Median, Quartiles & Mode — Exercise 26(B)

A life insurance agent found the following data for distribution of ages of 100 policy holders:

On a graph sheet draw an ogive using the given data. Take 2 cm = 5 years along one axis and 2 cm = 10 policy holders along the other axis. Use your graph to find:

(a) The median age.

(b) Number of policy holders whose age is above 52 years.

Answer

Steps of construction :

1. Take 2 cm = 5 years on x-axis.

2. Take 1 cm = 10 policy holders on y-axis.

3. Plot the points (20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96) and (60, 100).

4. Join the points by a free-hand curve.

Here, n = 100

Median = $\dfrac{n}{2} = \dfrac{100}{2}$ = 50th term

(a) Through J = 50 draw a horizontal line to meet the ogive at K. Through K, draw a vertical line to meet the x-axis at L. The abscissa of the point L represents 42.

Hence, the median age = 42 years.

(b) Through M = 52 draw a vertical line to meet the ogive at N. Through N, draw a horizontal line to meet the y-axis at O. The ordinate of the point O represents 91.

Hence, 91 people have their age less than or equal to 52.

∴ No. of people whose age is greater than 52 = 100 - 91 = 9.

Hence, number of policy holders whose age is above 52 years equal to 9.

The marks of 200 students in a test were recorded as follows :

Using graph sheet draw ogive for the given data and use it to find the,

(a) median

(b) number of students who obtained more than 65% marks

(c) number of students who did not pass, if the pass percentage was 35.

Answer

Steps :

1. Take 1 cm = 10 marks on x-axis.

2. Take 1 cm = 20 students on y-axis.

3. Plot the points (10, 5), (20, 12), (30, 23), (40, 43), (50, 83), (60, 135), (70, 171), (80, 186), (90, 195) and (100, 200).

4. Join the points by free hand curve.

(a) n = 200, which is even

Median = $\dfrac{n}{2} = \dfrac{200}{2}$ = 100^th term.

Through point L = 100 draw a horizontal line parallel to x-axis touching the graph at point M, through M draw a vertical line parallel to y-axis touching x-axis at point N = 53.

Hence, median = 53.

(b) Total marks = 100

65% of 100 = 65

Through point O = 65 draw a vertical line parallel to y-axis touching the graph at point P, through P draw a horizontal line parallel to x-axis touching y-axis at point Q = 154.

∴ 154 students score less than or equal to 65%.

∴ 46 (200 - 154) students score more than 65%.

Hence, 46 students score more than 65%.

(c) Total marks = 100

35% of 100 = 35

Through point R = 35 draw a vertical line parallel to y-axis touching the graph at point S, through S draw a horizontal line parallel to x-axis touching y-axis at point T = 33.

Hence, 33 students did not pass the exam.

Marks obtained by 200 students in an examination are given below:

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on other axis. From the graph, find:

(i) the median

(ii) the upper-quartile

(iii) number of students scoring more than 65 marks

(iv) if 10 students qualify for merit-scholarship, find the minimum marks required to qualify.

Answer

Cumulative frequency distribution table :

Here, n = 200, which is even.

Steps of construction:

1. Take 1 cm along x-axis = 10 marks

2. Take 2 cm along y-axis = 20 students

3. Plot the point (0, 0) as ogive starts from x- axis representing lower limit of first class.

4. Plot the points (10, 5), (20, 15), (30, 29), (40, 50), (50, 75), (60, 109), (70, 145), (80, 172), (90, 188), (100, 200).

5. Joint the points by a free hand curve.

(i) To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{200}{2}$ = 100.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the points M represents 57.5.

Hence, the median is 57.5.

(ii) By formula,

$\text{Upper quartile} = \Big(\dfrac{3\text{n}}{4}\Big)^ \text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = \Big(\dfrac{(3 \times 200)}{4}\Big)^\text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = \dfrac{600}{4}^\text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = 150^\text{ th}\text{ term} \\[1em]$

Draw a line parallel to x-axis from point R (Number of students) = 150, touching the graph at point Q. From point Q draw a line parallel to y-axis touching x-axis at point N.

From graph,

N = 72

Hence, the upper quartile is 72.

(iii) Total marks = 100.

Let E be the point on x-axis representing marks = 65.

Through E draw a vertical line to meet the ogive at B. Through B, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 128.

Hence, 128 students score less than or equal to 65, so, students scoring more than 65 = 200 - 128 = 72.

