Further Maths Central limit theorem Practice Questions

Free Further Maths Central limit theorem practice questions with full step-by-step worked solutions. Covers central-limit-theorem, standard-error, sampling-distribution-of-the-mean, distribution-of-a-sum. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The mass of a randomly selected ball bearing, in grams, is modelled by a random variable XX with mean μ=52.4\mu=52.4 and standard deviation σ=3.2\sigma=3.2. The distribution of XX is not known to be normal. A random sample of n=64n=64 ball bearings is taken. Let Xˉ\bar{X} denote the sample mean mass. Using the central limit theorem, state the standard deviation of the sampling distribution of Xˉ\bar{X} (the standard error of the mean), giving your answer to 44 decimal places.
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Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(52.4,0.16)\bar{X}\approx N\left(52.4,0.16\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Divide the population standard deviation by n\sqrt{n}

    σn=10.2464=0.4000\frac{\sigma}{\sqrt{n}}=\sqrt{\frac{10.24}{64}}=0.4000

    The standard error is σ/n\sigma/\sqrt{n}, not σ/n\sigma/n.

  3. State the final value

    sd⁡(Xˉ)=σn=0.4000\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=0.4000

    This is the required value, correct to 44 decimal places.

Answer
sd⁡(Xˉ)=σn=0.4000\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=0.4000
Question 2
2 markseasy
The processing time of a randomly selected transaction, in seconds, is modelled by a random variable XX with mean μ=500\mu=500 and standard deviation σ=40\sigma=40. The distribution of XX is not known to be normal. A random sample of n=400n=400 transactions is taken. Let Xˉ\bar{X} denote the sample mean processing time. Which of the following gives the approximate distribution of Xˉ\bar{X} predicted by the central limit theorem? In each option the second parameter is the variance.
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Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(500,4)\bar{X}\approx N\left(500,4\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=500,σ2=1600,n=400\mu=500,\quad\sigma^{2}=1600,\quad n=400

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=500E(\bar{X})=\mu=500

    The sample mean is an unbiased estimator of μ\mu.

  4. Select the option matching this distribution

    Xˉ≈N(500,4)\bar{X}\approx N\left(500,4\right)

    This is the distribution given by the central limit theorem.

Answer
Xˉ≈N(500,4)\bar{X}\approx N\left(500,4\right)
Question 3
4 marksintermediate
The duration of a randomly selected telephone call, in minutes, is modelled by a random variable XX with mean μ=9.6\mu=9.6 and standard deviation σ=1.5\sigma=1.5. The distribution of XX is not known to be normal. A random sample of n=125n=125 telephone calls is taken. Let Xˉ\bar{X} denote the sample mean duration. Which of the following statements best explains why Xˉ\bar{X} may be modelled by a normal distribution?
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Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(9.6,0.018)\bar{X}\approx N\left(9.6,0.018\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=9.6,σ2=2.25,n=125\mu=9.6,\quad\sigma^{2}=2.25,\quad n=125

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=9.6E(\bar{X})=\mu=9.6

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=2.25125=0.018\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{2.25}{125}=0.018

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.018=0.1342\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.018}=0.1342

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0120,σ2n=0.2012(both wrong)\frac{\sigma}{n}=0.0120,\qquad\frac{\sigma^{2}}{\sqrt{n}}=0.2012\qquad\text{(both wrong)}

    Neither equals σ/n=0.1342\sigma/\sqrt{n}=0.1342.

  7. Select the statement that correctly justifies the approximation

    Xˉ≈N(9.6,0.018)by the central limit theorem\bar{X}\approx N\left(9.6,0.018\right)\quad\text{by the central limit theorem}

    The central limit theorem applies because nn is large, whatever the distribution of XX.

Answer
Xˉ≈N(μ,σ2n) by the central limit theorem\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\text{ by the central limit theorem}
Question 4
6 markshard
The processing time of a randomly selected transaction, in seconds, is modelled by a random variable XX with mean μ=21.5\mu=21.5 and standard deviation σ=4\sigma=4. The distribution of XX is not known to be normal. A random sample of n=100n=100 transactions is taken. Let Xˉ\bar{X} denote the sample mean processing time. Using the central limit theorem, estimate P(21<Xˉ<22)P(21<\bar{X}<22), giving your answer to 44 decimal places. As the approximating normal distribution is continuous, strict and non-strict inequalities give the same value.
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Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(21.5,0.16)\bar{X}\approx N\left(21.5,0.16\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=21.5,σ2=16,n=100\mu=21.5,\quad\sigma^{2}=16,\quad n=100

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=21.5E(\bar{X})=\mu=21.5

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=16100=0.16\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{16}{100}=0.16

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.16=0.4000\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.16}=0.4000

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0400,σ2n=1.6000(both wrong)\frac{\sigma}{n}=0.0400,\qquad\frac{\sigma^{2}}{\sqrt{n}}=1.6000\qquad\text{(both wrong)}

    Neither equals σ/n=0.4000\sigma/\sqrt{n}=0.4000.

