Measures of Dispersion (Part 2)

Posted on April 21, 2020 by user

Measures of Dispersion (Part 2)
7.2b Interquartile Range 1

(A) Interquartile Range of Ungrouped Data

Example 1:

Find the interquartile range of the following data.

(a) 7, 5, 1, 3, 6, 11, 8

(b) 12, 4, 6, 18, 9, 16, 2, 14

Solution:
(a) Rearrange the data according to ascending order.

Interquartile Range

= upper quartile – lower quartile

= 8 – 3

= 5

(b) Rearrange the data according to ascending order.

Interquartile Range

= upper quartile – lower quartile

\frac{14 + 16}{2} - \frac{4 + 6}{2}

= 15 – 5
= 10

(B) Interquartile Range of Grouped Data (without Class Interval)

Example 2:

The table below shows the marks obtained by a group of Form 4 students in school mid-term science test.

Determine the interquartile range of the distribution.

Solution:

Lower quartile, Q₁= the

\frac{1}{4} {(24)}^{th}

observation

= the 6^thobservation

= 2

Upper quartile, Q₃= the

\frac{3}{4} {(24)}^{th}

observation

= the 18^thobservation

= 4

Interquartile Range

= upper quartile – lower quartile

= Q₃– Q₁

= 4 – 2

= 2

Measures of Dispersion (Part 1)

Posted on April 21, 2020 by user

Measures of Dispersion (Part 1)

7.2a Range

(A) Range of Ungrouped Data

Example 1:
Find the range for the set of data 2, 4, 7, 10, 13, 16 and 18.

Solution:

Range = largest value – smallest value

= 18 – 2

= 16

(B) Range of Grouped Data (with Class Interval)

Example 2:

The grouped frequency distribution was obtained from 100 students regarding the scores in their test shown as below.

Find the range of the data.

Solution:

\begin{array}{l} Range \\ = midpoint of the highest class - midpoint of the lowest class \\ Range = \frac{35 + 39}{2} - \frac{5 + 9}{2} \\ = 37 - 7 \\ = 30 \end{array}

7.1c Median

Posted on April 21, 2020 by user

7.1c Median

1. The median of a group of data refers to the value which is at the middle of the data after the data has been arranged according to grouped data and ungrouped data.

(A) Ungrouped Data

Example 1:

Find the median for each of the sets of data given below.

(a) 15, 18, 21, 25, 20, 18

(b) 13, 6, 9, 17, 11

Solution:
(a) Arrange the data in the ascending order

15, 18, 18, 20, 21, 25

Median = T_{\frac{n + 1}{2}} = T_{\frac{6 + 1}{2}} = T_{3 \frac{1}{2}} = \frac{18 + 20}{2} = 19

(b) 6, 9, 11, 13, 17

Median = T_{\frac{n + 1}{2}} = T_{\frac{5 + 1}{2}} = T_{3} = 11

(B) Grouped Data (without Class Interval)

Example 2:

The frequency table shows the marks obtained by 40 students in a biology test.

Solution:

Median = T_{\frac{n + 1}{2}} = T_{\frac{40 + 1}{2}} = T_{20 \frac{1}{2}} = 60 \leftarrow (20 \frac{1}{2} th term is 60)

(C) Grouped Data (with Class Interval)

m = median

L = Lower boundary of median class

N = Number of data

F = Total frequency before median class

f_m = Total frequency in median class

c = Class size = (Upper boundary – lower boundary)

1. The median can be determined from an accumulative frequency table and the ogive.

2. The ogive is an accumulative graph; the median, quartiles and the range between quartiles can be determined from it.

Example 3:

The grouped frequency distribution was obtained from 100 students regarding the scores in their test shown as below.

Find the median.

Solution:
Method 1: using formula

\begin{array}{l} S t e p 1 \\ Median class is given by T_{\frac{n}{2}} = T_{\frac{100}{2}} = T_{50} \\ ∴ Median class is the class 20 - 24 \\ S t e p 2 \\ m = L + (\frac{\frac{N}{2} - F}{f_{m}}) c \\ m = 19.5 + (\frac{\frac{100}{2} - 33}{26}) 5 \\ m = 19.5 + 3.269 = 22.77 \end{array}

7.1b Mode

Posted on April 21, 2020 by user

7.1b Mode

Mode is the observation which has the highest frequency in a set of data.

