Median, Quartiles & Mode — Multiple Choice Questions

A data has 35 observations arranged in a descending order. Which organization represents the median?

1. 16th

2. 17th

3. 18th

4. 19th

Answer

We know that,

The value of the middle-most observation obtained after arranging the data in an ascending order, is called median of the data.

Here no. of observations, n = 35

By formula,

Median = $\dfrac{\text{n} + 1}{2} \text{ th observation}$

$= \dfrac{35 + 1}{2} \text{ th observation} \\[1em] = \dfrac{36}{2} \text{ th observation} \\[1em] = 18 \text{ th observation}$

Since, given data is arranged in descending order,

Median = 18 th observation.

Hence, Option 3 is the correct option.

The median of first 8 prime numbers is:

1. 7

2. 9

3. 11

4. 13

Answer

Prime numbers : 2, 3, 5, 7, 11, 13, 17, 19

No. of terms (n) = 8, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

Substituting values we get:

$= \dfrac{\dfrac{8}{2} \text{ th observation} + \Big(\dfrac{8}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{4 \text{ th observation} + \Big(4 + 1\Big) \text{ rth observation}}{2} \\[1em] = \dfrac{4 \text{ th observation} + 5 \text{ th observation}}{2} \\[1em] = \dfrac{7 + 11}{2} \\[1em] = \dfrac{18}{2} \\[1em] = 9.$

Hence, Option 2 is the correct option.

If the median height of the students of a class is 105 cm, it means that

1. the average height of the students of the class is 105 cm

2. maximum number of students in the class are 105 cm tall

3. there are as many students in the class, taller than 105 cm as are shorter than 105 cm

4. none of these

Answer

We know that,

The value of the middle-most observation obtained after arranging the data in an ascending order, is called median of the data.

Hence, Option 3 is the correct option.

The median of the following observations arranged in ascending order is 64. Find the value of x :

27, 31, 46, 52, x, x + 4, 71, 79, 85, 90

1. 60

2. 61

3. 62

4. 66

Answer

No. of observations = 10, which is even.

Median = $\dfrac{\left(\dfrac{n}{2}\right)^\text{th}\text{term} + \left(\dfrac{n}{2} + 1\right)^\text{th}\text{term}}{2}$

$\left(\dfrac{n}{2}\right)^\text{th}\text{term} = \dfrac{10}{2}$ = 5^th term = x

$\left(\dfrac{n}{2} + 1\right)^\text{th}\text{term} = \dfrac{10}{2} + 1$ = 5 + 1 = 6^th term = x + 4

Given,

Median = 64

$\therefore \dfrac{x + x + 4}{2} = 64 \\[1em] \Rightarrow \dfrac{2x + 4}{2} = 64 \\[1em] \Rightarrow \dfrac{2(x + 2)}{2} = 64 \\[1em] \Rightarrow x + 2 = 64 \\[1em] \Rightarrow x = 64 - 2 \\[1em] \Rightarrow x = 62$

Hence, Option 3 is the correct option.

If 25 is removed from the data 20, 24, 25, 26, 27, 28, 29, 30, then the median increases by:

1. 0.5

2. 1

3. 1.5

4. 2

Answer

Set of observations = 20, 24, 25, 26, 27, 28, 29, 30

Here, n = 8, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

$= \dfrac{\dfrac{8}{2} \text{ th observation} + \Big(\dfrac{8}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{4 \text{ th observation} + \Big(4 + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{4 \text{ th observation} + 5 \text{ th observation}}{2} \\[1em] = \dfrac{26 + 27}{2} \\[1em] = \dfrac{53}{2} \\[1em] = 26.5$

If 25 is removed from given set, then we get:

Set of observations = 20, 24, 26, 27, 28, 29, 30

Here, n = 7, which is odd.

By formula,

Median = $\dfrac{\text{n} + 1}{2} \text{ th observation}$

$= \dfrac{7 + 1}{2} \text{ th observation} \\[1em] = \dfrac{8}{2} \text{ th observation} \\[1em] = 4 \text{ th observation} \\[1em] = 27$

Difference between median = 27 - 26.5 = 0.5

Hence, Option 1 is the correct option.

The mean of 4, 5, 1, 3, 7, 4 is x. The numbers 2, 4, 3, 2, 3, y, 3 have mean x - 1 and median z. Then, y + z = ?

options

1. 4

2. 5

3. 6

4. 7

Answer

Given,

Mean of 4, 5, 1, 3, 7, 4 is x.

Sum of observations = 4 + 5 + 1 + 3 + 7 + 4 = 24

Mean (x) = $\dfrac{\text{Sum of observations}}{\text{No. of observations}} = \dfrac{24}{6}$ = 4.

Given,

Numbers 2, 4, 3, 2, 3, y, 3 have mean x - 1 = 4 - 1 = 3.

