymean-variancemodel-choiceconditions
A-Level34 questionsStep-by-step solutions
Question 1
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)
Question 2
8 markschallenging
The random variable X∼B(250, 0.5)X\sim B(250,\ 0.5). Which statement best explains why a normal approximation is appropriate here?
Show worked solution

Worked solution

  1. Identify the parameters

    n=250, p=0.5n=250,\ p=0.5

    Read off n and p from the binomial model.

  2. Check the validity conditions

    np=125, n(1−p)=125np=125,\ n(1-p)=125

    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=250×0.5=125np=250\times 0.5=125

    The mean of the approximating normal is np.

  5. Compute the variance

    np(1−p)=250×0.5×0.5=62.5np(1-p)=250\times 0.5\times 0.5=62.5

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

  6. Find the standard deviation

    σ=62.5=7.9057\sigma=\sqrt{62.5}=7.9057

    The standard deviation is the square root of the variance.

  7. Check the first validity condition

    np=125>5np=125>5

    The expected number of successes should exceed 5.

  8. Check the second validity condition

    n(1−p)=125>5n(1-p)=125>5

    The expected number of failures should also exceed 5.

  9. Write the approximating distribution

    X≈N(125, 62.5)X\approx N(125,\ 62.5)

    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=62.5, σ=7.9057\sigma^2=62.5,\ \sigma=7.9057

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

  14. Confirm p is close to one half

    p=0.5p=0.5

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

  15. Select the correct explanation

    nn is large with np=125np=125 and n(1−p)=125n(1-p)=125 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=125np=125 and n(1−p)=125n(1-p)=125 both exceeding 55, and pp close to 0.50.5
Question 3
8 markschallenging
The random variable X∼B(5000, 0.5)X\sim B(5000,\ 0.5). Which approximation is most appropriate?
Show worked solution

Worked solution

  1. Identify the parameters

    n=5000, p=0.5n=5000,\ p=0.5

    Read off n and p.

  2. Assess the size of n and p

    np=2500, p=0.5np=2500,\ p=0.5

    Large n with p near 0.5 suggests normal; small p suggests Poisson.

  3. Compute the mean

    np=5000×0.5=2500np=5000\times 0.5=2500

    The mean helps decide which approximation fits.

  4. Compute the variance

    np(1−p)=1250np(1-p)=1250

    The variance measures the spread of the distribution.

  5. Recall when the normal approximation applies

    large n, p near 0.5\text{large }n,\ p\text{ near }0.5

    Then N(np,\ np(1-p)) is a good approximation.

  6. Recall when the Poisson approximation applies

    large n, small p, np moderate\text{large }n,\ \text{small }p,\ np\text{ moderate}

    Then \mathrm{Po}(np) is a good approximation.

  7. Compare p with one half

    p=0.5p=0.5

    Closeness to 0.5 favours the normal model.

  8. Consider the size of np

    np=2500np=2500

    A moderate np with small p points to Poisson.

  9. Note the shape of the distribution

    skewed for small p, symmetric near p=0.5\text{skewed for small }p,\ \text{symmetric near }p=0.5

    The shape indicates the more suitable model.

  10. State the normal candidate

    N(2500, 1250)N(2500,\ 1250)

    This is the normal option.

  11. State the Poisson candidate

    Po(2500)\mathrm{Po}(2500)

    This is the Poisson option.

  12. Check that n is large

    n=5000n=5000

    Both approximations require a large number of trials.

  13. Consider the variance-to-mean ratio

    np(1−p)np=1−p=0.5\frac{np(1-p)}{np}=1-p=0.5

    Near 1 (small p) is Poisson-like; well below 1 near p=0.5.

  14. Recall the exact model

    X∼B(5000, 0.5)X\sim B(5000,\ 0.5)

    This is the distribution before approximating.

  15. Select the appropriate model

    Normal approximation N(2500, 1250)N(2500,\ 1250)

    Choose the approximation whose conditions are satisfied.

