Measures of Central Tendency

Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.

Answer

The sum of terms = 5.7 + 6.6 + 7.2 + 9.3 + 6.2 = 35

Number of terms = 5

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No. of terms}} = \dfrac{∑x_i}{n}$

$\therefore A.M. = \dfrac{35}{5} = 7.$

Hence, the mean of 5.7, 6.6, 7.2, 9.3, 6.2 is 7.

The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20. Find :

(i) the mean of their marks.

(ii) the mean of their marks when the marks of each student are increased by 4.

(iii) the mean of their marks when 2 marks are deducted from the marks of each student.

(iv) the mean of their marks when the marks of each student are doubled.

Answer

(i) The sum of marks of all students = 12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20 = 207.

$\therefore \text{The mean of marks} = \dfrac{\text{Sum of marks of all students}}{\text{No. of students}} = \dfrac{∑x_i}{n} \\[1em] = \dfrac{207}{15} \\[1em] = 13.8$

Hence, the mean of marks of 15 students is 13.8

(ii) When the marks of each student are increased by 4, then the sum of their marks increases by 15 × 4 i.e. by 60.

∴ The new sum of marks of all students = 207 + 60 = 267.

$\therefore \text{The new mean of marks} = \dfrac{\text{ New sum of marks}}{\text{No. of students}} \\[1em] = \dfrac{267}{15} \\[1em] = 17.8$

Hence, the mean of marks of 15 students, when the marks of each student are increased by 4 is 17.8

(iii) When the marks of each student is decreased by 2, then the sum of their marks decreases by 15 × 2 i.e. by 30.

∴ The new sum of marks of all students = 207 - 30 = 177.

$\therefore \text{The new mean of marks} = \dfrac{\text{ New sum of marks}}{\text{No. of students}} \\[1em] = \dfrac{177}{15} \\[1em] = 11.8$

Hence, the mean of marks of 15 students, when the marks of each student is decreased by 2 is 11.8

(iv) When the marks of each student are doubled, then the sum of their marks will also be doubled.

∴ The new sum of marks of all students = 207 × 2 = 414.

$\therefore \text{The new mean of marks} = \dfrac{\text{ New sum of marks}}{\text{No.of students}} \\[1em] = \dfrac{414}{15} \\[1em] = 27.6$

Hence, the mean of marks of 15 students, when the marks of each student is doubled is 27.6

The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x.

Answer

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No. of terms}} = \dfrac{∑x_i}{n}$

$\therefore 8 = \dfrac{6 + y + 7 + x + 14}{5} \\[1em] \Rightarrow 8 = \dfrac{x + y + 27}{5} \\[1em] \Rightarrow 40 = x + y + 27 \\[1em] \Rightarrow x + y = 13 \\[1em] \Rightarrow y = 13 - x.$

Hence, the value of y = 13 - x.

The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15, find the 9th variate.

Answer

Let 9th variate be x.

Sum of terms = 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 + x = 89 + x

Number of terms = 9.

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No. of terms}}$

Given, A.M. = 11

$\therefore 11 = \dfrac{89 + x}{9} \\[1em] \Rightarrow 99 = 89 + x \\[1em] \Rightarrow x = 99 - 89 \\[1em] \Rightarrow x = 10.$

Hence, the value of 9th variate is 10.

The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes $12\dfrac{15}{16}$ years. What is the age of the girl ?

Answer

$\text{Arithmetic mean (A.M.)} = \dfrac{\text{Sum of age of students}}{\text{No.of students}} \\[1em] \Rightarrow 13 = \dfrac{\text{Sum of age of students}}{33} \\[1em] \Rightarrow \text{Sum of age of students} = 33 \times 13 \\[1em] \Rightarrow \text{Sum of age of students} = 429.$

Let the age of girl that leaves the class be x. So, sum of age of students becomes 429 - x and total no of students = 32. Given, new mean = $12\dfrac{15}{16}$.

$\therefore 12\dfrac{15}{16} = \dfrac{429 - x}{32} \\[1em] \dfrac{207}{16} = \dfrac{429 -x}{32} \\[1em] 207 \times 32 = 16(429 - x) \\[1em] 6624 = 6864 - 16x \\[1em] 16x = 6864 - 6624 \\[1em] 16x = 240 \\[1em] x = \dfrac{240}{16} = 15.$

Hence, the age of girl is 15 years.

In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that the marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.

Answer

$\text{ Mean of marks} = \dfrac{\text{Incorrect sum of marks}}{\text{No. of students}} \\[1em] \Rightarrow 18.2 = \dfrac{\text{Incorrect sum of marks}}{40} \\[1em] \Rightarrow \text{Incorrect sum of marks} = 40 \times 18.2 \\[1em] \Rightarrow \text{Incorrect sum of marks} = 728.$

As the marks of one student was wrongly copied as 21 instead of 29, correct sum of marks = 728 - 21 + 29 = 736.

$\therefore \text{Correct mean } = \dfrac{\text{Correct sum of marks}}{\text{No. of students}} \\[1em] = \dfrac{736}{40} = 18.4$

Hence, the correct mean is 18.4.

Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.

Answer

$\text{Arithmetic mean (A.M.)} = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore \text{Sum of terms} = \text{A.M.} \times \text{No. of terms}.$

Given, mean of 10 numbers is 13.

∴ Sum of 10 terms = 13 × 10 = 130.

Given, mean of 15 numbers is 18.

∴ Sum of 15 terms = 15 × 18 = 270.

Sum of 25 terms = 130 + 270 = 400.

$\therefore A.M. = \dfrac{400}{25} = 16$

Hence, the mean of 25 numbers is 16.

Find the mean of the following distribution :

Answer

We construct the following table:

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{390}{20}$ = 19.5

Hence, the mean of the following distribution is 19.5

The contents of 100 matchboxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.

(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean upto exactly 39 matches.

Answer

(i) We construct the following table:

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{3813}{100}$ = 38.1

Hence, the mean of number of matches per box is 38.1

(ii) Mean = $\dfrac{\text{No. of matches}}{\text{No. of boxes}}$

Let the no. of matches added to total contents of 100 boxes be x in order to bring mean to 39. So, total matches becomes 3813 + x.

$\therefore 39 = \dfrac{3813 + x}{100} \\[1em] \Rightarrow 3900 = 3813 + x \\[1em] \Rightarrow x = 3900 - 3813 = 87.$

Hence, 87 extra matches need to added to bring the mean upto exactly 39 matches.

Find the mean for the following distribution by short cut method:

Answer

We construct the following table as under taking the assumed mean, a = 63.

Mean = $a + \dfrac{∑f_id_i}{∑f_i} = 63 + \dfrac{-23}{60} = 63- 0.38$ = 62.62.

Hence, the mean of the following distribution is 62.62.

(i) Calculate done mean wages correct to the nearest rupees.

(ii) If the number of workers in each categories double what would be the new mean wage?

Answer

(i) We construct the following table:

Mean = $\dfrac{Σf_ix_i}{Σf_i}$

= $\dfrac{42400}{50}$

= 848.

Hence, the mean wage is ₹ 848.

(ii) If the number of workers in each category is doubled then total wage will also be doubled.

New total wage = 42400 × 2 = 84800 and number of workers = 50 × 2 = 100.

Mean = $\dfrac{\text{New total wage}}{\text{No. of workers}}$

= $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{84800}{100}$​ = 848.

Hence, the new mean wage is also ₹ 848.

The mean of the following data is 16. Calculate the value of f.

Answer

By formula; Mean = $\dfrac{Σf_ix_i}{Σf_i}$

Substituting the values, we get

$\Rightarrow 16 = \dfrac{415 + 15f}{25 + f}\\[1em] \Rightarrow 16(25 + f) = 415 + 15f\\[1em] \Rightarrow 400 + 16f = 415 + 15f\\[1em] \Rightarrow 16f - 15f = 415 - 400\\[1em] \Rightarrow f = 15.$

Hence, the value of f = 15.

Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data :

If the mean of the distribution is 7.2, find a and b.

Answer

We construct the following table:

Given, total no. of students = 40 and mean = 7.2

∴ 35 + a + b = 40
⇒ a + b = 5
⇒ a = 5 - b      .....(i)

$\text{Mean} = \dfrac{∑f_ix_i}{∑f_i} \\[1em] \therefore 7.2 = \dfrac{246 + 6a + 9b}{40} \\[1em] \Rightarrow 288 = 246 + 6a + 9b \\[1em] \Rightarrow 288 - 246 = 6a + 9b \\[1em] \Rightarrow 6a + 9b = 42$

Putting value of a from Eq (i)

$\Rightarrow 6(5 - b) + 9b = 42 \\[1em] \Rightarrow 30 - 6b + 9b = 42 \\[1em] \Rightarrow 30 + 3b = 42 \\[1em] \Rightarrow 3b = 12 \\[1em] \Rightarrow b = 4.$

a = 5 - b = 5 - 4 = 1.

Hence, the value of a = 1 and b = 4.

Calculate the mean of the following distribution:

Answer

$\text{Mean }= \dfrac{∑f_iy_i}{∑f_i}\\[1em] = \dfrac{780}{24}\\[1em] = 32.5$

Hence, mean of the following distribution is 32.5.

Calculate the mean of the following distribution :

Answer

We construct the following table:

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{3600}{100}$ = 36.

Hence, mean of the following distribution is 36.

Calculate the mean of the following distribution using step deviation method :

Answer

We construct the following table, taking assumed mean a = 25.
Here, c (width of each class) = 10.

$\therefore \text{ Mean} = a + c \times \dfrac{∑f_iu_i}{∑f_i} \\[1em] = 25 + 10 \times \dfrac{63}{100} \\[1em] = 25 + \dfrac{630}{100} \\[1em] = 25 + 6.3 \\[1em] = 31.3$

Hence, mean of the following distribution is 31.3

The data on the number of patients attending a hospital in a month is given below. Find the average (mean) number of patients attending the hospital in a month using the short cut method.

