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Secondary 4 Elementary Mathematics Statistics Probability Quiz

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Secondary 4 Elementary Mathematics Quiz - Statistics Probability

Name: __________________________
Class: __________________________
Date: __________________________
Score: ______ / 60

Duration: 60 Minutes
Total Marks: 60

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all necessary working clearly; no marks will be given for correct answers without working.
Give non-exact numerical answers correct to 3 significant figures, unless one decimal place is specified for money.
The use of an approved scientific calculator is expected.

Section A: Data Analysis and Measures of Spread (Questions 1–5)

[15 Marks]

1. The heights, in cm, of 8 students are recorded as follows: $155, \quad 162, \quad 158, \quad 170, \quad 165, \quad 158, \quad 172, \quad 160$

(a) Find the median height.
 Answer: _______________ cm [1]

(b) Find the interquartile range of the heights.
 Answer: _______________ cm [2]

2. A set of data consists of five numbers: $4, 7, x, 12, 15$ . The mean of this data set is $9$ .

(a) Find the value of $x$ .
 Answer: $x =$ _______________ [1]

(b) Hence, find the variance of this data set.
 Answer: _______________ [2]

3. The table below shows the distribution of marks obtained by 40 students in a Mathematics test.

Marks ( $x$ )	10	20	30	40	50
Frequency ( $f$ )	4	8	12	10	6

(a) Calculate the mean mark.
 Answer: _______________ [2]

(b) Calculate the standard deviation of the marks.
 Answer: _______________ [3]

4. Two sets of data, Set A and Set B, have the following statistics:

Set A: Mean = 50, Standard Deviation = 5
Set B: Mean = 50, Standard Deviation = 12

(a) Which set of data is more consistent? Give a reason for your answer.
 Answer: _______________
Reason: __________________________________________________________ [2]

(b) If every value in Set A is multiplied by 3, what is the new standard deviation?
 Answer: _______________ [1]

5. The cumulative frequency table below shows the time taken, $t$ minutes, by 50 runners to complete a race.

Time ( $t$ min)	$t \le 10$	$t \le 20$	$t \le 30$	$t \le 40$	$t \le 50$
Cumulative Frequency	5	18	35	45	50

(a) How many runners took between 20 and 30 minutes?
 Answer: _______________ [1]

(b) Estimate the median time using linear interpolation or a cumulative frequency curve.
 Answer: _______________ min [2]

Section B: Probability and Combined Events (Questions 6–12)

[25 Marks]

6. A bag contains 5 red balls, 3 blue balls, and 2 green balls. A ball is drawn at random.

(a) Find the probability that the ball is not blue.
 Answer: _______________ [1]

(b) Find the probability that the ball is red or green.
 Answer: _______________ [1]

7. Two fair six-sided dice are thrown. Let $S$ be the sum of the scores on the two dice.

(a) Draw a possibility diagram (sample space) to show all possible outcomes.
 [2]

(b) Find the probability that $S$ is a prime number.
 Answer: _______________ [2]

8. Events $A$ and $B$ are defined such that $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cap B) = 0.2$ .

(a) Are events $A$ and $B$ mutually exclusive? Give a reason.
 Answer: _______________
Reason: __________________________________________________________ [1]

(b) Find $P(A \cup B)$ .
 Answer: _______________ [1]

9. A box contains 4 white chocolates and 6 dark chocolates. Two chocolates are chosen at random without replacement.

(a) Complete the tree diagram below by filling in the probabilities.

       1st Choice          2nd Choice
      /        \          /          \
     W          D        W            D
    / \        / \      / \          / \
   W   D      W   D    W   D        W   D

[2]

(b) Find the probability that both chocolates are white.
 Answer: _______________ [2]

10. The probability that it rains on any given day in November is $0.3$ . The weather on consecutive days is independent.