Answer
Normal approximation N(2500, 1250)N(2500,\ 1250)
Question 4
8 markschallenging
The random variable X∼B(10000, 0.0003)X\sim B(10000,\ 0.0003). Which approximation is most appropriate?
Show worked solution

Worked solution

  1. Identify the parameters

    n=10000, p=0.0003n=10000,\ p=0.0003

    Read off n and p.

  2. Assess the size of n and p

    np=3, p=0.0003np=3,\ p=0.0003

    Large n with p near 0.5 suggests normal; small p suggests Poisson.

  3. Compute the mean

    np=10000×0.0003=3np=10000\times 0.0003=3

    The mean helps decide which approximation fits.

  4. Compute the variance

    np(1−p)=2.9991np(1-p)=2.9991

    The variance measures the spread of the distribution.

  5. Recall when the normal approximation applies

    large n, p near 0.5\text{large }n,\ p\text{ near }0.5

    Then N(np,\ np(1-p)) is a good approximation.

  6. Recall when the Poisson approximation applies

    large n, small p, np moderate\text{large }n,\ \text{small }p,\ np\text{ moderate}

    Then \mathrm{Po}(np) is a good approximation.

  7. Compare p with one half

    p=0.0003p=0.0003

    Closeness to 0.5 favours the normal model.

  8. Consider the size of np

    np=3np=3

    A moderate np with small p points to Poisson.

  9. Note the shape of the distribution

    skewed for small p, symmetric near p=0.5\text{skewed for small }p,\ \text{symmetric near }p=0.5

    The shape indicates the more suitable model.

  10. State the normal candidate

    N(3, 2.9991)N(3,\ 2.9991)

    This is the normal option.

  11. State the Poisson candidate

    Po(3)\mathrm{Po}(3)

    This is the Poisson option.

  12. Check that n is large

    n=10000n=10000

    Both approximations require a large number of trials.

  13. Consider the variance-to-mean ratio

    np(1−p)np=1−p=0.9997\frac{np(1-p)}{np}=1-p=0.9997

    Near 1 (small p) is Poisson-like; well below 1 near p=0.5.

  14. Recall the exact model

    X∼B(10000, 0.0003)X\sim B(10000,\ 0.0003)

    This is the distribution before approximating.

  15. Select the appropriate model

    Poisson approximation Po(3)\mathrm{Po}(3)

    Choose the approximation whose conditions are satisfied.

Answer
Poisson approximation Po(3)\mathrm{Po}(3)
Question 5
8 markschallenging
The random variable X∼B(300, 0.4)X\sim B(300,\ 0.4) is approximated by a normal variable YY. Which continuity-corrected statement correctly represents P(X=120)P(X= 120)?
Show worked solution

Worked solution

  1. State the discrete probability required

    P(X=120)P(X= 120)

    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=300×0.4=120np=300\times 0.4=120

    The mean of the approximating normal is np.

  5. Compute the variance

    np(1−p)=300×0.4×0.6=72np(1-p)=300\times 0.4\times 0.6=72

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

  6. Find the standard deviation

    σ=72=8.4853\sigma=\sqrt{72}=8.4853

    The standard deviation is the square root of the variance.

  7. Check the first validity condition

    np=120>5np=120>5

    The expected number of successes should exceed 5.

  8. Check the second validity condition

    n(1−p)=180>5n(1-p)=180>5

    The expected number of failures should also exceed 5.

  9. Write the approximating distribution

    X≈N(120, 72)X\approx N(120,\ 72)

    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=72, σ=8.4853\sigma^2=72,\ \sigma=8.4853

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

  14. Confirm p is close to one half

    p=0.4p=0.4

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

  15. Select the corrected statement

    P(119.5<Y<120.5)P(119.5< Y< 120.5)

    This applies the continuity correction in the right direction.

Answer
P(119.5<Y<120.5)P(119.5< Y< 120.5)

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