Measures of Central Tendency — Exercise 21.2

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 is:

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.