The sample mean and the central limit theorem Problem set A One question at a time

A normal population has mean 52 and variance 40. A random sample of 8 observations is taken. \(\bar X\) is the sample mean.

1. (a) State the distribution of \(\bar X\).

   • \(Var(\bar{X})\) = 40 ÷ 8 = 5
   • Mean stays 52

2. (b) Write down the standard error.

   • SE = √5 = 2.24…

A random variable \(X\sim N(150,\,20^2)\). A sample of 16 independent observations is taken. Find \(P(\bar X<145)\).

• \(\bar X\sim N(150,\,20^2/16)\) = \(N(150,\,25)\)
• From the calculator: \(P(\bar{X}<145)\) = 0.159 (normal cdf with μ = 150, σ = √25 = 5)

A random variable X has mean 28 and variance 49. Samples of independent observations of X are taken. Give the distribution of \(\bar X\) in each case, or say that it cannot be found.

1. (a) The distribution of X is not known and n = 45.

   • n = 45 > 30, so the central limit theorem applies
   • Variance = 49 ÷ 45

2. (b) The distribution of X is not known and n = 20.

   • n = 20 is not more than 30 and X is not known to be normal

3. (c) X is normally distributed and n = 20.

   • X is normal, so \(\bar X\) is normal for any n
   • Variance = 49 ÷ 20 = 2.45

The time for a kettle to boil is normally distributed with mean 3.4 minutes and standard deviation 0.5 minutes. A random sample of 4 boils is timed. Find the probability that the mean time is greater than 3.7 minutes.

• \(\bar X\sim N(3.4,\,0.5^2/4)\) = \(N(3.4,\,0.0625)\)
• From the calculator: \(P(\bar{X}>3.7)\) = 0.115 (normal cdf with μ = 3.4, σ = √0.0625 = 0.25)

The waiting time at a clinic has a skewed distribution with mean 18 minutes and standard deviation 11 minutes. A random sample of 60 patients is taken. Find the probability that the mean waiting time is less than 15 minutes.

• n = 60 > 30, so by the central limit theorem \(\bar X\approx N(18,\,11^2/60)\)
• From the calculator: \(P(\bar{X}<15)\) = 0.0173 (normal cdf with μ = 18, σ = 1.42…)

The mass of a parcel has mean 3.6 kg and standard deviation 1.8 kg. A van is loaded with 45 randomly chosen parcels. Find the probability that their total mass is more than 170 kg.

• n = 45 > 30, so the central limit theorem applies
• \(S\approx N(45\times 3.6,\,45\times 1.8^2)\) = \(N(162,\,145.8)\)
• From the calculator: \(P(S>170)\) = 0.254 (normal cdf with μ = 162, σ = √145.8 = 12.1…)

The score on a reaction test is normally distributed with mean 60 and standard deviation 15.

1. (a) Find the probability that one randomly chosen value is more than 66.

   • From the calculator: \(P(X>66)\) = 0.345 (normal cdf with μ = 60, σ = √225 = 15)

2. (b) Find the probability that the mean of a random sample of 25 values is more than 66.

   • \(\bar{X}\sim N(60,\,9)\)
   • From the calculator: \(P(\bar{X}>66)\) = 0.0228 (normal cdf with μ = 60, σ = √9 = 3)

3. (c) Explain why your two answers are so different.

   • The standard deviation of one value is 15. The standard error of the mean is 15 ÷ √25 = 3

A scale gives readings that are normally distributed with mean 100 g and standard deviation 20 g. A technician takes several readings and works with their mean.

1. (a) Find the probability that the sample mean is within 2 of 100 when the sample has 25 observations.

   • \(\bar{X}\sim N(100,\,16)\)
   • From the calculator: \(P(98<\bar{X}<102)\) = 0.383 (normal cdf, σ = 20 ÷ √25 = 4)

2. (b) Find the same probability when the sample has 100 observations.

   • \(\bar{X}\sim N(100,\,4)\)
   • From the calculator: \(P(98<\bar{X}<102)\) = 0.683 (normal cdf, σ = 20 ÷ √100 = 2)

3. (c) What happens to the probability as the sample size grows?

   • The standard error shrinks as n grows, so more of the distribution lies within the same distance of the mean.