Example:

Find the mode for each of the sets of data given below.

(a) 15, 18, 21, 25, 20, 18

(b) 3, 6, 9, 11, 17

(c) 0, 1, 2, 7, 3, 2, 1, 1, 2, 1, 2

Solution:

(a) The mode is 18. ← (18 occurs 2 times)

(b) Does not have a mode.

(c) The modes are 1 and 2. ← (Both numbers occur 4 times)

7.1a Mean

Posted on April 21, 2020 by user

7.1a Mean
1. The mean of the data is an average value obtained by using the formula:
Mean= Total data values Number of data

(A) Ungrouped Data

Example 1

(a) Find the mean for the set of data 2, 4, 7, 10, 13, 16 and 18.

(b) A value of xis added into the above set of data, the mean for this new data is 9.5. Determine the value of x.

Solution:
(a)

\begin{array}{l} \bar{x} = \frac{2 + 4 + 7 + 10 + 13 + 16 + 18}{7} \\ \bar{x} = \frac{70}{7} = 10 \end{array}

(b)

\begin{array}{l} New mean = 9.5 \\ \frac{70 + x}{8} = 9.5 \\ 70 + x = 76 \\ x = 6 \end{array}

(B) Grouped Data (Without Class Interval)

Example 2

The frequency table shows the marks obtained by 40 students in a Biology test.

Find the mean marks.

Solution:

\begin{array}{l} Mean marks, \bar{x} \\ \bar{x} = \frac{(50) (6) + (55) (8) + (60) (15) + (65) (10) + (70) (1)}{6 + 8 + 15 + 10 + 1} \\ \bar{x} = \frac{2360}{40} = 59 \end{array}

(C) Grouped Data (With Class Interval)

Example 3

The grouped frequency distribution was obtained from 100 students regarding the scores in their test shown as below.

Find the mean scores.

Solution:

\begin{array}{l} Mid-point = \frac{5 + 9}{2} = 7 \\ Mean scores, \bar{x} = \frac{\sum f x}{\sum f} = \frac{2290}{100} = 22.9 \end{array}

Long Questions (Question 4)

Posted on April 21, 2020 by user

Question 4:
Solutions by scale drawing will not be accepted.
Diagram below shows a triangle PRS. Side PR intersects the y-axis at point Q.

(a) Given PQ : QR = 2 : 3, find
(i) The coordinates of P,
(ii) The equation of the straight line PS,
(iii) The area, in unit², of triangle PRS.
(b) Point M moves such that its distance from point R is always twice its distance from point S.
Find the equation of the locus M.

Solution:
(a)(i)

\begin{array}{l} P = (\frac{2 (6) + 3 h}{2 + 3}, \frac{2 (12) + 3 k}{2 + 3}) \\ (0, 6) = (\frac{12 + 3 h}{5}, \frac{24 + 3 k}{5}) \\ \frac{12 + 3 h}{5} = 0 \\ 3 h = - 12 \\ h = - 4 \\ \frac{24 + 3 k}{5} = 6 \\ 3 k = 30 - 24 \\ k = 2 \\ P = (- 4, 2) \end{array}

(a)(ii)

\begin{array}{l} m_{P S} = \frac{2 - (- 6)}{- 4 - 2} \\ = - \frac{8}{6} \\ = - \frac{4}{3} \\ Equation of P S : \\ y - y_{1} = - \frac{4}{3} (x - 2) \\ y - (- 6) = - \frac{4}{3} x + \frac{8}{3} \\ 3 y + 18 = - 4 x + 8 \\ 3 y = - 4 x - 10 \end{array}

(a)(iii)

\begin{array}{l} Area of △ P R S \\ = \frac{1}{2} | \begin{array}{l} - 4      2       6 \\ 2 - 612 \end{array} \begin{array}{l} - 4 \\ 2 \end{array} | \\ = \frac{1}{2} | (24 + 24 + 12) - (4 - 36 - 48) | \\ = \frac{1}{2} | 60 - (- 80) | \\ = 70 {unit}^{2} \end{array}

