AI HL · AHL 8
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Mixed practice: linear combinations and the sample mean
AHL 8.6
The random variable \(U\) has mean 18 and variance 12.8. The random variable \(V\) is such that \(3U+4V=100\). Find the mean and variance of \(V\).
Have a go first. Then open the working one line at a time.
\(V=\dfrac{100-3U}{4}=25-0.75U\)
Now \(V\) is a linear transformation of \(U\), with \(a=-0.75\) and \(b=25\).
\(E(V)=25-0.75(18)=11.5\)
Use \(E(aU+b)=aE(U)+b\).
\(\mathrm{Var}(V)=(-0.75)^2(12.8)=7.2\)
Square the coefficient. The negative sign and the 25 do not matter.
The time a student takes to run 100 m has mean 15.2 s and standard deviation 1.4 s. Two students are chosen at random. Find the mean and standard deviation of the first student's time minus the second student's time.
Have a go first. Then open the working one line at a time.
\(D=X_1-X_2\), two independent observations of \(X\)
Two different students, so two separate observations.
\(E(D)=15.2-15.2=0\)
The difference between two people of the same population averages 0.
\(\mathrm{Var}(D)=1.4^2+1.4^2=3.92\), sd \(=\sqrt{3.92}=1.98\) s
A difference adds variances, because the minus sign is squared.
A trolley with mass exactly 18 kg is loaded with 6 suitcases and 4 rucksacks. A suitcase has mean mass 21.5 kg and standard deviation 3.2 kg. A rucksack has mean mass 8.4 kg and standard deviation 1.5 kg. All items are independent. Find the mean and standard deviation of the total mass.
Have a go first. Then open the working one line at a time.
\(T=18+S_1+\dots+S_6+R_1+\dots+R_4\)
Six different suitcases and four different rucksacks. The trolley is a constant.
\(E(T)=18+6(21.5)+4(8.4)=180.6\) kg
Add every mean.
\(\mathrm{Var}(T)=6(3.2^2)+4(1.5^2)=61.44+9=70.44\)
The constant 18 has no variance.
\(\sqrt{70.44}=8.39\) kg
Square root last.
A population has variance 6.4. A researcher wants the standard deviation of the sample mean to be less than 0.25. Find the smallest sample size \(n\) that works.
Have a go first. Then open the working one line at a time.
\(\sqrt{\dfrac{6.4}{n}}<0.25\)
The sd of the sample mean is \(\sigma/\sqrt{n}\), and here \(\sigma^2=6.4\).
\(\dfrac{6.4}{n}<0.0625\), so \(n>102.4\)
Square both sides, then rearrange. Check the direction: a larger \(n\) gives a smaller sd.
\(n=103\)
Always round up. A smaller whole number would not meet the bound.
A sample of size \(n\) is taken from a population with standard deviation \(\sigma\). When the sample size is increased by 60, the standard deviation of the sample mean becomes half of its original value. Find \(n\).
Have a go first. Then open the working one line at a time.
\(\dfrac{\sigma}{\sqrt{n+60}}=\dfrac12\cdot\dfrac{\sigma}{\sqrt{n}}\)
The new sd is half the old one.
\(\sqrt{n+60}=2\sqrt{n}\)
\(\sigma\) cancels.
\(n+60=4n\), so \(3n=60\) and \(n=20\)
Check: \(\sqrt{80}=2\sqrt{20}\).
A spinner has mean score 5 and standard deviation 1.2. Ana spins once and triples the score. Ben spins twice and adds the scores. Find the mean and standard deviation of Ana's result minus Ben's result.
Have a go first. Then open the working one line at a time.
\(D=3X-(X_1+X_2)\)
Ana: one observation, tripled. Ben: two different observations.
\(E(D)=3(5)-2(5)=5\)
Means follow the coefficients.
\(\mathrm{Var}(D)=3^2(1.2^2)+2(1.2^2)=12.96+2.88=15.84\)
Tripled: \(9\sigma^2\). Two spins: \(2\sigma^2\). Add them.
\(\sqrt{15.84}=3.98\)
Square root last.