(a) Find the probability that it rains on both Monday and Tuesday.
 Answer: _______________ [1]

(b) Find the probability that it rains on at least one of the two days.
 Answer: _______________ [2]

11. In a class of 30 students, 18 study Physics, 15 study Chemistry, and 5 study neither subject.

(a) Represent this information on a Venn diagram.
 [2]

(b) A student is chosen at random. Find the probability that the student studies both Physics and Chemistry.
 Answer: _______________ [2]

(c) Given that a student studies Physics, find the probability that they also study Chemistry.
 Answer: _______________ [2]

12. A spinner has 4 equal sectors labeled 1, 2, 3, and 4. The spinner is spun twice.

(a) List the sample space for the product of the two scores.
 [2]

(b) Find the probability that the product is an even number.
 Answer: _______________ [2]

Section C: Applications and Interpretation (Questions 13–20)

[20 Marks]

13. The masses of 100 apples are summarized in the following frequency table.

Mass ( $m$ grams)	$80 < m \le 100$	$100 < m \le 120$	$120 < m \le 140$	$140 < m \le 160$
Frequency	15	35	30	20

(a) State the modal class.
 Answer: _______________ [1]

(b) Calculate an estimate of the mean mass.
 Answer: _______________ g [3]

14. The box-and-whisker plot below summarizes the test scores of Class A.

Minimum: 20
Lower Quartile ( $Q_1$ ): 45
Median ( $Q_2$ ): 60
Upper Quartile ( $Q_3$ ): 75
Maximum: 95

(a) Calculate the Interquartile Range (IQR) for Class A.
 Answer: _______________ [1]

(b) Class B has a median of 65 and an IQR of 20. Compare the performance of Class A and Class B.
 Answer: __________________________________________________________
_________________________________________________________________ [2]

15. A survey was conducted on 200 people regarding their preference for Tea ( $T$ ) or Coffee ( $C$ ).

120 people like Tea.
90 people like Coffee.
30 people like neither.

(a) Find the number of people who like both Tea and Coffee.
 Answer: _______________ [2]

(b) Find the probability that a randomly selected person likes Tea but not Coffee.
 Answer: _______________ [2]

16. The standard deviation of a set of 5 numbers is 4. If each number is increased by 3, what is the new standard deviation?
 Answer: _______________ [1]

17. A bag contains tickets numbered 1 to 10. One ticket is drawn. Let $E$ be the event that the number is even. Let $F$ be the event that the number is greater than 6.

(a) List the outcomes in event $E \cap F$ .
 Answer: _______________ [1]

(b) Find $P(E \cup F)$ .
 Answer: _______________ [2]

18. The heights of students in a school are normally distributed (conceptual check). If the mean height is 160 cm and the standard deviation is 10 cm, approximately what percentage of students have heights between 150 cm and 170 cm? (Assume standard normal distribution properties: $\pm 1\sigma \approx 68\%$ ).
 Answer: _______________ % [1]

19. Two events $X$ and $Y$ are independent. $P(X) = 0.6$ and $P(Y) = 0.4$ . Find $P(X' \cap Y')$ .
 Answer: _______________ [2]

20. The table shows the number of hours ( $h$ ) 10 students spent studying and their exam scores ( $s$ ).

Student	A	B	C	D	E	F	G	H	I	J
Hours ( $h$ )	2	4	1	5	3	6	2	4	5	3
Score ( $s$ )	40	60	30	75	50	80	45	65	70	55

(a) Without calculating, state whether you expect the correlation between hours studied and score to be positive, negative, or zero.
 Answer: _______________ [1]

(b) Calculate the mean number of hours studied.
 Answer: _______________ hours [1]

*** End of Quiz ***

Answers

Secondary 4 Elementary Mathematics Quiz - Statistics Probability (Answer Key)

1. Data: $155, 158, 158, 160, 162, 165, 170, 172$ (Ordered) (a) Median is average of 4th and 5th terms: $\frac{160+162}{2} = 161$ . Answer: 161 cm [1]

(b) $Q_1$ (median of lower half $155, 158, 158, 160$ ) = $\frac{158+158}{2} = 158$ . $Q_3$ (median of upper half $162, 165, 170, 172$ ) = $\frac{165+170}{2} = 167.5$ . $IQR = Q_3 - Q_1 = 167.5 - 158 = 9.5$ . Answer: 9.5 cm [2]