Sum of observations = 2 + 4 + 3 + 2 + 3 + y + 3 = 17 + y

Mean (x) = $\dfrac{\text{Sum of observations}}{\text{No. of observations}} = \dfrac{17 + \text{y}}{7}$

$\Rightarrow 3 = \dfrac{17 + \text{y}}{7}$

⇒ 7 × 3 = 17 + y

⇒ 21 = 17 + y

⇒ y = 21 - 17

⇒ y = 4

Observations in ascending order are = 2, 2, 3, 3, 3, 4, 4

Here, n = 7, which is odd.

By formula,

Median = $\dfrac{\text{n} + 1}{2} \text{ th observation}$

$= \dfrac{7 + 1}{2} \text{ th observation} \\[1em] = \dfrac{8}{2} \text{ th observation} \\[1em] = 4 \text{ th observation} \\[1em] = 3$

∴ z = 3.

y + z = 4 + 3 = 7.

Hence, Option 4 is the correct option.

Which measure of central tendency would be the most appropriate for a shoe dealer to determine the quantity of different sizes that he should order?

1. Mean

2. Median

3. Mode

4. Any of these

Answer

Mode represents the value that occurs most frequently in a dataset.

The mode will determine the most frequently sold shoe size in the shop.

Hence, Option 3 is the correct option.

Which of the following cannot be determined graphically?

1. Mean

2. Median

3. Mode

4. None of these

Answer

We know that,

To determine mean we do not need any graphical method, since, it relies on the exact numerical value of every single observation in a dataset and graphs like histograms or frequency polygons represent data in ranges, specific individual values to determine mean is not possible.

Hence, Option 1 is the correct option.

The median of a frequency distribution is found graphically with the help of

1. Ogive

2. Histogram

3. Frequency polygon

4. Bar graph

Answer

The median of a grouped frequency distribution is typically found graphically using a cumulative frequency curve, also known as an ogive.

Hence, Option 1 is the correct option.

The median of 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6 is :

1. -1.5

2. 0

3. 2

4. 3.5

Answer

Given,

Set of observations = 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6

Arranging the numbers in ascending order :

-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6.

Here n = 14, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

$= \dfrac{\dfrac{14}{2} \text{ th observation} + \Big(\dfrac{14}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{7 \text{ th observation} + \Big(7 + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{7 \text{ th observation} + 8 \text{ th observation}}{2} \\[1em] = \dfrac{2 + 5}{2} \\[1em] = \dfrac{7}{2} \\[1em] = 3.5$

Hence, Option 4 is the correct option.

The median of the following data is:

1. 30

2. 31

3. 35

4. 40

Answer

Cumulative frequency distribution table is:

Here, n = 11, which is odd.

By formula,

Median = $\dfrac{\text{n} + 1}{2} \text{ th observation}$

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

From table,

6th observation corresponds to 30.

∴ Median = 30

Hence, Option 1 is the correct option.

Consider the following table :

The median of the above frequency distribution is :

1. 122 mm

2. 122.5 mm

3. 122.75 mm

4. 123 mm

Answer

Cumulative frequency distribution table is as follows :

Here n = 50, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

$= \dfrac{\dfrac{50}{2} \text{ th observation} + \Big(\dfrac{50}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{25 \text{ th observation} + \Big(25 + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{25 \text{ th observation} + 26 \text{ th observation}}{2} \\[1em] = \dfrac{122 + 122}{2} \\[1em] = \dfrac{244}{2} \\[1em] = 122 \text{ mm}.$

Since, all observations from 25th to 26th corresponds to 122 mm.

Hence, Option 1 is the correct option.

Consider the following table:

The upper limit of the median class is :

1. 10

2. 15

3. 20

4. 25

Answer

We construct the cumulative frequency distribution table as under :

Here n = 70, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

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

As observation from 19th to 37th lies in the class 10 - 15

∴ Median class = 10 - 15, with upper limit = 15

Hence, Option 2 is the correct option.

The marks secured (out of 10) by a student in 15 unit tests are as follows:

5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 8, 7, 9, 10, 8

The mode of the above data is :

1. 5

2. 7

3. 8

4. 10

Answer

From above set of numbers : 8 occurs most of the time.

Mode = 8.

Hence, Option 3 is the correct option.

Average value of the median of 2, 8, 3, 7, 4, 6, 1 and the mode of 2, 9, 3, 4, 9, 6, 9 is:

1. 6

2. 6.5

3. 8

4. 9

Answer

Arranging set of observations in ascending order : 1, 2, 3, 4, 6, 7, 8

Here, n = 7, which is odd.

By formula,

Median = $\dfrac{\text{n} + 1}{2} \text{ th observation}$

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

Given,

2, 9, 3, 4, 9, 6, 9

From above set of numbers : 9 occurs most of the time.

Mode = 9.

Average of median and mode = $\dfrac{9 + 4}{2} = \dfrac{13}{2}$ = 6.5

Hence, Option 2 is the correct option.

The mode of a frequency distribution can be determined graphically from :

1. Histogram

2. Frequency polygon

3. Ogive

4. Bar graph

Answer

The mode of a frequency distribution can be determined graphically from Histogram.

Hence, Option 1 is the correct option.

