Measures of Central Tendency — Multiple Choice Questions

Choose the correct answer from the given four options (1 to 8) :

If the classes of a frequency distribution are 1 - 10, 11 - 20, 21 - 30, ...., 51 - 60, then the size of each class is

1. 9

2. 10

3. 11

4. 5.5

Answer

On converting the class from discontinuous intervals into continuous intervals.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$= \dfrac{11 - 10}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

Class interval = 10.5 - 0.5 = 10.

Hence, Option 2 is the correct option.

If the classes of a frequency distribution are 1 - 10, 11 - 20, 21 - 30, ....., 61 - 70, then the upper limit of the class 11 - 20 is

1. 20

2. 21

3. 19.5

4. 20.5

Answer

On converting the class from discontinuous intervals into continuous intervals.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$= \dfrac{11 - 10}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

From table, the upper limit of the class 11 - 20 is 20.5

Hence, Option 4 is the correct option.

In a grouped frequency distribution, the mid-values of the classes are used to measure which of the following central tendency?

1. median

2. mode

3. mean

4. all of these

Answer

In a grouped frequency distribution, the mid-values of the classes are used to measure mean.

Hence, Option 3 is the correct option.

In the formula : x̄ = a + $\dfrac{Σ f_id_i}{Σ f_i}$ for finding the mean of the grouped data, d_i's are deviations from a (assumed mean) of

1. lower limits of the classes

2. upper limits of the classes

3. mid-points of the classes

4. frequencies of the classes

Answer

In the formula : x̄ = a + $\dfrac{Σ f_id_i}{Σ f_i}$ for finding the mean of the grouped data, d_i's are deviations from a (assumed mean) of mid-points of the classes.

Hence, Option 3 is the correct option.

Construction of a cumulative frequency distribution table is useful in determining the

1. mean

2. median

3. mode

4. all the three measures

Answer

Cumulative frequency distribution table is useful in determining the median.

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,

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

All observations from 7th to 8th are equal, each lies in the interval 20 - 30.

So, median class = 20 - 30

Hence, option 3 is the correct option.

Consider the following frequency distribution :

The upper limit of the median class is

1. 17

2. 17.5

3. 18

4. 18.5

Answer

Converting the discontinuous interval into continuous interval.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$= \dfrac{11 - 10}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

We construct the cumulative frequency distribution table as under :

Here n (total no. of observations) = 57.

As n is odd,

$\therefore \text{Median } = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{57 + 1}{2} \\[1em] = \dfrac{58}{2} \\[1em] = 29 \text{th observation}$

As observation from 24^th to 38^th lies in the class 11.5 - 17.5

∴ Median class = 11.5 - 17.5, with upper limit = 17.5

Hence, Option 2 is the correct option.

For the following distribution :

The sum of lower limits of the median class and 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,

∴ Median = ($\dfrac{n}{2}$th observation + ($\dfrac{n}{2} + 1$)th observation) / 2

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

As observation from 26^th to 37^th 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 modal class of a given distribution always corresponds to the:

1. Interval with highest frequency

2. Interval with lowest frequency

3. The first interval

4. The last interval

Answer

The "modal class" refers to the interval in a frequency distribution that has the highest number of observations, meaning it's the class with the most occurrences.

Hence, option 1 is the correct option.

An ogive curve is used to determine

1. range

2. mean

3. mode

4. median

Answer

An ogive curve is used to determine median.

Hence, Option 4 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 = $\dfrac{n}{2}\text{th term} = \dfrac{10}{2}$ = 5th term.

Given,

Median = 64

∴ x + 4 = 64

⇒ x = 64 - 4 = 60.

Hence, Option 1 is the correct option.