Home / 9709_s23_qp_52

Question 1

Topic – ALV: 5.4

The random variable X takes the values -2, 2 and 3. It is given that

$$P(X = x) = k(x^2 – 1),$$

where k is a constant.

(a) Draw up the probability distribution table for X, giving the probabilities as numerical fractions.

(b) Find E(X) and Var(X).

▶️Answer/Explanation

Solution :-

(a) $quad [3k + 3k + 8k = 1, so] k = \frac{1}{14}$

$E(X) = -2 \frac{3}{14} + 2 \frac{3}{14} + 3 \frac{8}{14} = \frac{-6+6+24}{14}$

$Var(X) = (-2)^2 \times \frac{3}{14} + 2^2 \times \frac{3}{14} + 3^2 \times \frac{8}{14} – (their E(X))^2 $

$ = 4 \times \frac{3}{14} + 4 \times \frac{3}{14} + 9 \times \frac{8}{14} – \left(their \frac{12}{7}\right)^2 $

$ = \left[\frac{12+12+72}{14} – \left(their \frac{12}{7}\right)^2\right] $

$E(X) = \frac{12}{7}, 1.71, 1 \frac{5}{7} $

$Var(X) = \frac{192}{49}, 3.92, 3 \frac{45}{49} $

Question 2

Topic – ALV: 5.3

2. A sports event is taking place for 4 days, beginning on Sunday. The probability that it will rain on Sunday is 0.4. On any subsequent day, the probability that it will rain is 0.7 if it rained on the previous day and 0.2 if it did not rain on the previous day.

(a) Find the probability that it does not rain on any of the 4 days of the event.

(b) Find the probability that the first day on which it rains during the event is Tuesday.

(c) Find the probability that it rains on exactly one of the 4 days of the event.

▶️Answer/Explanation

solution:-

(a) $P(\text{no rain}) = 0.6 \times (0.8)^3 = 0.3072 \approx \frac{192}{625}$

(b) $0.6 \times 0.8 \times 0.2 = 0.096 \approx \frac{12}{125}$

(c) $P(RDDD) = 0.4 \times 0.3 \times 0.8 \times 0.8 = 0.0768 \approx \frac{48}{625}$

$P(DRDD) = 0.6 \times 0.2 \times 0.3 \times 0.8 = 0.0288 \approx \frac{18}{625}$

$P(DDRD) = 0.6 \times 0.8 \times 0.2 \times 0.3 = 0.0288 \approx \frac{18}{625}$

$P(DDDR) = 0.6 \times 0.8 \times 0.8 \times 0.2 = 0.0768 \approx \frac{48}{625}$

$0.2112 \approx \frac{132}{625}$

Question 3

Topic – ALV: 5.1

The following back-to-back stem-and-leaf diagram represents the monthly salaries, in dollars, of 27 employees at each of two companies, A and B.

(a) Find the median and the interquartile range of the monthly salaries of employees in company A.

The lower quartile, median, and upper quartile for company B are $2600, $2690, and $2780, respectively.

(b) Draw two box-and-whisker plots in a single diagram to represent the information for the salaries of employees at companies A and B.

(c) Comment on whether the mean would be a more appropriate measure than the median for comparing the given information for the two companies.

▶️Answer/Explanation

Solution :-

(a) $Median = 2710$

$2840 – 2610$

$230$

(b) $\text{Box-and-whisker plot on provided grid.}$

$\text{B: } 2540 \quad 2600 \quad 2690 \quad 2780 \quad 3090$

$\text{A: } 2500 \quad 2610 \quad 2710 \quad 2840 \quad 3010$

(c) $\text{Examples:}$

$\text{Mean less appropriate than median because of extreme value for company B}$

$\text{[at \$3090].}$

$\text{No, extreme value in company B.}$

$\text{No, \$3090 is an anomaly.}$

Question 4

Topic – ALV: 5.3

4. A fair 5-sided spinner has sides labelled 1, 2, 3, 4, 5. The spinner is spun repeatedly until a 2 is obtained on the side on which the spinner lands. The random variable X denotes the number of spins required.

(a) Find P(X = 4).

(b) Find P(X < 6).

Two fair 5-sided spinners, each with sides labelled 1, 2, 3, 4, 5, are spun at the same time. If the numbers obtained are equal, the score is 0. Otherwise, the score is the higher number minus the lower number.

(c) Find the probability that the score is greater than 0 given that the score is not equal to 2.

The two spinners are spun at the same time repeatedly.

(d) For 9 randomly chosen spins of the two spinners, find the probability that the score is greater than 2 on at least 3 occasions.