Take assumed mean as 45. Give your answer correct to 2 decimal places.

Answer

Construct table as under, taking assumed mean a = 45.

$\therefore \text{ Mean} = a + \dfrac{∑f_id_i}{∑f_i} \\[1em] = 45 + \dfrac{-140}{30} \\[1em] = 45 + (-4.67) \\[1em] = 40.33$

Hence, mean of the following distribution is 40.33

The following table keep the daily wages of worker in a factory

Calculate the mean by short cut method.

Answer

We construct the following table as under taking the assumed mean, a = 625.

By formula,

$\text{Mean }= a + \dfrac{∑f_id_i}{∑f_i}\\[1em] = 625 + \dfrac{-250}{100}\\[1em] = 625 - 2.5\\[1em] = 622.5$

Hence, mean of the following distribution is ₹622.5.

Calculate the mean of the distribution given below using the short cut method.

Answer

Construct the table as under, taking assumed mean as 45.5

$\therefore \text{ Mean} = a + \dfrac{∑f_id_i}{∑f_i} \\[1em] = 45.5 + \dfrac{70}{50} \\[1em] = 45.5 + 1.4 \\[1em] = 46.9$

Hence, mean of the following distribution is 46.9 marks

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Answer

We construct the following table :

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{499}{40}$ = 12.475

Hence, mean of the following distribution is 12.475

If the mean of the following distribution is 24, find the value of a :

Answer

We construct the following table :

$\therefore \text{Mean} = \dfrac{∑f_iy_i}{∑f_i} \\[1em] \Rightarrow 24 = \dfrac{810 + 15a}{30 + a} \\[1em] \Rightarrow 24(30 + a) = 810 + 15a \\[1em] \Rightarrow 720 + 24a = 810 + 15a \\[1em] \Rightarrow 24a - 15a = 810 - 720 \\[1em] \Rightarrow 9a = 90 \\[1em] \Rightarrow a = 10.$

Hence, the value of a = 10.

The mean of the following distribution is 50. Find the unknown frequency.

Answer

By formula,

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

Substituting values we get :

$\Rightarrow 50 = \dfrac{2020 + 30f}{34 + f}\\[1em] \Rightarrow 50(34 + f) = 2020 + 30f\\[1em] \Rightarrow 1700 + 50f = 2020 + 30f\\[1em] \Rightarrow 50f - 30f = 2020 - 1700\\[1em] \Rightarrow 20f = 320\\[1em] \Rightarrow f = \dfrac{320}{20}\\[1em] \Rightarrow f = 16.$

Hence, the value of f = 16.

The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q :

Answer

We construct the following table :

Given,

The sum of frequencies = 50.

∴ 32 + p + q = 50
⇒ p + q = 18
⇒ p = 18 - q      .....(i)

$\therefore \text{Mean} = \dfrac{∑f_iy_i}{∑f_i} \\[1em] \Rightarrow 57.6 = \dfrac{1940 + 30p + 70q}{50} \\[1em] \Rightarrow 57.6 \times 50 = 1940 + 30p + 70q \\[1em] \Rightarrow 2880 = 1940 + 30p + 70q \\[1em] \Rightarrow 30p + 70q = 2880 - 1940 \\[1em] \Rightarrow 30p + 70q = 940$

Putting value of p from (i) in above equation,

$\Rightarrow 30(18 - q) + 70q = 940 \\[1em] \Rightarrow 540 - 30q + 70q = 940 \\[1em] 540 + 40q = 940 \\[1em] 40q = 940 - 540 \\[1em] 40q = 400 \\[1em] q = 10$

Using (i),

⇒ p = 18 - q = 18 - 10 = 8.

Hence, the value of p = 8 and q = 10.

The following table gives the life time in days of 100 electricity tubes of a certain make :

Find the mean lifetime of electricity tubes.

Answer

We construct the following table :

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{14500}{100}$ = 145

Hence, mean of the following distribution is 145 days

The following table gives the duration of movies in minutes.

Using step–deviation method, find the mean duration of the movies.

Answer

In the given table i is the class interval which is equal to 10.

Mean = A + $\dfrac{Σfu}{Σf} \times i = 135 + \dfrac{-38}{50} \times 10$

= $135 + \dfrac{-380}{50}$

= 135 - 7.6

= 127.4

Hence, mean = 127.4

Shown below is a table illustrating the monthly income distribution in a company with 100 employees.

Using step-deviation method, find the mean monthly income of an employee.

Answer

In the given table,

Class size (i) = 4.

By formula,

Mean = A + $\dfrac{Σfu}{Σf} \times i$

= 6 + $\dfrac{19}{100} \times 4$

= 6 + $\dfrac{76}{100}$

= 6 + 0.76

= 6.76

Hence, mean = 6.76

Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.

Answer

We construct the following table :

∴ Mean = $\dfrac{∑f_iy_i}{∑f_i} = \dfrac{1545}{33}$ = 46.8

Hence, mean of the following distribution is 46.8

A student scored the following marks in 11 questions of a question paper :

3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7.

Find the median marks.

