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AI HL · AHL 8

Statistics 2

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AHL 8.8

The sample mean and the central limit theorem

AHL 8.8

The sample mean and the central limit theorem

Target: I can write the distribution of the sample mean of n observations from a normal population as N(μ, σ²/n). (+4 more)

By the end of this section

Key points

The sample mean of a normal population

  • If \(X\sim N(\mu,\sigma^2)\) and \(X_1,\dots,X_n\) are independent observations of X, then \(\bar X\sim N\left(\mu,\dfrac{\sigma^2}{n}\right)\). This is exact for every n.
  • The mean stays the same. The variance is divided by n, so the standard deviation of \(\bar X\) is \(\dfrac{\sigma}{\sqrt n}\), the standard error.
  • A larger sample gives a narrower distribution for \(\bar X\): means of big samples stay close to μ.
  • For the calculator, enter the standard deviation \(\sigma/\sqrt n\), not the variance.

The central limit theorem

  • If the population is not normal, and n > 30, then approximately \(\bar X\sim N\left(\mu,\dfrac{\sigma^2}{n}\right)\) and \(\sum X_i\sim N(n\mu,\,n\sigma^2)\).
  • Check n > 30 first. This is the condition for the theorem in this course.
  • If n is 30 or less and the population is not stated to be normal, you cannot give the distribution of \(\bar X\).
  • The theorem works for any population you will meet: skewed, bimodal or anything else, as long as it has a mean and a variance.

Keep these apart

  • One observation X and the mean \(\bar X\) of a sample have different distributions. Read the question: "a randomly chosen" is X, "the mean of a sample of n" is \(\bar X\).
  • The total of n observations has variance \(n\sigma^2\), not \(n^2\sigma^2\) and not \(\sigma^2\).
  • In a simulation that takes many samples of the same size, the sample means centre on μ and their spread matches \(\sigma/\sqrt n\). That is what the theorem says.

On the calculator

Probability for a sample mean

  • Find \(\sigma/\sqrt n\) first. It is the standard deviation of \(\bar X\).
  • Use the normal cumulative distribution with mean μ and standard deviation \(\sigma/\sqrt n\).
  • Check the answer against a sketch. A sample mean is rarely far from μ, so the probability of a mean far from μ should be small.

Screenshot and video to come

Explore: the sample mean

Choose the shape of a population and a sample size n. Take samples, and each sample mean is added to the histogram. The curve is N(μ, σ²/n).

Samples taken
Mean of the sample means
μ, the population mean
Standard deviation of the sample means
σ/√n, the standard error

Take a sample to start.

  1. With n = 1, take 100 samples. What does the histogram look like? Why?
  2. Choose the two-humped population and make n larger. At what size does the histogram first look like the curve?
  3. Compare the mean of the sample means with μ, and their standard deviation with σ/√n. Do they agree better after 10 samples or after 1000?
  4. To halve the standard error, how many times larger must n be? Check it with the slider.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A sample from a normal population

The mass of a newborn lamb is normally distributed with mean 4.2 kg and standard deviation 0.6 kg. A random sample of 6 newborn lambs is weighed. Find the probability that the mean mass of the sample is greater than 4.5 kg.

Working
  1. The population

    \(X\sim N(4.2,\,0.36)\)

    X is the mass of one lamb. The second number is the variance, 0.6² = 0.36.

  2. The sample mean

    \(\bar{X}\sim N\left(4.2,\,\dfrac{0.36}{6}\right)=N(4.2,\,0.06)\)

    The population is normal, so \(\bar X\) is normal for any sample size.

  3. From the calculator

    From the calculator: \(P(\bar{X}>4.5)\) = 0.110 (normal cdf with μ = 4.2, σ = √0.06 = 0.245…)

    Enter the standard deviation √0.06, not 0.06.

Video walkthrough coming
Variance and standard error

A population has mean 70 and variance 36. Samples of size 9 are taken. Find the variance of the sample mean \(\bar X\). Then find the standard error when the sample size is 36.

Working
  1. Variance for n = 9

    \(Var(\bar{X})\) = 36 ÷ 9 = 4

    The mean of \(\bar X\) is still 70.

  2. Standard error for n = 9

    SE = √4 = 2

    This equals \(\sigma/\sqrt n = 6/3\).

  3. Standard error for n = 36

    SE = 6 ÷ √36 = 1

    Four times the sample size halves the standard error.

Video walkthrough coming
The central limit theorem for a mean

The time a visitor spends on a website has a skewed distribution with mean 12.5 minutes and standard deviation 8 minutes. A random sample of 50 visitors is taken. Use the central limit theorem to find the probability that the mean time for the sample is more than 14 minutes.

Working
  1. Check the condition

    n = 50 > 30

    The population is not normal, so the theorem is needed, and n > 30 lets us use it.

  2. Distribution of the mean

    \(\bar{X}\approx N\left(12.5,\,\dfrac{64}{50}\right)=N(12.5,\,1.28)\)

    Approximately normal, with the variance divided by 50.

  3. From the calculator

    From the calculator: \(P(\bar{X}>14)\) = 0.0924 (normal cdf with μ = 12.5, σ = √1.28 = 1.13…)

    About one sample in eleven has a mean this high.

Video walkthrough coming
The central limit theorem for a total

The mass of an adult passenger has mean 74 kg and standard deviation 12 kg. The shape of the distribution is not known. A lift carries 36 randomly chosen passengers. Find the probability that their total mass is more than 2700 kg.

Working
  1. Check the condition

    n = 36 > 30

    The theorem applies to the total as well.

  2. Distribution of the total S

    \(S\approx N(36\times 74,\,36\times 144)=N(2664,\,5184)\)

    Mean \(n\mu\) and variance \(n\sigma^2\), with \(\sigma^2=12^2=144\).

  3. From the calculator

    From the calculator: \(P(S>2700)\) = 0.309 (normal cdf with μ = 2664, σ = √5184 = 72)

    Here the standard deviation is √5184 = 72.

Video walkthrough coming