Hence, the number of students who scored more than 65 marks is 72.

(iv) Given, 10 students qualify for merit-scholarship. This corresponds to 190th student. Since, 200 - 10 = 190

Draw a line parallel to x-axis from point F = 190, touching the graph at point G. From point G draw a line parallel to y-axis touching x-axis at point S.

From graph,

S = 91

Hence, the minimum marks required to qualify is 91.

The table below shows the distribution of the scores obtained by 120 shooters in shooting competition. Using a graph sheet, draw an ogive for the distribution.

Use your ogive to estimate :

(i) the median

(ii) the inter-quartile range

(iii) the number of shooters who obtained more than 75% score.

Answer

Cumulative frequency distribution table :

Here, n = 120, which is even.

(i) Steps of construction:

1. Take 1 cm along x-axis = 10 scores

2. Take 2 cm along y-axis = 20 shooters

3. Plot the point (0, 0) as ogive starts from x- axis representing lower limit of first class.

4. Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120)

5. Joint the points by a free hand curve.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{120}{2}$ = 60.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the points M represents 43.

Hence, the median is 43.

(ii) To find lower quartile:

Let B be the point on y-axis representing frequency = $\dfrac{\text{n}}{4} = \dfrac{120}{4}$ = 30.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 30.

To find upper quartile:

Let C be the point on y-axis representing frequency = $\dfrac{3\text{n}}{4} = \dfrac{3 \times 120}{4} = \dfrac{360}{4}$ = 90.

Through C, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at S. The abscissa of the point S represents 56.

Inter-quartile range = Upper quartile - Lower quartile = 56 - 30 = 26

Hence, the inter quartile range is = 26.

(iii) Total score = 100

So, more than 75% score mean more than 75 score.

Let T be the point on x-axis representing scores = 75

Through T, draw a vertical line to meet the ogive at G. Through G, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 110

Shooters who have scored less than 75% = 110

So, students scoring more than 75% = Total students - Students who have scored less = 120 - 110 = 10

Hence, there are 10 number of shooters who obtained more than 75% score.

The daily wages of 80 workers in a project are given below:

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = ₹ 50 on x-axis and 2 cm = 10 workers on y-axis). Use your ogive to estimate:

(i) the median wage of the workers.

(ii) the lower quartile wage of the workers.

(iii) the number of workers who earn more than ₹ 625 daily.

Answer

Cumulative frequency distribution table :

Here, n = 80, which is even.

Steps of construction:

1. Take 2 cm along x-axis = ₹ 50

2. Take 2 cm along y-axis = 10 workers

3. Since, scale on x-axis starts at 400, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 400.

4. Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75), (750, 80) representing upper class limits and the respective cumulative frequencies. Also plot the point representing lower limit of the first class i.e, 400 - 450.

5. Joint the points by a free hand curve.

(i) Here, n = 80

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{80}{2}$ = 40.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the points M represents 600.5.

Hence, the median is ₹ 600.5.

(ii) To find lower quartile:

Let B be the point on y-axis representing frequency = $\dfrac{\text{n}}{4} = \dfrac{80}{4}$ = 20.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 550.

Hence, lower quartile wage = ₹ 550.

(iii) Let T be the point on x-axis representing wage = ₹ 625.

Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at C. The ordinate of the point C. The ordinate of point C represents 51.

Workers who earn less than ₹ 625 = 51.

So, workers earning more than ₹ 625 = Total workers - workers who earn less than ₹ 625 = 80 - 51 = 29.

Hence, there are 29 workers earning more than ₹ 625 daily.

Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students.

Use your ogive to estimate the following :

(i) the percentage of students weighing 55 kg or more

(ii) the weight above which the heaviest 30% of the students fall

(iii) the number of students who are (a) under weight and (b) Over-weight, if 55.70 kg is considered as standard weight.

Answer

Cumulative frequency distribution table :

Here, n = 200, which is even.

Steps of construction:

1. Take 2 cm along x-axis = 5 kg

2. Take 2 cm along y-axis = 20 units.

3. Since, scale on x-axis starts at 40, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 40.

4. Plot the point (40, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200).

6. Join the points by a free hand curve.

7. Draw a line parallel to y-axis from point J(weight) = 55, touching the graph at point Q. From point Q draw a line parallel to x-axis touching y-axis at point K.

From graph, K = 44.

Hence, 44 students weight 55 kg or less.

