1. [Maximum mark: 3]
For \( n \) values of the variable \( x \), it is given that:
\[ \sum (x – 200) = 446 \]
and
\[ \sum x = 6846. \]
Find the value of \( n \).
▶️Answer/Explanation
Solution:
Given:
\[ \sum (x – 200) = \sum x – \sum 200 = 446 \]
\[ \sum 200 = 200n \]
Substituting the given values:
\[ 6846 – 200n = 446 \]
\[ 200n = 6846 – 446 = 6400 \]
\[ n = \frac{6400}{200} = 32 \]
2. [Maximum mark: 6]
A fair 6-sided die has the numbers 1, 2, 2, 3, 3, 3 on its faces. The die is rolled twice. The random variable \( X \) denotes the sum of the two numbers obtained.
(a) Draw up the probability distribution table for \( X \).
(b) Find \( E(X) \) and \( \text{Var}(X) \).
▶️Answer/Explanation
(a) Probability distribution table:
| \( X \) | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|
| \( P(X) \) | \(\frac{1}{36}\) | \(\frac{4}{36}\) | \(\frac{10}{36}\) | \(\frac{12}{36}\) | \(\frac{9}{36}\) |
(b) Calculations:
\[ E(X) = \frac{1 \times 2 + 4 \times 3 + 10 \times 4 + 12 \times 5 + 9 \times 6}{36} = \frac{168}{36} = \frac{14}{3} \approx 4.67 \]
\[ \text{Var}(X) = \frac{1 \times 4 + 4 \times 9 + 10 \times 16 + 12 \times 25 + 9 \times 36}{36} – \left(\frac{14}{3}\right)^2 = \frac{824}{36} – \frac{196}{9} = \frac{10}{9} \approx 1.11 \]
3. [Maximum mark: 7]
The back-to-back stem-and-leaf diagram shows the diameters, in cm, of 19 cylindrical pipes produced by each of two companies, \( A \) and \( B \).
(a) Find the median and interquartile range of the pipes produced by company \( A \).
(b) Draw box-and-whisker plots for companies \( A \) and \( B \) on the grid below.
(c) Make one comparison between the diameters of the pipes produced by companies \( A \) and \( B \).
▶️Answer/Explanation
(a) For company \( A \):
Median = 0.355 cm
Lower quartile = 0.348 cm, Upper quartile = 0.366 cm
Interquartile range = 0.366 – 0.348 = 0.018 cm
(b) Box-and-whisker plots:
– For company \( A \): Min = 0.331 cm, Q1 = 0.348 cm, Median = 0.355 cm, Q3 = 0.366 cm, Max = 0.382 cm.
– For company \( B \): Min = 0.342 cm, Q1 = 0.346 cm, Median = 0.352 cm, Q3 = 0.370 cm, Max = 0.388 cm.
(c) Comparison:
The diameters of pipes produced by company \( A \) are generally larger and more spread out than those produced by company \( B \).
4. [Maximum mark: 7]
The weights, in kg, of bags of rice produced by Anders have the distribution \( N(2.02, 0.03^2) \).
(a) Find the probability that a randomly chosen bag of rice produced by Anders weighs between 1.98 and 2.03 kg.
(b) The weights of bags of rice produced by Binders are normally distributed with mean 2.55 kg and standard deviation \( \sigma \) kg. In a random sample of 5000 of these bags, 134 weighed more than 2.6 kg. Find the value of \( \sigma \).
▶️Answer/Explanation
(a) Probability calculation:
\[ P(1.98 < X < 2.03) = P\left(\frac{1.98 – 2.02}{0.03} < Z < \frac{2.03 – 2.02}{0.03}\right) = P(-1.333 < Z < 0.333) \]
Using standard normal tables:
\[ = \Phi(0.333) – \Phi(-1.333) = 0.6304 – 0.0912 = 0.539 \]
(b) Solving for \( \sigma \):
Given \( P(X > 2.6) = \frac{134}{5000} = 0.0268 \), so \( P(X < 2.6) = 0.9732 \).
Using standard normal tables, \( z \) for \( P(Z < z) = 0.9732 \) is approximately 1.93.
\[ \frac{2.6 – 2.55}{\sigma} = 1.93 \]
\[ \sigma = \frac{0.05}{1.93} \approx 0.0259 \text{ kg} \]
5. [Maximum mark: 8]
In a large college, 28% of the students do not play any musical instrument, 52% play exactly one musical instrument, and the remainder play two or more musical instruments.