Answer

On arranging the marks in ascending order, we get

1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8.

Here, n (no. of observations) = 11, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{11 + 1}{2} \\[1em] = \dfrac{12}{2} \\[1em] = 6 \text{th observation}$

6th observation = 5.

Hence, the median marks are 5.

For the following set of numbers, find the median :

10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.

Answer

On arranging the numbers in ascending order we get,

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81.

Here, n (no. of observations) = 12, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{12}{2} \text{th observation} + \big(\dfrac{12}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 6th observation + 7th observation}}{2} \\[1em] = \dfrac{15 + 17}{2} \\[1em] = \dfrac{32}{2} \\[1em] = 16.$

Hence, median = 16.

Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5.

Answer

On arranging the numbers in ascending order we get,

0, 1, 1, 2, 2, 3, 3, 3, 4, 5.

Sum of terms = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24.

By definition,

$\text{Mean} = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] = \dfrac{24}{10} \\[1em] = 2.4$

Here, n (no. of observations) = 10, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \big(\dfrac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 5th observation + 6th observation}}{2} \\[1em] = \dfrac{2 + 3}{2} \\[1em] = \dfrac{5}{2} \\[1em] = 2.5$

Hence, mean = 2.4 and median = 2.5

The median of the observations 11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.

Answer

Observations arranged in ascending order are :

11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47.

Here, n (no. of observations) = 9, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{9 + 1}{2} \\[1em] = \dfrac{10}{2} \\[1em] = 5 \text{th observation}$

Given, median = 24 = 5th observation = x + 4.

⇒ 24 = x + 4
⇒ x = 24 - 4 = 20.

Putting the value of x in observations :

11, 12, 14, 18, 24, 29, 32, 38, 47.

Sum of terms = 11 + 12 + 14 + 18 + 24 + 29 + 32 + 38 + 47 = 225.

By definition,

$\text{Mean } = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] = \dfrac{225}{9} \\[1em] = 25.$

Hence, the value of x = 20 and mean = 25.

The mean of the numbers 1, 7, 5, 3, 4, 4 is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m - 1 and median q. Find (i) p (ii) q (iii) the mean of p and q.

Answer

(i) Given, the mean of the numbers 1, 7, 5, 3, 4, 4 is m.

Sum of terms = 1 + 7 + 5 + 3 + 4 + 4 = 24.

By definition,

$\text{Mean } = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore \text{m} = \dfrac{24}{6} = 4.$

Given, the mean of the numbers 3, 2, 4, 2, 3, 3, p is (m - 1) i.e. 3.

Sum of terms = 3 + 2 + 4 + 2 + 3 + 3 + p = 17 + p.

By definition,

$\text{Mean } = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore 3 = \dfrac{17 + p}{7} \\[1em] \Rightarrow 21 = p + 17 \\[1em] \Rightarrow p = 21 - 17 = 4.$

Hence, the value of p is 4.

(ii) Given, the median of the numbers 3, 2, 4, 2, 3, 3, 4 is q.

Arranging the numbers in ascending order we get,

2, 2, 3, 3, 3, 4, 4.

Here, n (no. of observations ) = 7, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{7 + 1}{2} \\[1em] = \dfrac{8}{2} \\[1em] = 4 \text{th observation}$

Given, median = q = 4th observation = 3.

⇒ q = 3.

Hence, the value of q is 3.

(iii) Mean of p and q i.e. 4 and 3 is,

$\text{Mean } = \dfrac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] = \dfrac{4 + 3}{2} \\[1em] = \dfrac{7}{2} \\[1em] = 3.5$

Hence, mean of p and q is 3.5

Find the median for the following distribution:

Answer

The given variates (wages) are already in ascending order. We construct the cumulative frequency table as under :

Here, n (total no. of workers) = 47, which is odd.

$\text{Median }= \Big(\dfrac{n + 1}{2}\Big)\text{th observation}\\[1em] = \Big(\dfrac{47 + 1}{2}\Big)\text{th observation}\\[1em] = \Big(\dfrac{48}{2}\Big)\text{th observation}\\[1em] = 24\text{th observation}\\[1em]$

All observations from 23rd to 29th are equal, each = 480.

Median = ₹480

Hence, median = ₹480.

Marks obtained by 70 students are given below :

Calculate the median marks.

Answer

On arranging the given variates (marks) in ascending order, we construct the cumulative frequency table as under :

Here, n (total no. of students) = 70, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{70}{2} \text{th observation} + \big(\dfrac{70}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 35th observation + 36th observation}}{2} \\[1em]$

All observations from 21st to 38th are equal, each = 50.

Hence, median

$= \dfrac{50 + 50}{2} \\[1em] = \dfrac{100}{2} \\[1em] = 50$

Hence, median marks = 50.

Calculate the mean and the median for the following distribution :

Answer

The given numbers are already in ascending order. We construct the cumulative frequency table as under :

Here, n (no. of observations) = 20, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{20}{2} \text{th observation} + \big(\dfrac{20}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 10th observation + 11th observation}}{2} \\[1em]$

All observations from 9th to 14th are equal, each = 20.