(c) Mean $\bar{x} = \frac{1290}{8} = 161.25$ . $\sum x^2 = 155^2 + ... + 172^2 = 208366$ . $\sigma = \sqrt{\frac{\sum x^2}{n} - (\bar{x})^2} = \sqrt{\frac{208366}{8} - 161.25^2} = \sqrt{26045.75 - 26001.5625} = \sqrt{44.1875} \approx 6.65$ . Answer: 6.65 cm [2]

2. (a) Mean $= \frac{4+7+x+12+15}{5} = 9 \Rightarrow 38+x = 45 \Rightarrow x=7$ . Answer: $x = 7$ [1]

(b) Data: $4, 7, 7, 12, 15$ . Mean $= 9$ . Variance $\sigma^2 = \frac{(4-9)^2 + (7-9)^2 + (7-9)^2 + (12-9)^2 + (15-9)^2}{5}$ $= \frac{25 + 4 + 4 + 9 + 36}{5} = \frac{78}{5} = 15.6$ . Answer: 15.6 [2]

3. (a) $\sum fx = 10(4)+20(8)+30(12)+40(10)+50(6) = 40+160+360+400+300 = 1260$ . Mean $= \frac{1260}{40} = 31.5$ . Answer: 31.5 [2]

(b) $\sum fx^2 = 100(4)+400(8)+900(12)+1600(10)+2500(6) = 400+3200+10800+16000+15000 = 45400$ . $\sigma = \sqrt{\frac{45400}{40} - 31.5^2} = \sqrt{1135 - 992.25} = \sqrt{142.75} \approx 11.9$ . Answer: 11.9 [3]

4. (a) Set A is more consistent because it has a smaller standard deviation (less spread). Answer: Set A; Lower SD indicates less variability. [2]

(b) New SD $= 3 \times \text{Old SD} = 3 \times 5 = 15$ . Answer: 15 [1]

5. (a) $CF(30) - CF(20) = 35 - 18 = 17$ . Answer: 17 [1]

(b) Median is at $\frac{50}{2} = 25$ th value. This lies in the interval $20 < t \le 30$ . Lower boundary $= 20$ , Frequency in interval $= 35-18=17$ , Cumulative freq before $= 18$ . Median $= 20 + \left( \frac{25-18}{17} \right) \times 10 = 20 + \frac{70}{17} \approx 20 + 4.12 = 24.1$ . Answer: 24.1 min [2]

6. Total balls $= 10$ . (a) Not Blue $= 5 \text{ (Red)} + 2 \text{ (Green)} = 7$ . $P = \frac{7}{10}$ . Answer: $\frac{7}{10}$ or 0.7 [1]

(b) Red or Green $= 7$ . $P = \frac{7}{10}$ . Answer: $\frac{7}{10}$ or 0.7 [1]

7. (a) Diagram should show 36 outcomes (1,1) to (6,6). [2]

(b) Prime sums: 2, 3, 5, 7, 11. Outcomes: 2: (1,1) -> 1 3: (1,2), (2,1) -> 2 5: (1,4), (2,3), (3,2), (4,1) -> 4 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) -> 6 11: (5,6), (6,5) -> 2 Total favorable $= 1+2+4+6+2 = 15$ . $P = \frac{15}{36} = \frac{5}{12}$ . Answer: $\frac{5}{12}$ [2]

(c) Sum $> 9$ : 10, 11, 12. 10: (4,6), (5,5), (6,4) -> 3 11: (5,6), (6,5) -> 2 12: (6,6) -> 1 Total $= 6$ . $P = \frac{6}{36} = \frac{1}{6}$ . Answer: $\frac{1}{6}$ [1]

8. (a) No, because $P(A \cap B) = 0.2 \neq 0$ . Mutually exclusive events cannot happen together. Answer: No; Intersection is not zero. [1]

(b) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$ . Answer: 0.7 [1]