(a) A random sample of 12 students from the college is chosen. Find the probability that more than 9 of these students play at least one musical instrument
(b) A random sample of 90 students from the college is now chosen. Use an approximation to find the probability that fewer than 40 of these students play exactly one musical instrument.
▶️Answer/Explanation
(a) Probability calculation:
Let \( p = 0.72 \) (probability of playing at least one instrument).
\[ P(X > 9) = P(10) + P(11) + P(12) \]
\[ = \binom{12}{10} (0.72)^{10} (0.28)^2 + \binom{12}{11} (0.72)^{11} (0.28)^1 + \binom{12}{12} (0.72)^{12} (0.28)^0 \]
\[ \approx 0.1937 + 0.0906 + 0.0194 = 0.3037 \]
(b) Normal approximation:
Mean \( \mu = 90 \times 0.52 = 46.8 \), Variance \( \sigma^2 = 90 \times 0.52 \times 0.48 = 22.464 \).
Using continuity correction:
\[ P(X < 40) \approx P\left(Z < \frac{39.5 – 46.8}{\sqrt{22.464}}\right) = P(Z < -1.540) \approx 0.0618 \]
6. [Maximum mark: 10]
(a) Find the number of different arrangements of the 9 letters in the word CROCODILE.
(b) Find the number of different arrangements of the 9 letters in the word CROCODILE in which there is a C at each end and the two Os are not together.
(c) Four letters are selected from the 9 letters in the word CROCODILE. Find the number of selections in which the number of Cs is not the same as the number of Os.
(d) Find the number of ways in which the 9 letters in the word CROCODILE can be divided into three groups, each containing three letters, if the two Cs must be in different groups.
▶️Answer/Explanation
(a) Total arrangements:
\[ \frac{9!}{2! \times 2!} = 90720 \]
(b) Arrangements with C at each end and Os not together:
Fix C at both ends: \( \frac{7!}{2!} = 2520 \) total arrangements.
Subtract arrangements where Os are together: \( 6! = 720 \).
Total valid arrangements: \( 2520 – 720 = 1800 \).
(c) Selections where number of Cs ≠ number of Os:
Total selections: \( \binom{9}{4} = 126 \).
Subtract cases where number of Cs = number of Os (0 Cs and 0 Os, or 1 C and 1 O):
\[ 126 – \left(\binom{5}{4} + \binom{2}{1} \times \binom{2}{1} \times \binom{5}{2}\right) = 126 – (5 + 4 \times 10) = 126 – 45 = 81 \]
(d) Groupings with Cs in different groups:
Total ways to divide into groups: \( \frac{\binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}}{3!} = 280 \).
Subtract cases where both Cs are in the same group: \( \binom{7}{1} \times \binom{6}{3} \times \binom{3}{3} / 2! = 70 \).
Total valid groupings: \( 280 – 70 = 210 \).
7. [Maximum mark: 9]
Hanna buys 12 hollow chocolate eggs that each contain a sweet. The eggs look identical but Hanna knows that 3 contain a red sweet, 4 contain an orange sweet, and 5 contain a yellow sweet. Each of Hanna’s three children in turn randomly chooses and eats one of the eggs, keeping the sweet it contained.
(a) Find the probability that all 3 eggs chosen contain the same colour sweet.
(b) Find the probability that all 3 eggs chosen contain a yellow sweet, given that all three children have the same colour sweet.
(c) Find the probability that at least one of Hanna’s three children chooses an egg that contains an orange sweet.
▶️Answer/Explanation
(a) Probability all same colour:
\[ P(\text{all red}) = \frac{3}{12} \times \frac{2}{11} \times \frac{1}{10} = \frac{6}{1320} \]
\[ P(\text{all orange}) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} = \frac{24}{1320} \]
\[ P(\text{all yellow}) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} \]
Total probability: \( \frac{6 + 24 + 60}{1320} = \frac{90}{1320} = \frac{3}{44} \approx 0.0682 \).
(b) Conditional probability:
\[ P(\text{all yellow} \mid \text{same colour}) = \frac{P(\text{all yellow})}{P(\text{same colour})} = \frac{\frac{60}{1320}}{\frac{90}{1320}} = \frac{2}{3} \approx 0.667 \].
(c) Probability at least one orange:
\[ P(\text{no orange}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{336}{1320} \]
\[ P(\text{at least one orange}) = 1 – \frac{336}{1320} = \frac{984}{1320} = \frac{41}{55} \approx 0.745 \].