Hence, median

$= \dfrac{20 + 20}{2} \\[1em] = \dfrac{40}{2} \\[1em] = 20.$

Now calculating mean,

$\text{ Mean} = \dfrac{∑f_ix_i}{∑f_i} \\[1em] = \dfrac{390}{20} \\[1em] = 19.5$

Hence, mean = 19.5 and median = 20.

The daily output of 19 workers are :

41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.

Find :

(i) the median

(ii) lower quartile

(iii) upper quartile

(iv) inter quartile range

Answer

On arranging the given wages in ascending order we get,

21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53.

Here, n (no. of observations) = 19, which is odd.

(i) As n is odd,

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{19 + 1}{2} \\[1em] = \dfrac{20}{2} \\[1em] = 10 \text{th observation}$

∴ Median = 10th observation = 31.

Hence, median = 31.

(ii) As n is odd,

$\therefore \text{Lower quartile} (Q_1) = \dfrac{n + 1}{4} \text{th observation} \\[1em] = \dfrac{19 + 1}{4} \\[1em] = \dfrac{20}{4} \\[1em] = 5 \text{th observation}$

∴ Lower quartile (Q₁) = 5th observation = 27.

Hence, lower quartile = 27.

(iii) As n is odd,

$\therefore \text{Upper quartile} (Q_3) = \dfrac{3(n + 1)}{4} \text{th observation} \\[1em] = \dfrac{3(19 + 1)}{4} \\[1em] = \dfrac{60}{4} \\[1em] = 15 \text{th observation}$

∴ Upper quartile (Q₃) = 15th observation = 41.

Hence, upper quartile = 41.

(iv) Inter quartile range = Q₃ - Q₁ = 41 - 27 = 14.

Hence, inter quartile range = 14.

From the following frequency distribution, find :

(i) the median

(ii) lower quartile

(iii) upper quartile

(iv) inter quartile range

Answer

The given variates are already in ascending order. We construct the cumulative frequency table as under

Here, n (no. of observations) = 48, which is even.

(i) As n is even,

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{48}{2} \text{th observation} + \big(\dfrac{48}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 24th observation + 25th observation}}{2} \\[1em]$

All observations from 19th to 27th are equal, each = 22.

Hence, median

$= \dfrac{22 + 22}{2} \\[1em] = \dfrac{44}{2} \\[1em] = 22.$

Hence, median = 22.

(ii) As n is even,

$\therefore \text{Lower quartile} (Q_1) = \dfrac{n}{4} \text{th observation} \\[1em] = \dfrac{48}{4} \\[1em] = 12 \text{th observation}$

∴ Lower quartile (Q₁) = 12th observation = 20.

Hence, lower quartile = 20.

(iii) As n is even,

$\therefore \text{Upper quartile} (Q_3) = \dfrac{3n}{4} \text{th observation} \\[1em] = \dfrac{3 × 48}{4} \\[1em] = \dfrac{144}{4} \\[1em] = 36 \text{th observation}$

∴ Upper quartile (Q₃) = 36th observation = 27.

Hence, upper quartile = 27.

(iv) Inter quartile range = Q₃ - Q₁ = 27 - 20 = 7.

Hence, inter quartile range = 7.

For the following frequency distribution, find :

(i) the median

(ii) lower quartile

(iii) upper quartile

Answer

The given numbers are already in ascending order. We construct the cumulative frequency table as under

Here, n (no. of observations) = 63, which is odd.

(i) As n is odd,

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{63 + 1}{2} \\[1em] = \dfrac{64}{2} \\[1em] = 32 \text{nd observation}$

∴ Median = 32nd observation = 40.

Hence, median = 40.

(ii) As n is odd,

$\therefore \text{Lower quartile} (Q_1) = \dfrac{n + 1}{4} \text{th observation} \\[1em] = \dfrac{63 + 1}{4} \\[1em] = \dfrac{64}{4} \\[1em] = 16 \text{th observation}$

∴ Lower quartile (Q₁) = 16th observation = 34.

Hence, lower quartile = 34.

(iii) As n is odd,

$\therefore \text{Upper quartile} (Q_3) = \dfrac{3(n + 1)}{4} \text{th observation} \\[1em] = \dfrac{3(63 + 1)}{4} \\[1em] = \dfrac{3 \times 64}{4} \\[1em] = 48 \text{th observation}$

∴ Upper quartile (Q₃) = 48th observation = 48.

Hence, upper quartile = 48.

Find the mode of the following sets of numbers :

(i) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8

(ii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

Answer

(i) In the given data : 5, 7, 6, 8, 9, 0, 6, 8, 1, 8

8 is repeated more number of times than any other number,

∴ mode = 8.

(ii) In the given data : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

5 is repeated more number of times than any other number,

∴ mode = 5.

Find the mean, median and mode of the following distribution :

8, 10, 7, 6, 10, 11, 6, 13, 10.

Answer

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No.of terms}} = \dfrac{∑x_i}{n}$

Sum of terms = 8 + 10 + 7 + 6 + 10 + 11 + 6 + 13 + 10 = 81.