9. Total 10 chocolates (4W, 6D). (a) Tree: 1st W: 4/10. 2nd W: 3/9, 2nd D: 6/9. 1st D: 6/10. 2nd W: 4/9, 2nd D: 5/9. [2]

(b) $P(WW) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$ . Answer: $\frac{2}{15}$ [2]

(c) $P(WD) + P(DW) = (\frac{4}{10} \times \frac{6}{9}) + (\frac{6}{10} \times \frac{4}{9}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$ . Answer: $\frac{8}{15}$ [2]

10. $P(R) = 0.3, P(NR) = 0.7$ . (a) $P(RR) = 0.3 \times 0.3 = 0.09$ . Answer: 0.09 [1]

(b) $P(\text{At least one}) = 1 - P(\text{None}) = 1 - (0.7 \times 0.7) = 1 - 0.49 = 0.51$ . Answer: 0.51 [2]

11. Total 30. Neither 5. So $n(P \cup C) = 25$ . $n(P) + n(C) - n(P \cap C) = 25 \Rightarrow 18 + 15 - n(P \cap C) = 25 \Rightarrow 33 - 25 = 8$ . (a) Venn: Intersection 8. P only $18-8=10$ . C only $15-8=7$ . Outside 5. [2]

(b) $P(P \cap C) = \frac{8}{30} = \frac{4}{15}$ . Answer: $\frac{4}{15}$ [2]

12. (a) Sample space for product (16 outcomes): 1, 2, 3, 4 2, 4, 6, 8 3, 6, 9, 12 4, 8, 12, 16 [2]

(b) Even products: All except odd $\times$ odd. Odd numbers: 1, 3. Odd $\times$ Odd outcomes: (1,1), (1,3), (3,1), (3,3) -> 4 outcomes. Even outcomes $= 16 - 4 = 12$ . $P = \frac{12}{16} = \frac{3}{4}$ . Answer: $\frac{3}{4}$ [2]

13. (a) Highest frequency is 35. Answer: $100 < m \le 120$ [1]

(b) Midpoints: 90, 110, 130, 150. $\sum fx = 90(15) + 110(35) + 130(30) + 150(20) = 1350 + 3850 + 3900 + 3000 = 12100$ . Mean $= \frac{12100}{100} = 121$ . Answer: 121 g [3]

14. (a) $IQR = 75 - 45 = 30$ . Answer: 30 [1]

(b) Class B has a higher median (65 vs 60), suggesting generally higher scores. Class B has a smaller IQR (20 vs 30), suggesting more consistent performance than Class A. Answer: Class B performed better on average and was more consistent. [2]

15. Total 200. Neither 30. Union $= 170$ . $n(T \cup C) = n(T) + n(C) - n(T \cap C) \Rightarrow 170 = 120 + 90 - n(T \cap C) \Rightarrow n(T \cap C) = 40$ . (a) Answer: 40 [2]

(b) Like Tea only $= 120 - 40 = 80$ . $P = \frac{80}{200} = \frac{2}{5}$ or 0.4. Answer: 0.4 [2]

16. Adding a constant does not change the spread. Answer: 4 [1]

17. $S = \{1..10\}$ . $E = \{2,4,6,8,10\}$ . $F = \{7,8,9,10\}$ . (a) $E \cap F = \{8, 10\}$ . Answer: $\{8, 10\}$ [1]

(b) $E \cup F = \{2,4,6,7,8,9,10\}$ . Count = 7. $P = \frac{7}{10}$ . Answer: 0.7 [2]

18. Range 150 to 170 is $\mu - \sigma$ to $\mu + \sigma$ . Answer: 68% [1]

19. Independent. $P(X') = 0.4, P(Y') = 0.6$ . $P(X' \cap Y') = 0.4 \times 0.6 = 0.24$ . Answer: 0.24 [2]

20. (a) As hours increase, scores generally increase. Answer: Positive [1]

(b) Sum hours $= 2+4+1+5+3+6+2+4+5+3 = 35$ . Mean $= \frac{35}{10} = 3.5$ . Answer: 3.5 hours [1]