Elementary Statistics MCQ Quiz - Objective Question with Answer for Elementary Statistics - Download Free PDF

Calculation:

To find the median, we first need to arrange the data in ascending order:

1, 1, 3, 4, 4, 6, 7, 8, 9, 11

There are 10 data points (an even number). For an even number of data points, the median is the average of the two middle values.

The two middle values are the 5th and 6th values in the sorted list, which are 4 and 6.

Median = (4 + 6) / 2 = 10 / 2 = 5

Therefore, the median of the given data is 5.

Calculation:

Formula for Estimated Mean (x̄):

x̄ = Σ(f × x) / Σf

Calculation:

x̄ = 1288 / 21

x̄ ≈ 61.33

The estimated mean of the data is approximately 61.33.

Calculation:

To find the mode, we need to identify the number that appears most frequently in the data set.

Let's list the data and count the occurrences of each number:

53: 1

55: 1

56: 2

58: 2

59: 2

60: 1

61: 2

62: 3

64: 1

65: 2

67: 1

68: 2

70: 1

From the counts, the number 62 appears 3 times, which is more than any other number.

Therefore, the mode of the given data is 62.

Given:

Assertion (A): The mode of the data: 5, 9, 11, 7, 12, 18, 32, 39, 19, 52, 27, 21, 23, 52, 18, 34, 41, 18, 40, 17 is 52.

Reason (R): Mode is that value of the variable in the data which has maximum frequency.

Analysis:

The data set is: 5, 9, 11, 7, 12, 18, 32, 39, 19, 52, 27, 21, 23, 52, 18, 34, 41, 18, 40, 17.

The frequencies of the numbers are as follows:

- 5: 1 time

- 9: 1 time

- 11: 1 time

- 7: 1 time

- 12: 1 time

- 18: 3 times

- 32: 1 time

- 39: 1 time

- 19: 1 time

- 52: 2 times

- 27: 1 time

- 21: 1 time

- 23: 1 time

- 34: 1 time

- 41: 1 time

- 40: 1 time

- 17: 1 time

Clearly, the number 18 occurs 3 times, which is more frequent than any other number in the data set. Therefore, the mode is 18, not 52.

The reason (R) is correct because the mode is indeed the value that occurs most frequently in a data set. However, the assertion (A) is false because the mode is not 52, but 18.

Conclusion:

The correct answer is (d) A is false and R is true.

Formula used:

The formula for the mode of a grouped frequency distribution is:

Mode = L + [(f₁ - f₀) / (2f₁ - f₀ - f₂)] × h

Where: L is the lower boundary of the modal class,

f₁ is the frequency of the modal class,  

f₀ is the frequency of the class preceding the modal class,

f₂ is the frequency of the class succeeding the modal class,

h is the class width.

Calculation:

The mode is given as 3.6, so the modal class is 3–5 (since the mode lies between 3 and 5). In this class: 

L = 3 (lower boundary of the class),

f₁ = 22 (frequency of the modal class),

f₀ = 16 (frequency of the class preceding the modal class),

f₂ = k (frequency of the class succeeding the modal class),

h = 2 (class width).

We are given that the mode = 3.6, so substituting the values into the mode formula:

3.6 = 3 + [(22 - 16) / (2 × 22 - 16 - k)] × 2

3.6 = 3 + [6 / (44 - 16 - k)] × 2

3.6 = 3 + [6 / (28 - k)] × 2

3.6 - 3 = [6 / (28 - k)] × 2

0.6 = [6 / (28 - k)] × 2

0.6 / 2 = 6 / (28 - k)

0.3 = 6 / (28 - k)

0.3 × (28 - k) = 6

28 - k = 6 / 0.3

28 - k = 20

k = 28 - 20 = 8

Substitute k = 8 into (2k + 1):

(2k + 1) = 2 × 8 + 1 = 16 + 1 = 17

∴ The value of (2k + 1) is 17.

Given:

If mode = 8 and mean – median = 12

Formula used

Mode = mean – 3 (mean - median)

Mode = 3median - 2mean

Calculation

We know that, Mode = mean – 3(mean -median)

Put the value, 8 = mean – 3 (12)

Mean = 36 + 8 = 44

Concept:

Relation between mode, median and mean is given by:

Mode = 3 × median – 2 × mean

Calculation:

Given:

Mode – median = 2

As we know

Mode = 3 × median – 2 × mean

Now, Mode = median + 2

⇒ (2 + median) = 3median – 2mean   

⇒ 2Median - 2Mean = 2

⇒ Median - Mean = 1

Concept:

The mode is the value that appears most often in a set of data values.

