Central limit theorem Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Central limit theorem questions. See exactly how to solve problems on central-limit-theorem, standard-error, sampling-distribution-of-the-mean, distribution-of-a-sum.

central-limit-theoremstandard-errorsampling-distribution-of-the-meandistribution-of-a-sumidentifying-the-distributionconditions
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.

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 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. 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.

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

    σn=25100=0.5000\frac{\sigma}{\sqrt{n}}=\sqrt{\frac{25}{100}}=0.5000

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

  3. 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.

  4. State the final value

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

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

Answer
sd⁡(Xˉ)=σn=0.5000\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=0.5000
Question 3
2 markseasy
The volume of a randomly selected bottle of juice, in millilitres, is modelled by a random variable XX with mean μ=330\mu=330 and standard deviation σ=6\sigma=6. The distribution of XX is not known to be normal. A random sample of n=50n=50 bottles of juice is taken. Let Xˉ\bar{X} denote the sample mean volume. Using the central limit theorem, state the variance of the sampling distribution of Xˉ\bar{X}, giving your answer to 44 decimal places.

Worked solution

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

    Xˉ≈N(330,0.72)\bar{X}\approx N\left(330,0.72\right)

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

  2. Divide the population variance by the sample size

    σ2n=3650=0.72\frac{\sigma^{2}}{n}=\frac{36}{50}=0.72

    The variance of Xˉ\bar{X} is σ2/n\sigma^{2}/n.

  3. State the final value

    Var⁡(Xˉ)=σ2n=0.7200\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=0.7200

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

Answer
Var⁡(Xˉ)=σ2n=0.7200\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=0.7200
Question 4
2 markseasy
The lifetime of a randomly selected light bulb, in hours, is modelled by a random variable XX with mean μ=1200\mu=1200 and standard deviation σ=80\sigma=80. The distribution of XX is not known to be normal. A random sample of n=40n=40 light bulbs is taken. Let T=∑i=1nXiT=\sum_{i=1}^{n}X_i denote the total lifetime of the sample. Using the central limit theorem, state the standard deviation of the distribution of TT, giving your answer to 44 decimal places.

Worked solution

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

    T≈N(48000,256000)T\approx N\left(48000,256000\right)

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

  2. Take the square root of the variance of the total

    nσ2=40×6400=505.9644\sqrt{n\sigma^{2}}=\sqrt{40\times 6400}=505.9644

    The standard deviation of the total is σn\sigma\sqrt{n}.

  3. Write down the population parameters and the sample size

    μ=1200,σ2=6400,n=40\mu=1200,\quad\sigma^{2}=6400,\quad n=40

    These are the three quantities the central limit theorem needs.

  4. State the final value

    sd⁡(T)=nσ2=505.9644\operatorname{sd}(T)=\sqrt{n\sigma^{2}}=505.9644

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

Answer
sd⁡(T)=nσ2=505.9644\operatorname{sd}(T)=\sqrt{n\sigma^{2}}=505.9644
Question 5
2 markseasy
The waiting time of a randomly selected customer, in minutes, is modelled by a random variable XX with mean μ=8.4\mu=8.4 and standard deviation σ=2.5\sigma=2.5. The distribution of XX is not known to be normal. A random sample of n=50n=50 customers is taken. Let T=∑i=1nXiT=\sum_{i=1}^{n}X_i denote the total waiting time of the sample. Using the central limit theorem, state the mean of the distribution of TT, giving your answer to 44 decimal places.

Worked solution

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

    T≈N(420,312.5)T\approx N\left(420,312.5\right)

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

  2. Multiply the population mean by the sample size

    nμ=50×8.4=420n\mu=50\times 8.4=420

    Expectation is additive over the sample.

  3. State the final value

    E(T)=nμ=420.0000E(T)=n\mu=420.0000

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

Answer
E(T)=nμ=420.0000E(T)=n\mu=420.0000

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