$\therefore \text{A.M.} = \dfrac{81}{9} = 9.$

∴ Mean = 9.

On arranging the marks in ascending order, we get

6, 6, 7, 8, 10, 10, 10, 11, 13.

Here, n (no. of observations) = 9, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{9 + 1}{2} \\[1em] = \dfrac{10}{2} \\[1em] = 5 \text{th observation}$

5th observation = 10.

∴ Median = 10.

In the given data : 8, 10, 7, 6, 10, 11, 6, 13, 10

10 is repeated more number of times than any other number,

∴ Mode = 10.

Hence, mean = 9, median = 10 and mode = 10.

Calculate the mean, median and the mode of the following numbers :

3, 1, 5, 6, 3, 4, 5, 3, 7, 2.

Answer

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No.of terms}} = \dfrac{∑x_i}{n}$

Sum of terms = 3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2 = 39

$\therefore \text{A.M.} = \dfrac{39}{10} = 3.9$

∴ Mean = 3.9

On arranging the numbers in ascending order we get,

1, 2, 3, 3, 3, 4, 5, 5, 6, 7.

Here, n (no. of observations) = 10, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \big(\dfrac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{5th observation + 6th observation}}{2} \\[1em] = \dfrac{3 + 4}{2} \\[1em] = \dfrac{7}{2} \\[1em] = 3.5$

∴ Median = 3.5

In the given data : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2.

3 is repeated more number of times than any other number,

∴ Mode = 3.

Hence, mean = 3.9, median = 3.5 and mode = 3.

The marks of 10 students of a class in an examination arranged in ascending order are as follows :

13, 35, 43, 46, x, x + 4, 55, 61, 71, 80.

If the median marks is 48, find the value of x. Hence, find the mode of the given data.

Answer

Here, n (no. of observations) = 10, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \big(\dfrac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{5th observation + 6th observation}}{2} \\[1em] = \dfrac{x + (x + 4)}{2} \\[1em] = \dfrac{2x + 4}{2} \\[1em] = x + 2.$

Given, median marks = 48.

∴ x + 2 = 48
⇒ x = 46.

Putting value of x in data we get,

13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data 46 is repeated more number of times than any other number.

Hence, the value of x = 46 and mode = 46.

Find the mode and median of the following frequency distribution :

Answer

The variates are already in ascending order. We construct the cumulative frequency table as under :

Total number of observations = 29, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{29 + 1}{2} \\[1em] = \dfrac{30}{2} \\[1em] = 15 \text{th observation}$

All observations from 13th to 17th are equal, each = 13, so median = 13.

As the variate 14 has maximum frequency 9, so mode = 14.

Hence, median = 13 and mode = 14.

In a class of 40 students marks obtained by the students in a class test (out of 10) are given below :

Calculate the following for the given distribution :

(i) median

(ii) mode.

Answer

(i) The variates are already in ascending order. We construct the cumulative frequency table as under :

Total number of observations = 40, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \dfrac{\dfrac{40}{2} \text{th observation} + \big(\dfrac{40}{2} + 1\big)\text{th observation}}{2} \\[1em] = \dfrac{\text{ 20th observation + 21st observation}}{2} \\[1em]$

All observations from 16th to 25th are equal, each = 6.

Hence, median

$= \dfrac{6 + 6}{2} \\[1em] = \dfrac{12}{2} \\[1em] = 6.$

Hence, median = 6.

(ii) As the variate 6 has maximum frequency 10, so mode = 6.

Hence, mode = 6.

The marks obtained by 30 students in a class assessment of 5 marks is given below :

Calculate the mean, median and mode of the above distribution.

Answer

The variates (marks) are already in ascending order. We construct the cumulative frequency table as under :

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{90}{30}$ = 3.

Total number of observations = 30, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \dfrac{\dfrac{30}{2} \text{th observation} + \big(\dfrac{30}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 15th observation + 16th observation}}{2} \\[1em]$

All observations from 11th to 20th are equal, each = 3.

Hence, median

$= \dfrac{3 + 3}{2} \\[1em] = \dfrac{6}{2} \\[1em] = 3.$

As the variate 3 has maximum frequency 10, so mode = 3.

Hence, mean = 3, median = 3 and mode = 3.

The distribution table given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Answer

The variates (marks) are already in ascending order. We construct the cumulative frequency table as under :

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{171}{25}$ = 6.84.

Total number of observations = 25, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{25 + 1}{2} \\[1em] = \dfrac{26}{2} \\[1em] = 13 \text{th observation}$

All observations from 13th to 18th are equal, each = 7, so median = 7.

As the variate 6 has maximum frequency 9, so mode = 6.

Hence, mean = 6.84, median = 7 and mode = 6.

The following table gives the weekly wages (in ₹) of workers in a factory:

Calculate:

(i) the mean.

(ii) the modal class.

(iii) the number of workers getting weekly wages below ₹800

(iv) the number of workers getting ₹650 or more but less than ₹850 as weekly wages.

