2021 H2 Mathematics Paper 2 Question 11

The Binomial Distribution
Sampling Theory

Answers

(ai)
The light fittings should be selected randomly: that is, each light fitting in the population has the same probability of being selected into the sample. We should select the light fittings this way to remove bias from our sample so that the results from the sample approximates the population.
(aii)
p≈0.03{p \approx 0.03}
(aiii)
34.1{34.1}
(bi)
  • The probability of a heating element is faulty is the same for each heating element
  • Whether a heating element is faulty is independent of any other heating element
(bii)
0.779{0.779}
(biii)
0.0505{0.0505}
(biv)
0.593{0.593}

Full solutions

(ai)
The light fittings should be selected randomly: that is, each light fitting in the population has the same probability of being selected into the sample. We should select the light fittings this way to remove bias from our sample so that the results from the sample approximates the population. ■{\blacksquare}
(aii)
x‾=∑xn=0+19+2×38+…+7×44+19+…+4=3\begin{align*} \overline{x} &= \frac{\sum x}{n} \\ &= \frac{0 + 19 + 2 \times 38 + \ldots + 7 \times 4}{4 + 19 + \ldots + 4} \\ &= 3 \end{align*}
E(X)≈x‾100p≈3p≈0.03  ■\begin{align*} \textrm{E}(X) &\approx \overline{x} \\ 100p &\approx 3 \\ p &\approx 0.03 \; \blacksquare \end{align*}
(aiii)
P(X=3)=0.22747\begin{align*} \mathrm{P}(X=3) &= 0.22747 \end{align*}
Expected number of days=150(0.22747)=34.1 (3 sf)  ■\begin{align*} & \textrm{Expected number of days} \\ & = 150 (0.22747) \\ & = 34.1 \textrm{ (3 sf)} \; \blacksquare \end{align*}
(bi)
  • The probability of a heating element is faulty is the same for each heating element   ■{\; \blacksquare}
  • Whether a heating element is faulty is independent of any other heating element   ■{\; \blacksquare}
(bii)
X∼B(80,0.02)X \sim \textrm{B}\left(80, 0.02\right)
P(1≤X≤4)=P(X≤4)−P(X≤0)=0.779 (3sf)  ■\begin{align*} & \mathrm{P}(1 \leq X \leq 4) \\ &= \mathrm{P}(X \leq 4) - \mathrm{P}(X \leq 0) \\ &= 0.779 \textrm{ (3sf)} \; \blacksquare \end{align*}
(biii)
P(X>3)=1−P(X≤3)=0.076855\begin{align*} \mathrm{P}(X > 3) &= 1 - \mathrm{P}(X\leq 3) \\ &= 0.076855 \end{align*}
Let Y{Y} denote the r.v. of the number of days in which there are more than 3{3} faulty heating elements, our of 5{5} days
Y∼B(5,0.076855)Y \sim \textrm{B}\left(5, 0.076855\right)
P(Y≥2)=1−P(Y≤1)=0.0505  ■\begin{align*} \mathrm{P}(Y \geq 2) &= 1 - \mathrm{P}(Y\leq 1) \\ &= 0.0505 \; \blacksquare \end{align*}
(biv)
Let W{W} denote the r.v. of the total number of faulty heating elements in 5{5} days
W∼B(400,0.02)W \sim \textrm{B}\left(400, 0.02\right)
P(W≤8)=0.593 (3 sf)  ■\begin{align*} \mathrm{P}(W \leq 8) &= 0.593 \textrm{ (3 sf)} \; \blacksquare \end{align*}