Question 9:
(a) It is found that 75% of the students in SMK Permai go to school by bus.
If 8 students are chosen at random from the school, find the probability that at most 2 of them do not go to school by bus.
[3 marks]
(b) The mass of fish reared in a pond is normally distributed with a mean of 1.5 kg and a standard deviation of 0.4 kg . The fish are graded by mass, as shown in Table 1, such that k is a constant.
(i) A fish is caught at random from the pond, find the probability that it is of grade A.
(ii) It is given that 72.84% of the fish in the pond are of grade B.
Represent the situation by sketching the standard normal distribution graph and express the shaded region in probability notation.
Hence, find the value of k.
[7 marks]
Answer:
(a)
$$ \begin{aligned} &\text { Let } X=\text { Student not go by bus }\\ &\begin{aligned} n=8, q=0.75, p & =1-0.75 \\ & =0.25 \end{aligned} \end{aligned} $$
$$ \begin{aligned} P(x \leqslant 2) & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^8 C_0(0.25)^0(0.75)^8+{ }^8 C_1(0.25)^1(0.75)^7+{ }^8 C_2(0.25)^2(0.75)^6 \\ & =0.1001+0.2670+0.3115 \\ & =0.6786 \end{aligned} $$
(b)(i)
$$ \begin{aligned} &X \sim N\left(1.5,0.4^2\right)\\ &\begin{aligned} P(X>2) & =P\left(Z>\frac{2-1.5}{0.4}\right) \\ & =P(Z>1.25) \\ & =0.1056 \end{aligned} \end{aligned} $$
(b)(ii)
$$ \begin{array}{r} P(k \leqslant X \leqslant 2)=72.84 \% \\ P\left(\frac{k-1.5}{0.4} \leqslant Z \leqslant \frac{2-1.5}{0.4}\right)=0.7284 \\ P\left(\frac{k-1.25}{0.4} \leqslant Z \leqslant 1.25\right)=0.7284 \\ P\left(Z \geqslant \frac{k-1.5}{0.4}\right)-P(Z \geqslant 1.25)=0.7284 \end{array} $$
$$ \begin{aligned} P\left(Z \geqslant \frac{k-1.5}{0.4}\right)-0.1056 & =0.7284 \\ P\left(Z \geqslant \frac{k-1.5}{0.4}\right) & =0.8340 \\ 1-P\left(Z<\frac{k-1.5}{0.4}\right) & =0.8340 \\ P\left(Z<\frac{k-1.5}{0.4}\right) & =0.1660 \\ P\left(Z>\frac{k-1.5}{0.4}\right) & =0.1660 \\ \frac{k-1.5}{0.4} & =-0.97 \\ k-1.5 & =-0.388 \\ k & =1.112 \end{aligned} $$
(a) It is found that 75% of the students in SMK Permai go to school by bus.
If 8 students are chosen at random from the school, find the probability that at most 2 of them do not go to school by bus.
[3 marks]
(b) The mass of fish reared in a pond is normally distributed with a mean of 1.5 kg and a standard deviation of 0.4 kg . The fish are graded by mass, as shown in Table 1, such that k is a constant.
(i) A fish is caught at random from the pond, find the probability that it is of grade A.
(ii) It is given that 72.84% of the fish in the pond are of grade B.
Represent the situation by sketching the standard normal distribution graph and express the shaded region in probability notation.
Hence, find the value of k.
[7 marks]
Answer:
(a)
$$ \begin{aligned} &\text { Let } X=\text { Student not go by bus }\\ &\begin{aligned} n=8, q=0.75, p & =1-0.75 \\ & =0.25 \end{aligned} \end{aligned} $$
$$ \begin{aligned} P(x \leqslant 2) & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^8 C_0(0.25)^0(0.75)^8+{ }^8 C_1(0.25)^1(0.75)^7+{ }^8 C_2(0.25)^2(0.75)^6 \\ & =0.1001+0.2670+0.3115 \\ & =0.6786 \end{aligned} $$
(b)(i)
$$ \begin{aligned} &X \sim N\left(1.5,0.4^2\right)\\ &\begin{aligned} P(X>2) & =P\left(Z>\frac{2-1.5}{0.4}\right) \\ & =P(Z>1.25) \\ & =0.1056 \end{aligned} \end{aligned} $$
(b)(ii)
$$ \begin{array}{r} P(k \leqslant X \leqslant 2)=72.84 \% \\ P\left(\frac{k-1.5}{0.4} \leqslant Z \leqslant \frac{2-1.5}{0.4}\right)=0.7284 \\ P\left(\frac{k-1.25}{0.4} \leqslant Z \leqslant 1.25\right)=0.7284 \\ P\left(Z \geqslant \frac{k-1.5}{0.4}\right)-P(Z \geqslant 1.25)=0.7284 \end{array} $$
$$ \begin{aligned} P\left(Z \geqslant \frac{k-1.5}{0.4}\right)-0.1056 & =0.7284 \\ P\left(Z \geqslant \frac{k-1.5}{0.4}\right) & =0.8340 \\ 1-P\left(Z<\frac{k-1.5}{0.4}\right) & =0.8340 \\ P\left(Z<\frac{k-1.5}{0.4}\right) & =0.1660 \\ P\left(Z>\frac{k-1.5}{0.4}\right) & =0.1660 \\ \frac{k-1.5}{0.4} & =-0.97 \\ k-1.5 & =-0.388 \\ k & =1.112 \end{aligned} $$