(b)

\begin{array}{l} Let P = (x, y) \\ M R = 2 M S \\ \sqrt{{(x - 6)}^{2} + {(y - 12)}^{2}} = 2 \sqrt{{(x - 2)}^{2} + {(y + 6)}^{2}} \\ {(x - 6)}^{2} + {(y - 12)}^{2} = 4 [{(x - 2)}^{2} + {(y + 6)}^{2}] \\ x^{2} - 12 x + 36 + y^{2} - 24 y + 144 = 4 [x^{2} - 4 x + 4 + y^{2} + 12 y + 36] \\ x^{2} - 12 x + y^{2} - 24 y + 180 = 4 x^{2} - 16 x + 4 y^{2} + 48 y + 160 \\ 3 x^{2} + 3 y^{2} - 4 x + 72 y - 20 = 0 \end{array}

Long Questions (Question 3)

Posted on April 21, 2020 by user

Question 3:

The diagram shows a triangle LMN where L is on the y-axis. The equation of the straight line LKN and MK are 2y – 3x + 6 = 0 and 3y + x– 13 = 0 respectively. Find

(a) the coordinates of K

(b) the ratio LK:KN

Solution:
(a)

2y – 3x + 6 = 0 ----(1)

3y + x – 13 = 0 ----(2)

x = 13 – 3y ----(3)

Substitute equation (3) into (1),

2y – 3 (13 – 3y) + 6 = 0

2y – 39 + 9y + 6 = 0

11y = 33

y = 3

Substitute y = 3 into equation (3),

x = 13 – 3 (3)

x = 4

Coordinates of K = (4, 3).

(b)

Given equation of LKN is 2y – 3x + 6 = 0

At y – axis, x = 0,

x coordinates of point L = 0.

\begin{array}{l} Ratio L K : K N \\ Equating the x coordinates, \\ \frac{L K (10) + K N (0)}{L K + K N} = 4 \\ 10 L K = 4 L K + 4 K N \\ 6 L K = 4 K N \\ \frac{L K}{K N} = \frac{4}{6} \\ \frac{L K}{K N} = \frac{2}{3} \\ Ratio L K : K N = 2 : 3 \end{array}

Long Questions (Question 2)

Posted on April 21, 2020 by user

Question 2:

In the diagram, PRS and QRT are straight lines. Given R is the midpoint of PS and
QR : RT = 1 : 3, Find

(a) the coordinates of R,

(b) the coordinates of T,

(c) the coordinates of the point of intersection between lines PQ and ST produced.

Solution:
(a)

Given R is the midpoint of PS.

\begin{array}{l} R = (\frac{3 + 7}{2}, \frac{2 + 6}{2}) \\ R = (5, 4) \end{array}

(b)

\begin{array}{l} Q R : R T = 1 : 3 \\ Lets coordinates of T = (x, y) \\ (\frac{(1) (x) + (3) (4)}{1 + 3}, \frac{(1) (y) + (3) (5)}{1 + 3}) = (5, 4) \\ \frac{x + 12}{4} = 5 \\ x + 12 = 20 \\ x = 8 \\ \frac{y + 15}{4} = 4 \\ y + 15 = 16 \\ y = 1 \\ T = (8, 1) \end{array}

(c)

\begin{array}{l} Gradient of P Q = \frac{5 - 2}{4 - 3} = 3 \\ Equation of P Q, \\ y - 2 = 3 (x - 3) \\ y - 2 = 3 x - 9 \\ y = 3 x - 7 - - - - (1) \\ Gradient of S T = \frac{6 - 1}{7 - 8} = - 5 \\ Equation of S T, \\ y - 1 = - 5 (x - 8) \\ y - 1 = - 5 x + 40 \\ y = - 5 x + 41 - - - - (2) \end{array}

Substitute (1) into (2),

3x – 7 = –5x + 41

8x = 48

x = 6

From (1),

y = 3(6) – 7 = 11

The coordinates of the point of intersection between lines PQ and ST = (6, 11).