Students weighing more than 55 kg = 200 - 44 = 156

Percentage of students weighing more than 55 kg = $\dfrac{156}{200} \times 100$ = 78%

Hence, percentage of students weighing more than 55 kg = 78%.

(ii) 30% of students = $\dfrac{30}{100} \times 200$ = 60.

Total students = 200

No. of students not in heaviest 30% = 200 - 60 = 140.

Draw a line parallel to x-axis from point I (no. of students) = 140, touching the graph at point R. From point R draw a line parallel to y-axis touching x-axis at point P.

From graph, P = 65

Hence, above 65 kg the heaviest 30% of the students fall.

(iii) Draw a line parallel to y-axis from point L (weight) = 55.70 kg, touching the graph at point M. From point M draw a line parallel to x-axis touching y-axis at point N.

(a) From graph,

N = 50.

∴ 50 students have weight less than 55.70 kg

Hence, 50 students are underweight.

(b) Since, 50 students have weight less than 55.70 kg

∴ 150 (200 - 50) students have weight more than 55.70 kg.

Hence, 150 students are overweight.

Using a graph paper, draw an ogive for the distribution which shows the marks obtained on the General knowledge paper by 100 students.

Use the ogive to estimate:

(i) the median

(ii) the number of students whose score is above 65.

Answer

Cumulative frequency distribution table :

Here, n = 100, which is even.

(i) Steps of construction:

1. Take 1 cm along x-axis = 10 marks

2. Take 2 cm along y-axis = 20 students

3. Plot the point (0, 0) as ogive starts from x- axis representing lower limit of first class.

4. Plot the points (10, 5), (20, 15), (30, 35), (40, 60), (50, 75), (60, 87), (70, 96), (80, 100).

5. Joint the points by a free hand curve.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{100}{2}$ = 50.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the points M represents 36.

Hence, the median is 36.

(ii) Total marks = 100.

Let E be the point on x-axis representing marks = 65.

Through E draw a vertical line to meet the ogive at B. Through B, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 93.

Hence, 93 students score less than or equal to 65, so, students scoring more than 65 = 100 - 93 = 7.

Hence, the number of students who scored more than 65 marks is 7.

The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution (take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis.)

Use your graph to estimate the following:

(i) the median

(ii) the inter-quartile range

(iii) the number of shooters who obtained a score of more than 85%

Answer

Cumulative frequency distribution table :

Here, n = 160, which is even.

(i) Steps of construction:

1. Take 1 cm along x-axis = 10 scores

2. Take 2 cm along y-axis = 20 shooters

3. Plot the point (0, 0) as ogive starts from x- axis representing lower limit of first class.

4. Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153), (100, 160)

5. Joint the points by a free hand curve.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{160}{2}$ = 80.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the points M represents 44.

Hence, the median score is 44.

(ii) To find lower quartile:

Let B be the point on y-axis representing frequency = $\dfrac{\text{n}}{4} = \dfrac{160}{4}$ = 40.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 29.

To find upper quartile:

Let C be the point on y-axis representing frequency = $\dfrac{3\text{n}}{4} = \dfrac{3 \times 160}{4} = \dfrac{480}{4}$ = 120.

Through C, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at S. The abscissa of the point S represents 60.

Inter-quartile range = Upper quartile - Lower quartile = 60 - 29 = 31.

Hence, the inter quartile range is = 31.

(iii) Total score = 100

So, more than 85% score mean more than 85 score.

Let T be the point on x-axis representing scores = 85

Through T, draw a vertical line to meet the ogive at E. Through E, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 149.

Shooters who have scored less than 85% = 149

So, students scoring more than 85% = Total students - Students who have scored less = 160 - 149 = 11.

Hence, there are 11 number of shooters who obtained more than 85% score.

A survey regarding height (in cm) of 60 boys belonging to class 10 of a school was conducted. The following data was recorded:

Taking 2 cm = height of 10 cm along one axis and 2 cm = 10 boys along the other axis, draw an ogive of the above distribution. Use the graph to estimate the following :

(i) the median

(ii) the lower quartile

(iii) if above 158 cm is considered as the tall boys of the class, find the number of boys in the class who are tall.

Answer

The cumulative frequency table for the given continuous distribution is :

1. Take 2 cm along x-axis = 5 cm (height)

2. Take 2 cm along y-axis = 10 (No. of boys)

3. Since, scale on x-axis starts at 135, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 135.

4. Plot the points (140, 4), (145, 12), (150, 32), (155, 46), (160, 53), (165, 59) and (170, 60) representing upper class limits and the respective cumulative frequencies.