▶️Answer/Explanation

Solution :-

(a) $[P(X=4) = (0.8)^3 (0.2) =] 0.1024, \frac{64}{625}$

(b)

$[P(X < 6) =] 1 – 0.8^5 = 0.672, \frac{2101}{3125}$

$\text{Alternative Method for Question 4(b)}$

$[P(X < 6) =] \left[1 + 4\left(\frac{1}{5}\right) – 4\left(\frac{1}{5}\right)^2 + \left(\frac{1}{5}\right)^3 – 4\left(\frac{1}{5}\right)^4 + \left(\frac{1}{5}\right)^5\right]$

$= 0.672, \frac{2101}{3125}$

(c)

$\left[P(X > 0 \cap X \neq 2) = \frac{P(X > 0 \cap X \neq 2)}{P(X \neq 2)}\right] = \frac{14}{25} \times \frac{19}{25} = \frac{14}{19}, 0.737$

$\text{Alternative Method for Question 4(c)}$

\begin{matrix}
1 & 2 & 3 & 4 & 5 \\
2 & 0 & 1 & 2 & 3 & 4 \\
3 & 1 & 0 & 1 & 2 & 3 \\
4 & 2 & 1 & 0 & 1 & 2 \\
5 & 3 & 2 & 1 & 0 & 1 \\
\end{matrix}

$\left[P(X > 0 \cap X \neq 2) = \frac{\text{Number of outcome (X > 0 ∩ X ≠ 2)}}{\text{Number of outcomes X ≠ 2}}\right] = \frac{14}{19}, 0.737$

(d)

$[P(X > 2) = 1 – P(0, 1, 2) \text{ with } p = \frac{6}{25}]$

$1 – \left[^{9}C_{0}\left(\frac{19}{25}\right)^{9}+^{9}C_{1}\left(\frac{6}{25}\right)^{1}\left(\frac{19}{25}\right)^{8}+^{9}C_{2}\left(\frac{6}{25}\right)^{2}\left(\frac{19}{25}\right)^{7}\right]$

$[1 – (0.08459 + 0.2404 + 0.3037)]$

$0.371$

$\text{Alternative Method for Question 4(d)}$

$[P(X > 2) = P(3, 4, 5, 6, 7, 8, 9) \text{ with } p = \frac{6}{25}]$

$^{9}C_{3}\left(\frac{6}{25}\right)^{3}\left(\frac{19}{25}\right)^{6}+^{9}C_{4}\left(\frac{6}{25}\right)^{4}\left(\frac{19}{25}\right)^{5}+…+^{9}C_{8}\left(\frac{6}{25}\right)^{8}\left(\frac{19}{25}\right)^{1}+^{9}C_{9}\left(\frac{6}{25}\right)^{9}$

$[0.2238 + 0.1060 + … + 7.529 \times 10^{-5} + 2.642 \times 10^{-6}]$

$0.371$

Question 5

Topic – ALV: 5.5

5. The lengths of Western bluebirds are normally distributed with mean 16.5 cm and standard deviation 0.6 cm.

A random sample of 150 of these birds is selected.

(a) How many of these 150 birds would you expect to have length between 15.4 cm and 16.8 cm?

The lengths of Eastern bluebirds are normally distributed with mean 18.4 cm and standard deviation σ cm. It is known that 72% of Eastern bluebirds have length greater than 17.1 cm.

(b) Find the value of σ.

A random sample of 120 Eastern bluebirds is chosen.

(c) Use an approximation to find the probability that fewer than 80 of these 120 bluebirds have length greater than 17.1 cm.

▶️Answer/Explanation

Solution :-

(a) \[[P(15.4<X<16.8)=]P(\frac{15.4-16.5}{0.6}<Z<\frac{16.8-16.5}{0.6})\]
\[[=P(-1.833<Z<0.5)]\]

\[[= \Phi(0.5)+\Phi(1.833)-1=]\]
$0.6915 +0.9666-1$

= 0.658

$[Expected number =] 0.6581 × 150$
$= 98, 99$

(b) \[[P(Z>\frac{17.1-18.4}{\sigma})=0.72]\]
\[\frac{17.1-18.4}{\sigma}=-0.583\]

\[\sigma=2.23\]

5(c) \[[Mean = 120 \times 0.72 =] 86.4\]
\[[Var = 120 \times 0.72 \times 0.28 =] 24.192\]

\[P(X < 80) = P(Z < \frac{79.5 – 86.4}{\sqrt{24.192}})\]

\[[P(Z < -1.4029) = 1 – \Phi(1.403)]\]
1 – 0.9196

0.0804

Question 6

Topic – ALV: 5.2

In a group of 25 people there are 6 swimmers, 8 cyclists and 11 runners. Each person competes in only one of these sports. A team of 7 people is selected from these 25 people to take part in a competition.

(a) Find the number of different ways in which the team of 7 can be selected if it consists of exactly 1 swimmer, at least 4 cyclists and at most 2 runners.

For another competition, a team of 9 people consists of 2 swimmers, 3 cyclists and 4 runners. The team members stand in a line for a photograph.

(b) How many different arrangements are there of the 9 people if the swimmers stand together, the cyclists stand together and the runners stand together?

(c) How many different arrangements are there of the 9 people if none of the cyclists stand next to each other? 

▶️Answer/Explanation

Solution :-

6(a) $S + 4C + 2R \quad ^6C_{1} \times ^8C_{4} \times ^4C_{2} [= 6 \times 70 \times 55] = 23100$
$S + 5C + 1R \quad ^6C_{1} \times ^8C_{5} \times ^5C_{1} [= 6 \times 56 \times 11] = 3696$
$S + 6C + [0R] \quad ^6C_{1} \times ^8C_{6} \times ^6C_{0} [= 6 \times 28] = 168$

$[Total =] 26964$

6(b) $2! \times 3! \times 4! \times 6$

$= 1728$

6(c) \text{ Method 1}

$6! \times 7 \times 6 \times 5$

151 200

$\text{ Method 2}$

$6! \times ^7P_{3}$

151 200

Scroll to Top