Long Questions (Question 1)

Posted on April 21, 2020 by user

Question 1:

The diagram shows a trapezium PQRS. Given the equation of PQ is 2y – x – 5 = 0, find

(a) The value of w,

(b) the equation of PS and hence find the coordinates of P.

(c) The locus of M such that triangle QMS is always perpendicular at M.

Solution:
(a)

\begin{array}{l} Equation of P Q \\ 2 y - x - 5 = 0 \\ 2 y = x + 5 \\ y = \frac{1}{2} x + \frac{5}{2} \\ ∴ m_{P Q} = \frac{1}{2} \\ In a trapizium, m_{P Q} = m_{S R} \\ \frac{1}{2} = \frac{0 - (- 3)}{w - 4} \\ w - 4 = 6 \\ w = 10 \end{array}

(b)

\begin{array}{l} m_{P Q} = \frac{1}{2} \\ m_{P S} = - \frac{1}{m_{P Q}} = - \frac{1}{\frac{1}{2}} = - 2 \end{array}

Point S = (4, –3), m = –2

y – y₁ = m (x– x₁)

y – (–3) = –2 (x – 4)

y + 3 = –2x + 8

y = –2x + 5

Equation of PS is y = –2x + 5

PS is y = –2x + 5-----(1)

PQ is 2y = x + 5-----(2)

Substitute (1) into (2)

2 (–2x + 5) = x + 5

–4x + 10 = x + 5

–5x = –5

x = 1

From (1), y = –2(1) + 5

y = 3

Coordinates of point P = (1, 3).

(c)

\begin{array}{l} Let M = (x, y) \\ Given that △ Q M S is perpendicular at M \\ Thus △ Q M S = 90^{\circ} \\ (m_{Q M}) (m_{M S}) = - 1 \\ (\frac{y - 5}{x - 5}) (\frac{y - (- 3)}{x - 4}) = - 1 \\ (y - 5) (y + 3) = - 1 (x - 5) (x - 4) \\ y^{2} + 3 y - 5 y - 15 = - 1 (x^{2} - 4 x - 5 x + 20) \\ y^{2} - 2 y - 15 = - x^{2} + 9 x - 20 \\ x^{2} + y^{2} - 9 x - 2 y + 5 = 0 \end{array}

Hence, the equation of locus of the moving point M is

x² + y²– 9x – 2y + 5 = 0.

Short Questions (Question 6 & 7)

Posted on April 21, 2020 by user

Question 6:

The point M is (–3, 5) and the point N is (4, 7). The point P moves such that PM: PN = 2: 3. Find the equation of the locus of P.

Solution:

\begin{array}{l} Let P = (x, y) \\ P M : P N = 2 : 3 \\ \frac{P M}{P N} = \frac{2}{3} \\ 3 P M = 2 P N \\ 3 \sqrt{{(x - (- 3))}^{2} + {(y - 5)}^{2}} = 2 \sqrt{{(x - 4)}^{2} + {(y - 7)}^{2}} \end{array}

Square both sides to eliminate the square roots.

9[x² + 6x + 9 + y² – 10y + 25] = 4 [x²– 8x + 16 + y² – 14y + 49]

9x² + 54x + 9y² – 90y + 306 = 4x² – 32x + 4y² – 56y + 260

5x² + 5y²+ 86x – 34y + 46 = 0

Hence, the equation of the locus of point P is

5x² + 5y² + 86x – 34y + 46 = 0

Question 7:

Given the points A(0, 2) and B (6, 5). Find the equation of the locus of a moving point P such that the triangle APB always has a right angle at P.

Solution:

Let P = (x, y)

Given that triangle APB= 90^o, thus AP is perpendicular to PB.

Hence, (m_AP)(m_PB) = –1.

(m_AP)(m_PB) = –1

(\frac{y - 2}{x - 0}) (\frac{y - 5}{x - 6}) = - 1

(y – 2)(y – 5) = – x(x – 6)

y² – 7y + 10 = –x² + 6x

y² + x² – 6x – 7y + 10 = 0

Hence, the equation of the locus of point P is

y² + x² – 6x – 7y + 10 = 0.