2020 H2 Mathematics Paper 2 Question 9

The Binomial Distribution
Sampling Theory

Answers

(i)

A random sample means that each pen in the box of 100 (the population) has an equal probability of being selected into the sample of 10

(ii)

0.981{0.981}

(iii)

0.0535{0.0535}

(iv)

0.983{0.983}

(v)

While both methods have similar probability of accepting a chosen box, the alternative testing procedure test less pens overall and could be cheaper to implement as a result

Full solutions

(i)

A random sample means that each pen in the box of 100 (the population) has an equal probability of being selected into the sample of 10 ■{\blacksquare}

(ii)

Let X{X} denote the r.v. of the number of faulty pens out of 10{10}
X∼B(10,0.06)X \sim \textrm{B}\left(10, 0.06\right)
P(X≤2)=0.98116=0.981 (3 sf)  ■\begin{align*} \mathrm{P}(X \leq 2) &= 0.98116 \\ & = 0.981 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

Let Y{Y} denote the r.v. of the number of boxes that are rejected out of 75{75}
Y∼B(75,1−0.98116)Y \sim \textrm{B}(75, 1-0.98116)
P(Y>0.05(75))=P(Y>3.75)=1−P(Y≤3)=0.0535 (3 sf)  ■\begin{align*} & \textrm{P}\Big(Y > 0.05(75)\Big) \\ & = \textrm{P}(Y > 3.75) \\ & = 1 - \textrm{P}(Y \leq 3) \\ & = 0.0535 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iv)

Let W{W} denote the r.v. of the number of faulty pens out of 5{5}
P(accepted)=P(W=0)=  +P(W1=1)⋅P(W2≤1)=+P(W1=2)⋅P(W2=0)=0.73390+0.23422(0.96813)+0.029901(0.73390)=0.983 (3 sf)  ■\begin{align*} & \textrm{P}(\textrm{accepted}) \\ & = \textrm{P}(W=0) \\ & \phantom{=} \; + \textrm{P}(W_1=1) \cdot \textrm{P}(W_2 \leq 1) \\ & \phantom{=} \quad + \textrm{P}(W_1=2) \cdot \textrm{P}(W_2 = 0) \\ & = 0.73390 + 0.23422 (0.96813) + 0.029901 (0.73390) \\ &= 0.983 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(v)

While both methods have similar probability of accepting a chosen box, the alternative testing procedure test less pens overall and could be cheaper to implement as a result ■{\blacksquare}