The sample mean and its standard error Problem set B Worked solutions

A random variable \(X\) has the mean and variance given in each part. Find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\) for a sample of size \(n\).

1. (a) \(E(X)=3.4,\ \mathrm{Var}(X)=5.6,\ n=14\)

   • \(E(\bar{X})=E(X)=3.4\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{5.6}{14}=0.4\)

2. (b) \(E(X)=-2.3,\ \mathrm{Var}(X)=1.6,\ n=25\)

   • \(E(\bar{X})=E(X)=-2.3\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{1.6}{25}=0.064\)

Find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\) for the mean of a sample of size \(n\) from

1. (a) \(X\sim N(64,\,25),\ n=10\)

   • \(E(X)=64,\ \mathrm{Var}(X)=25\)
   • \(E(\bar{X})=E(X)=64\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{25}{10}=2.5\)

2. (b) \(X\sim N(0.8,\,0.04),\ n=16\)

   • \(E(X)=0.8,\ \mathrm{Var}(X)=0.04\)
   • \(E(\bar{X})=E(X)=0.8\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{0.04}{16}=0.0025\)

Find \(E(X)\) and \(\mathrm{Var}(X)\). Then find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\) for the mean of a sample of size \(n\) of observations of \(X\).

1. (a) \(X\sim B(8,\,0.25),\ n=6\)

   • \(E(X)=8(0.25)=2\)
   • \(\mathrm{Var}(X)=8(0.25)(0.75)=1.5\)
   • \(E(\bar{X})=E(X)=2\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{1.5}{6}=0.25\)

2. (b) \(X\sim B(20,\,0.6),\ n=16\)

   • \(E(X)=20(0.6)=12\)
   • \(\mathrm{Var}(X)=20(0.6)(0.4)=4.8\)
   • \(E(\bar{X})=E(X)=12\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{4.8}{16}=0.3\)

The table shows the probability distribution of the random variable \(X\).

1. (a) Find \(E(X)\).

   • \(E(X)=\sum xP(X=x)=1(0.3)+2(0.2)+3(0.4)+4(0.1)=2.3\)

2. (b) The random variable \(Y\) is given by \(Y=3X+4\). Find \(E(Y)\).

   • \(E(Y)=3E(X)+4=3(2.3)+4=10.9\)

3. (c) Given that \(\mathrm{Var}(X)=1.01\), find \(\mathrm{Var}(Y)\).

   • \(\mathrm{Var}(Y)=3^2(1.01)=9.09\)

The random variable \(W\) has \(P(W=w)=\dfrac{k}{w}\) for \(w=1, 2, 3, 4\).

1. (a) Find the value of \(k\).

   • \(k(\tfrac{1}{1}+\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4})=1\)
   • \(k=0.48\)

2. (b) Find \(E(W)\).

   • \(E(W)=\sum w\cdot\dfrac{0.48}{w}=4(0.48)=1.92\)

3. (c) Given that \(\mathrm{Var}(W)=1.1136\), find the variance of \(V=5W+7\).

   • \(\mathrm{Var}(V)=5^2(1.1136)=27.84\)

The life of a certain battery has mean 36 hours and standard deviation 5 hours. A tester measures a sample of 16 batteries and finds their mean life \(\bar{X}\).

1. (a) Find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\).

   • \(E(\bar{X})=36\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{5^2}{16}=1.5625\)

2. (b) Find the standard error of the mean.

   • \(SE=\dfrac{\sigma}{\sqrt{n}}=\dfrac{5}{\sqrt{16}}=1.25\)

3. (c) The tester uses a sample of 64 batteries instead. Find the new standard error and say how it compares with part (b).

   • \(SE=\dfrac{5}{\sqrt{64}}=0.625\)
   • The sample is four times larger, so the standard error is halved.

A call centre takes 40 calls each day. Each call is cancelled with probability 0.15, independently. Let \(X\) be the number of cancelled calls in one day. The mean number of cancelled calls per day is found over a sample of 25 days.

1. (a) Find \(E(X)\) and \(\mathrm{Var}(X)\).

   • \(E(X)=40(0.15)=6\)
   • \(\mathrm{Var}(X)=40(0.15)(0.85)=5.1\)

2. (b) Find the expected value and variance of the mean number of cancelled calls per day over the 25 days.

   • \(E(\bar{X})=E(X)=6\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{\mathrm{Var}(X)}{n}=\dfrac{5.1}{25}=0.204\)

The height of a seedling after two weeks is modelled by \(X\sim N(8.4,\,1.5^2)\), in cm. The mean height \(\bar{X}\) of 15 seedlings is recorded.

1. (a) Find \(E(\bar{X})\) and \(\mathrm{Var}(\bar{X})\).

   • \(\mathrm{Var}(X)=1.5^2=2.25\)
   • \(E(\bar{X})=8.4\)
   • \(\mathrm{Var}(\bar{X})=\dfrac{2.25}{15}=0.15\)

2. (b) Find the standard error of \(\bar{X}\).

   • \(SE=\sqrt{0.15}=0.387\)

An electrician charges £38 per hour plus a fixed call-out fee of £25. The time \(X\) hours of a job has mean 1.8 and variance 0.36. The cost of a job is \(C=38X+25\). The mean cost \(\bar{C}\) of a sample of 9 jobs is found. Find \(E(\bar{C})\), \(\mathrm{Var}(\bar{C})\) and the standard error of \(\bar{C}\).

• \(E(C)=38(1.8)+25=93.4\)
• \(\mathrm{Var}(C)=38^2(0.36)=519.84\)
• \(E(\bar{C})=93.4\)
• \(\mathrm{Var}(\bar{C})=\dfrac{519.84}{9}=57.76\)
• \(SE=\sqrt{57.76}=7.6\)