Also plot the point (135, 0) representing lower limit of the first class i.e. 135 - 140.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 60.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{60}{2}$ = 30.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 149.

Hence, the median height = 149 cm.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\dfrac{\text{n}}{4} = \dfrac{60}{4}$ = 15.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 146.

Hence, lower quartile = 146 cm.

(iii) Let S be the point on x-axis representing height = 158 cm.

Through S, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 51.

No. of boys shorter than 158 cm = 51

So, no. of boys taller than 158 cm = Total boys - boys shorter than 158 cm = 60 - 51 = 9.

Hence, there are 9 tall boys in the class.

40 students enter for a game of shot put competition. The distance thrown (in metres) is recorded below.

Use a graph paper to draw an ogive for the above distribution.

Use a scale of 2 cm = 1 m on one axis and 2 cm = 5 students on the other axis. Hence using your graph, find

(i) the median

(ii) Upper quartile

(iii) Number of students who cover a distance which is above 16 $\dfrac{1}{2}$ m.

Answer

Cumulative frequency distribution table :

Steps of construction:

1. Since, the scale on x-axis starts at 12, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 12.

2. Take 2 cm along x-axis = 1 m.

3. Take 2 cm along y-axis = 5 students.

4. Plot the point (12, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39) and (19, 40).

6. Join the points by a free hand curve.

(i) The total number of students is N = 40. The median position is found at $\dfrac{\text{N}}{2} = \dfrac{40}{2}$ = 20.

Draw a line parallel to x-axis from point A (number of students) = 20, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 14.6

Hence, the median = 14.6 m.

(ii) Here, n = 40, which is even.

By formula,

Upper quartile = $\dfrac{3\text{N}}{4} = \dfrac{3 \times 40}{4} = \dfrac{120}{4}$ = 30.

Draw a line parallel to x-axis from point J (number of students) = 30, touching the graph at point K. From point K draw a line parallel to y-axis touching x-axis at point L.

From graph, L = 15.6

Hence, the upper quartile = 15.6 m.

(iii) Draw a line parallel to y-axis from point D (Distance) = $16\dfrac{1}{2}$ m = 16.5 m, touching the graph at point E. From point E draw a line parallel to x-axis touching y-axis at point F.

From graph, F = 35.

It means there are 35 students who cover a distance either less or equal to $16\dfrac{1}{2}$ m.

Number of student who cover a distance which is above $16\dfrac{1}{2}$ m = 40 - 35 = 5.

Hence, number of students who cover a distance above $16\dfrac{1}{2}$ m = 5.

Use graph paper to answer this question:

During a medical checkup of 60 students in a school, weights were recorded as follows:

Taking 2 cm = 2 kg along one axis and 2 cm = 10 students along the other axis, draw an ogive. Use your graph to find the:

(i) Median

(ii) Upper quartile

(iii) Number of students whose weight is above 37 kg

Answer

Cumulative frequency distribution table :

Here, n = 60, which is even.

(a) Median = $\dfrac{\text{n}}{2} \text{ th term} = \dfrac{60}{2}$ = 30th term.

Steps of construction :

1. Take 2 cm = 2 kg on x-axis.

2. Take 2 cm = 10 students on y-axis.

3. Since, x axis starts at 28 hence, a kink is drawn at the starting of x-axis. Plot the point (28, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58) and (44, 60).

5. Join the points by a free-hand curve.

6. Draw a line parallel to x-axis from point A (no. of students) = 30, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 36.2

Hence, median = 36.2 kg

(ii) Here, n = 60, which is even.

By formula,

Upper quartile = $\dfrac{3\text{n}}{4} = \dfrac{3 \times 60}{4} = \dfrac{180}{4}$ = 45 th term.

Draw a line parallel to x-axis from point D (no. of students) = 45, touching the graph at point E. From point E draw a line parallel to y-axis touching x-axis at point F.

From graph, F = 38.2 kg

Hence, upper quartile = 38.2 kg.

(iii) Draw a line parallel to y-axis from point G (weight) = 37 kg, touching the graph at point H. From point H draw a line parallel to x-axis touching y-axis at point I.

From graph, I = 36.

∴ 36 students have weight less than or equal to 36 kg.

No. of students whose weight is more than 36 kg = 60 - 36 = 24.

Hence, no. of students whose weight is more than 36 kg = 24.