  7. Contrast the sample mean with the sample total

    Var⁡(Xˉ)=0.16≠1600=nσ2=Var⁡(∑i=1nXi)\operatorname{Var}(\bar{X})=0.16\neq 1600=n\sigma^{2}=\operatorname{Var}\left(\sum_{i=1}^{n}X_{i}\right)

    The question asks about the mean, so σ2/n\sigma^{2}/n is required.

  8. Standardise the lower endpoint

    z1=21−21.50.4000=−1.2500z_{1}=\frac{21-21.5}{0.4000}=-1.2500

    Subtract the mean and divide by the standard deviation.

  9. Standardise the upper endpoint

    z2=22−21.50.4000=1.2500z_{2}=\frac{22-21.5}{0.4000}=1.2500

    The same standardisation is applied to the other endpoint.

  10. Write the probability as a difference of cumulative probabilities

    P(21<Xˉ<22)=Φ(1.2500)−Φ(−1.2500)P(21<\bar{X}<22)=\Phi(1.2500)-\Phi(-1.2500)

    An interval probability is the difference of two values of Φ\Phi.

  11. State the final probability

    P(21<Xˉ<22)≈0.7887P(21<\bar{X}<22)\approx 0.7887

    This is the approximate probability, correct to 44 decimal places.

Answer
P(21<Xˉ<22)≈0.7887P(21<\bar{X}<22)\approx 0.7887
Question 5
9 markschallenging
The length of a randomly selected steel rod, in millimetres, is modelled by a random variable XX with mean μ=120\mu=120 and standard deviation σ=5\sigma=5. The distribution of XX is not known to be normal. A random sample of n=100n=100 steel rods is taken. Let Xˉ\bar{X} denote the sample mean length. Which of the following statements best explains why Xˉ\bar{X} may be modelled by a normal distribution?
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Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(120,0.25)\bar{X}\approx N\left(120,0.25\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=120,σ2=25,n=100\mu=120,\quad\sigma^{2}=25,\quad n=100

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=120E(\bar{X})=\mu=120

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=25100=0.25\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{25}{100}=0.25

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.25=0.5000\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.25}=0.5000

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0500,σ2n=2.5000(both wrong)\frac{\sigma}{n}=0.0500,\qquad\frac{\sigma^{2}}{\sqrt{n}}=2.5000\qquad\text{(both wrong)}

    Neither equals σ/n=0.5000\sigma/\sqrt{n}=0.5000.

  7. Contrast the sample mean with the sample total

    Var⁡(Xˉ)=0.25≠2500=nσ2=Var⁡(∑i=1nXi)\operatorname{Var}(\bar{X})=0.25\neq 2500=n\sigma^{2}=\operatorname{Var}\left(\sum_{i=1}^{n}X_{i}\right)

    The question asks about the mean, so σ2/n\sigma^{2}/n is required.

  8. Recall the statement of the central limit theorem

    Xˉ≈N(μ,σ2n)for large n\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\quad\text{for large }n

    For a large sample the sample mean is approximately normal whatever the distribution of XX.

  9. Recall the corresponding result for a sample total

    ∑i=1nXi≈N(nμ,nσ2)for large n\sum_{i=1}^{n}X_{i}\approx N\left(n\mu,n\sigma^{2}\right)\quad\text{for large }n

    The total has mean nμn\mu and variance nσ2n\sigma^{2}, not μ\mu and σ2/n\sigma^{2}/n.

  10. Distinguish the variance of the sample mean from its standard deviation

    Var⁡(Xˉ)=σ2n,sd⁡(Xˉ)=σn\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n},\qquad\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}

    The standard error is σ/n\sigma/\sqrt{n}; it is neither σ/n\sigma/n nor σ2/n\sigma^{2}/\sqrt{n}.

  11. Justify the variance of the sample mean

    Var⁡(1n∑i=1nXi)=1n2⋅nσ2=σ2n\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)=\frac{1}{n^{2}}\cdot n\sigma^{2}=\frac{\sigma^{2}}{n}

    Independence makes the variances add, and the factor 1/n1/n is squared.

  12. Justify the mean of the sample mean

    E(Xˉ)=1n∑i=1nE(Xi)=nμn=μE(\bar{X})=\frac{1}{n}\sum_{i=1}^{n}E(X_{i})=\frac{n\mu}{n}=\mu

    The sample mean is an unbiased estimator of the population mean.

  13. Write down the standardising transformation

    Z=Xˉ−μσ/n∼N(0,1)approximatelyZ=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)\quad\text{approximately}

    Subtract the mean and divide by the standard error.

  14. Recall the symmetry of the standard normal distribution

    Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z)

    Negative zz-values are handled by symmetry about 00.

  15. Recall the complement rule for the upper tail

    P(Z>z)=1−Φ(z)P(Z>z)=1-\Phi(z)

    The upper tail is one minus the cumulative probability.

  16. Select the statement that correctly justifies the approximation

    Xˉ≈N(120,0.25)by the central limit theorem\bar{X}\approx N\left(120,0.25\right)\quad\text{by the central limit theorem}

    The central limit theorem applies because nn is large, whatever the distribution of XX.

Answer
Xˉ≈N(μ,σ2n) by the central limit theorem\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\text{ by the central limit theorem}

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