If the mode of the following data is 7, the value of k is :

2, 4, 6, 7, 5, 6, 10, 6, 7, 2k + 1, 9, 7, 13

1. 2

2. 3

3. 4

4. 7

Answer

Given, mode = 7

From above set of numbers : 6 and 7 occurs most of the time (3 times each). For 7 to be the mode, its frequency must be greater than that of 6.

Therefore, the unknown expression 2k + 1 must be equal to 7.

⇒ 2k + 1 = 7

⇒ 2k = 7 - 1

⇒ 2k = 6

⇒ k = $\dfrac{6}{2}$

⇒ k = 3.

Hence, Option 2 is the correct option.

The lower limit of the modal class of the following data is:

1. 10

2. 20

3. 30

4. 40

Answer

Since the class 20 - 30 has highest frequency i.e. 13.

∴ Modal class = 20 - 30.

The lower limit of the modal class = 20.

Hence, Option 2 is the correct option.

Consider the following distribution :

The sum of lower limits of the median class and the modal class is :

1. 15

2. 25

3. 30

4. 35

Answer

We construct the cumulative frequency distribution table as under :

Here n (total no. of observations) = 66.

As n is even,

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{th observation}}{2}$

$= \dfrac{\dfrac{66}{2} \text{th observation} + \Big(\dfrac{66}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \dfrac{33 \text{th observation} + \Big(33 + 1\Big) \text{th observation}}{2} \\[1em] = \dfrac{33 \text{th observation} + 34 \text{th observation}}{2}$

As observation from 26th to 37th lie in the class 10 - 15,

∴ Median class = 10 - 15.

Since the class 15 - 20 has highest frequency i.e. 20.

∴ Modal class = 15 - 20.

Sum of lower limit of median and modal class = 10 + 15 = 25.

Hence, Option 2 is the correct option.

The relation connecting the measures of central tendency is:

1. Mode = 2 Median - 3 Mean

2. Mode = 3 Median - 2 Mean

3. Mode = 2 Median + 3 Mean

4. Mode = 3 Median + 2 Mean

Answer

By formula,

Mode = 3 Median - 2 Mean

Hence, Option 2 is the correct option.

Find the median of the data it being given that mode = 12.3 and mean = 10.5

1. 10.6

2. 10.8

3. 11.1

4. 11.4

Answer

By formula,

Mode = 3 Median - 2 Mean

⇒ 12.3 = 3 × Median - 2 × 10.5

⇒ 12.3 = 3 × Median - 21

⇒ 3 × Median = 12.3 + 21

⇒ 3 × Median = 33.3

⇒ Median = $\dfrac{33.3}{3}$

⇒ Median = 11.1

Hence, Option 3 is the correct option.

Find the mean of the data when it is given that mode = 50.5 and median = 45.5.

1. 43

2. 43.2

3. 43.5

4. 44

Answer

By formula,

Mode = 3 Median - 2 Mean

⇒ 50.5 = 3 × 45.5 - 2 Mean

⇒ 50.5 = 136.5 - 2 Mean

⇒ 2 Mean = 136.5 - 50.5

⇒ 2 Mean = 86

⇒ Mean = $\dfrac{86}{2}$

⇒ Mean = 43.

Hence, Option 1 is the correct option.

Find the mode of the data, it is given that median = 41.25 and mean = 33.75.

1. 54.85

2. 55.75

3. 56.25

4. 57.5

Answer

By formula,

Mode = 3 Median - 2 Mean

= 3 × 41.25 - 2 × 33.75

= 123.75 - 67.5

= 56.25.

Hence, Option 3 is the correct option.

If the difference between the mode and median of a certain data is 24, then the difference between median and mean is :

1. 8

2. 12

3. 24

4. 36

Answer

Given,

Mode - Median = 24

Mode = 24 + Median

By formula,

Mode = 3 Median - 2 Mean

⇒ 24 + Median = 3 Median - 2 Mean

⇒ 3 Median - Median - 2 Mean = 24

⇒ 2 Median - 2 Mean = 24

⇒ 2 (Median - Mean) = 24

⇒ Median - Mean = $\dfrac{24}{2}$

⇒ Median - Mean = 12.

Hence, Option 2 is the correct option.

The median class for the given distribution is :

1. 0 - 10

2. 10 - 20

3. 20 - 30

4. 30 - 40

Answer

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

Here, Cumulative frequency = 14, which is even.

By formula,

Median = $\dfrac{\dfrac{\text{n}}{2} \text{ th observation} + \Big(\dfrac{\text{n}}{2} + 1\Big) \text{ th observation}}{2}$

$= \dfrac{\dfrac{14}{2} \text{ th observation} + \Big(\dfrac{14}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{7 \text{ th observation} + \Big(7 + 1\Big) \text{ th observation}}{2} \\[1em] = \dfrac{7 \text{ th observation} + 8 \text{ th observation}}{2}$

All observations from 7th to 9th are equal, each lies in the interval 20 - 30.

So, median class = 20 - 30

Hence, Option 3 is the correct option.