Worksheet Ch 8: Measures of Dispersion

Section A • Bahagian A [12 Marks]

Question 1 [4 Marks]

The marks obtained by 9 students in a Chemistry quiz are:

14, 18, 22, 25, 27, 29, 32, 36, 40

(a) Find the range of the marks. [1 mark]
(b) Determine the interquartile range (IQR). [3 marks]

(a) Range $= 40 - 14$

$= \mathbf{26}$

(b) Median $Q_2 = 27$. Lower half: $14, 18, 22, 25 \implies Q_1 = \frac{18 + 22}{2} = 20$.

Upper half: $29, 32, 36, 40 \implies Q_3 = \frac{32 + 36}{2} = 34$.

$$\text{IQR} = Q_3 - Q_1 = 34 - 20 = \mathbf{14}$$

K2 N1

Question 2 [4 Marks]

A dataset has mean $\bar{x} = 15$ and standard deviation $\sigma = 3.2$. Calculate the new mean and new standard deviation if each value is:

(a) Subtracted by 4. [2 marks]

(b) Multiplied by 3. [2 marks]

(a) Subtraction of constant: Mean changes, $\sigma$ is unaffected

New Mean $= 15 - 4 = \mathbf{11}$; New $\sigma = \mathbf{3.2}$

(b) Multiplication by constant 3: Both mean and $\sigma$ multiplied by 3

New Mean $= 15 \times 3 = \mathbf{45}$; New $\sigma = 3.2 \times 3 = \mathbf{9.6}$

Question 3 [4 Marks]

A box plot has lower quartile $Q_1 = 30$ and upper quartile $Q_3 = 46$. Determine whether the value $72$ is an outlier. Show your calculation. [4 marks]

Step 1: Calculate $\text{IQR} = 46 - 30 = 16$

Step 2: Upper boundary limit $= Q_3 + 1.5(\text{IQR}) = 46 + 1.5(16) = 46 + 24 = 70$

Step 3: Comparison and conclusion

Since $72 > 70$, 72 is an outlier (pencilan).

Section B • Bahagian B (HOTS / KBAT) [18 Marks]

Question 4 (Variance & Standard Deviation from $\sum x, \sum x^2$) [9 Marks]

The table summarizes the mass in kg of 10 players in a school badminton squad: $\sum x = 580$ and $\sum x^2 = 34090$.

(a) Calculate the mean and standard deviation of the mass of the players. [4 marks]
(b) A new player with a mass of $62\text{ kg}$ joins the squad. Determine the new standard deviation. [5 marks]

(a) Mean: $\bar{x} = \frac{580}{10} = 58\text{ kg}$

$$\sigma = \sqrt{\frac{34090}{10} - (58)^2} = \sqrt{3409 - 3364} = \sqrt{45} = \mathbf{6.71}\text{ kg}$$

K2 N2

(b) With 11 players: New $\sum x = 580 + 62 = 642$; New $\sum x^2 = 34090 + 62^2 = 34090 + 3844 = 37934$

New Mean $\bar{x}_{\text{new}} = \frac{642}{11} = 58.36\text{ kg}$

$$\sigma_{\text{new}} = \sqrt{\frac{37934}{11} - (58.36)^2} = \sqrt{3448.55 - 3405.89} = \sqrt{42.66} = \mathbf{6.53}\text{ kg}$$

K3 N2

Question 5 (Consistency Analysis Between Two Archery Competitors) [9 Marks]

Two archers, Haris and Kevin, each shot 8 arrows during training. Their scores are shown below:

Haris: 9, 8, 9, 7, 8, 9, 10, 8

Kevin: 10, 6, 10, 7, 9, 5, 10, 7

(a) Calculate the mean score of Haris and Kevin. [3 marks]
(b) Calculate the standard deviation of scores for both archers. [4 marks]
(c) Based on your calculations, decide who should be chosen to represent the school in the state championship. Justify your answer. [2 marks]

(a) Mean calculations

$$\text{Mean Haris} = \frac{68}{8} = \mathbf{8.5}$$

$$\text{Mean Kevin} = \frac{64}{8} = \mathbf{8.0}$$

(b) Standard deviation

Haris: $\sum x^2 = 584 \implies \sigma_H = \sqrt{\frac{584}{8} - 8.5^2} = \sqrt{73 - 72.25} = \sqrt{0.75} = \mathbf{0.866}$

Kevin: $\sum x^2 = 540 \implies \sigma_K = \sqrt{\frac{540}{8} - 8.0^2} = \sqrt{67.5 - 64} = \sqrt{3.5} = \mathbf{1.871}$

K2 N2

Haris should be chosen because his standard deviation is smaller ($\sigma_H = 0.866 < 1.871$), indicating that his shooting performance is more consistent (lebih konsisten), and his mean score is higher.

K1 N1