Answer

(i) We construct the following table :

By formula,

$\text{Mean }= \dfrac{∑f_iu_i}{∑f_i}\\[1em] = \dfrac{55200}{80}\\[1em] = 690.$

Hence, mean = ₹ 690.

(ii) The class 550 - 600 has maximum frequency 20.

Hence, modal class = 550-600.

(iii) From table,

The number of workers getting weekly wages below ₹800 = 60.

(iv) From table,

The number of workers getting ₹650 or more but less than ₹850 as weekly wages = 72 - 35 = 37.

Draw a histogram for the following frequency distribution and find the mode from the graph :

Answer

Steps :

1. Take 1 cm along x-axis = 5 units and 1 cm along y-axis = 4 units.
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 14.

Hence, the required mode = 14.

A mathematics aptitude test of 50 students was recorded as follows :

Draw a histogram for the above data using a graph paper and locate the mode.

Answer

Steps :

1. Take 1 cm along x-axis = 10 marks and 1 cm along y-axis = 4 (students).
2. Since, the scale on x-axis starts at 50, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 50 and not at origin itself.
3. Construct rectangles corresponding to the given data.
4. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
5. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 82.5.

Hence, the required mode = 82.5

Draw a histogram and estimate the mode for the following frequency distribution :

Answer

Steps :

1. Take 1 cm along x-axis = 10 units and 1 cm along y-axis = 2 units.
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 23.

Hence, the required mode = 23.

Using a graph paper, draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data :

Answer

Steps :

1. Take 1 cm along x-axis = 1000 runs and 1 cm along y-axis = 4 (batsman).
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 4600.

Hence, the required mode = 4600.

Draw a histogram for the given data, using a graph paper:

Estimate the mode from the graph.

Answer

Steps :

1. Take 1 cm along x-axis = 1000 rupees and 1 cm along y-axis = 2 (No. of people).

2. Construct rectangles corresponding to the given data.

3. In highest rectangle, draw two st. lines AD and BC from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AD and BC.

4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 5400.

Hence, mode = ₹ 5,400.

Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below :

Draw a histogram representing the above distribution and estimate the mode from the graph.

Answer

Steps :

1. Take 1 cm along x-axis = ₹5 and 1 cm along y-axis = 5 students.
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents ₹21.50.

Hence, the required mode = ₹21.50.

Draw a histogram for the following distribution :

Hence, estimate the modal weight.

Answer

Steps :

1. The given frequency distribution is discontinuous, to convert it into continuous distribution,

     Adjustment factor = $\dfrac{45 - 44}{2} = \dfrac{1}{2}$ = 0.5

We construct the continuous frequency table for the given data :

2. Take 2 cm along x-axis = 5 kg and 1 cm along y-axis = 2 (students).

3. Since, the scale on x-axis starts at 39.5, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 39.5 and not at origin itself.

4. Construct rectangles corresponding to the given data.

5. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

6. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 52.75 kg.

Hence, the required mode = 52.75 kg.

Find the mode of the following distribution by drawing a histogram.

Also state the modal class.

Answer

Size of each class = difference between two consecutive mid-values = 18 - 12 = 6.

Constructing the table as under :

Steps :

1. Take 1 cm along x-axis = 6 units and 1 cm along y-axis = 4 units.

2. Since, the scale on x-axis starts at 9, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 9 and not at origin itself.

3. Construct rectangles corresponding to the given data.

4. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

5. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 30.5.

∴ Mode = 30.5

As the class 27 - 33 has the highest frequency, hence modal class = 27 - 33.

Hence, the required mode = 30.5 and modal class = 27 - 33.

The table given below shows the runs scored by a cricket team during the overs of a match.

Use graph sheet for this question. Take 2 cm = 10 overs along one axis and 2 cm = 10 runs along the other axis.

(a) Draw a histogram representing the above distribution.

(b) Estimate the modal runs scored.

Answer

Steps :

1. Take 2 cm along x-axis = 10 overs and 1 cm along y-axis = 10 runs.

2. Since, the scale on x-axis starts at 20, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 20 and not at origin itself.

3. Construct rectangles corresponding to the given data.

4. In highest rectangle, draw two st. lines KN and LI from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let Z be the point of intersection of KN and LI.

5. Through Z, draw a vertical line to meet the x-axis at A. The abscissa of the point A represents 57.

Hence, mode = 57.

Draw an ogive for the following frequency distribution :

Answer

To draw an ogive :

1. The cumulative frequency table for the given continuous distribution is :

2. Since, the scale on x-axis starts at 150, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 150.

3. Take 2 cm along x-axis = 10 cm (height)

4. Take 1 cm along y-axis = 5 students

5. Plot the points (160, 8), (170, 11), (180, 15), (190, 25), (200, 27) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 150 - 160.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

Draw an ogive for the following frequency distribution :

Answer

To draw an ogive :

1. The given frequency distribution is discontinuous, to convert it into continuous distribution,
   Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2
   $= \dfrac{11 - 10}{2} = \dfrac{1}{2} = 0.5$

The cumulative frequency table for the given data is :

2. Since, the scale on x-axis starts at 0.5, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 0.5

3. Take 2 cm along x-axis = 10

4. Take 1 cm along y-axis = 5

5. Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23), (50.5, 29), (60.5, 31) representing upper class limits and the respective cumulative frequencies.

Also plot the point representing lower limit of the first class i.e. 0.5 - 10.5

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

Draw a cumulative frequency curve for the following data :

Answer

To draw an ogive :

1. The cumulative frequency table for the given continuous distribution is :

2. Since, the scale on x-axis starts at 24, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 24.

3. Take 1 cm along x-axis = 5 marks

4. Take 1 cm along y-axis = 5 (students)

5. Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18), (54, 21) and (59, 23) representing upper class limits and the respective cumulative frequencies.

Also plot the point representing lower limit of the first class i.e. 24 - 29.

6. Join these points by a freehand drawing.

The required cumulative frequency curve is shown in figure above.

The weight of 50 workers is given below :

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use a graph to estimate the following :

(i) the upper and lower quartiles.

(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight.

Answer

1. The cumulative frequency table for the given continuous distribution is :

2. Take 1 cm along x-axis = 10 kg

3. Take 1 cm along y-axis = 5 (workers)

4. Since, scale on x-axis starts at 50, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 50.

5. Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 50 - 60.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) To find lower quartile :

Let A be the point on y-axis representing frequency = $\dfrac{n}{4} = \dfrac{50}{4}$ = 12.5

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 71 kg.

To find upper quartile :

Let B be the point on y-axis representing frequency = $\dfrac{3n}{4} = \dfrac{150}{4}$ = 37.5

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 93 kg.

Hence, lower quartile = 71 kg and upper quartile = 93 kg.

(ii) Let O be the point on x-axis representing 95 kg. Through O draw a vertical line to meet the ogive R. Through R, draw a horizontal line to meet the y-axis at point C. The ordinate of the point C represents 39.

So, the number of people whose weight is less than 95 kg = 39. So, overweight people = 50 - 39 = 11 kg.

Hence, the number of workers who are overweight are 11.

The table shows the distribution of 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

1. The cumulative frequency table for the given continuous distribution is :

2. Take 2 cm along x-axis = 10 scores

3. Take 2 cm along y-axis = 20 (shooters)

4. Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153) and (100, 160) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 160.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{n}{2} = \dfrac{160}{2}$ = 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 point M represents 44.

Hence, the required median score = 44.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\dfrac{n}{4} = \dfrac{160}{4}$ = 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 = $\dfrac{3n}{4} = \dfrac{480}{4}$ = 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 O. The abscissa of the point O represents 60.

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

Hence, the inter quartile range = 31 scores.

(iii) Total marks = 100.

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

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

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

Students 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 shooters who obtained a score of more than 85%.

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

1. The cumulative frequency table for the given continuous distribution is :

2. Take 2 cm along x-axis = 50 rupees

3. Take 1 cm along y-axis = 10 workers

4. 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.

5. Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75) and (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.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 80.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{n}{2} = \dfrac{80}{2}$ = 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 point M represents 604.

Hence, the required median wage = ₹604.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\dfrac{n}{4} = \dfrac{80}{4}$ = 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 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 more than ₹625 daily.

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 the other axis. Using the graph, determine :

(i) The median marks

(ii) The number of students who failed if minimum marks required to pass is 40.

(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.

Answer

1. The cumulative frequency table for the given continuous distribution is :

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

3. Take 1 cm along y-axis = 20 (students)

4. Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194) and (100, 200) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 200.

To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{n}{2} = \dfrac{200}{2}$ = 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 point M represents 57.

Hence, the required median marks = 57.

(ii) Let N be the point on x-axis representing marks = 40.

Through N, draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet the y-axis at B. The ordinate of the point B represents 46.

Students who scored less than 40 = 46.

Hence, 46 students failed in the examination.

(iii) Let O be the point on x-axis representing marks = 85.

Through O, 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 187.

Students who scored less than 85 = 187.

So, students scoring more than 85 = Total students - students scoring less than 85 = 200 - 187 = 13.

Hence, 13 students secured grade one in examination.

Use graph paper for this question.

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 on 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) median

(ii) lower quartile

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

Answer

1. The cumulative frequency table for the given continuous distribution is :

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

3. Take 2 cm along y-axis = 10 (boys)

4. 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.

5. 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 representing lower limit of the first class i.e. 135 - 140.

6. 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 = $\dfrac{n}{2} = \dfrac{60}{2}$ = 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 149.5.

Hence, the median height = 149.5 cm.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\dfrac{n}{4} = \dfrac{60}{4}$ = 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 O be the point on x-axis representing height = 158 cm.

Through O, 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 a shot put competition. The distance thrown in metre is recorded below:

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

Uses scale of 2 cm = 1 m on one axis and 2 cm = 5 students on other axis.

Hence, using your graph, find:

(i) the median

(ii) the quartile

(iii) no. of student who cover a distance which is above $16\dfrac{1}{2}$ m.

Answer

Cumulative frequency distribution table :