Calculation:

32 occurred 8 times

14 occurred 4 times

59 occurred 12 times

41 occurred 8 times 

28 occurred 10 times

7 occurred 16 times 

34 occurred 15 times

20 occurred 9 times

∴ Mode will be 7

Mean is the average of the given numbers,

⇒ Mean = (36 + 28 + 45 + 51)/4 = 160/4 = 40

Variance is calculated by taking the average of the squares of the difference between each term and the mean,

⇒ Variance = [(36 - 40)² + (28 - 40)² + (45 - 40)² + (51 - 40)²]/4

= [16 + 144 + 25 + 121]/4 = 306/4 = 76.5

Given:

Data is 3, 10, 10, 4, 7, 10, 5 

Formula used:

Average deviation about the mean 

\(∑\rm \frac{|x_{i} - x̅|}{n}\) where x̅ = Mean

xi = individual term 

n = total number of terms

Mean = Sum of all the terms/Total number of terms

Calculation:

n = total numbers in a data = 7

Mean x̅ = (3 + 10 + 10 + 4 + 7 + 10 + 5)/7 = 7

Mean deviation from mean = \(∑\rm \frac{|x_{i} - x̅|}{n}\)

Mean deviation from mean = (1/7) × [4 + 3 + 3 + 3 + 0 + 3 + 2]

∴ Mean deviation = 18/7

Given:

Mean of five consecutive even numbers = 16

Formula used:

\({\rm{V}} = \frac{{∑ {{\left| {{\rm{x}} - {\rm{m}}} \right|}^2}}}{{\rm{n}}}\)

\({\rm{Mean\;}}\left( {\rm{m}} \right) = \;\frac{{\left\{ {2{\rm{a\;}} + \left( {{\rm{n\;}} - 1} \right){\rm{d}}} \right\}}}{2}\)

V = variance

∑ = summation

x = observation

n = number of observations

a = 1^st term of the numbers

d = common difference

Calculation:

\(\frac{{\left\{ {2{\rm{a\;}} + \left( {{\rm{n\;}} - 1} \right){\rm{d}}} \right\}}}{2} = 16\)

⇒ 2a + (5 – 1)2 = 32

⇒ 2a + 4 × 2 = 32

⇒ 2a = 32 – 8

⇒ 2a = 24

⇒ a = 12

1^st term = 12

Other terms are 14, 16, 18, 20

\({\rm{V}} = {\rm{\;}}\frac{{{{\left( {12{\rm{\;}} - 16} \right)}^2} + {{\left( {14{\rm{\;}} - 16} \right)}^2} + {{\left( {16{\rm{\;}} - 16} \right)}^2} + {{\left( {18{\rm{\;}} - 16} \right)}^2} + {{\left( {20{\rm{\;}} - 16} \right)}^2}}}{5}\)

⇒ \({\rm{\;}}\frac{{16{\rm{\;}} + {\rm{\;}}4{\rm{\;}} + {\rm{\;}}0{\rm{\;}} + {\rm{\;}}4{\rm{\;}} + 16}}{5}\)

⇒ \({\rm{\;}}\frac{{40}}{5}\)

⇒ 8

⇒ V = 8

∴ The variance of the numbers is 8

Given

3, 4, 5, 7, 10, 10, 10

Concept used

Mean = Average

Deviation is the difference with the given number in the series.

Calculation

Mean = \(\frac{{3 + 4 + 5 + 7 + 10 + 10 + 10}}{7}\)

Mean = 49/7

Mean = 7

Checking the mean deviation with all the numbers given in the series.

Mean deviation 

⇒ |7 - 3|, |7 - 4|, |7 - 5|, |7 - 7|, |7 - 10|, |7 - 10|, |7 - 10|

⇒ 4, 3, 2, 0, 3, 3, 3

Mean deviation = \(\frac{{3 + 4 + 2 + 3 + 3 + 3}}{7}\)

Mean deviation = 18/7

GIVEN :

The standard deviation of a data set is given as 34.

CONCEPT :

The value of variance is the square of standard deviation.

FORMULA USED :
Standard Deviation = √Variance

CALCULATION :  

Using the formula :

Given:

The mid value of a class = 12

Width = 6

Formula used:

Lower limit = Mid value – width/2

Calculation:

Lower limit = 12 – 6/2

⇒ 12 – 3

⇒ 9

∴ The lower limit of the class is 9

Given:

7, 13, 15, 11, 4

Formula used:

 \({\rm{S}}.{\rm{D}} = √ {\frac{{∑|{\rm{x}} - {\rm{\;m|^2}}}}{{\rm{n}}}} \)

Mean (m) = Total of observations/number of observations

S.D = standard deviation

∑ = summation

x = observation

m = mean of the observations

n = number of observation

Calculation:

Mean of 7, 13, 15, 11, 4

⇒ 50/5

⇒ 10

\({\rm{S}}.{\rm{D}} = √ {\frac{{{{\left( {7 - 10} \right)}^2} + {{\left( {13 - 10} \right)}^2} + {{\left( {15\; - \;10} \right)}^2} + {{\left( {11 - 10} \right)}^2} + \;{{\left( {4 - 10} \right)}^2}}}{5}} \)

⇒ \(√ {\frac{{9 + \;9 + 25 + 1 + 36}}{5}} \)

⇒ \(√ {\frac{{80}}{5}} \)

⇒ √16

⇒ 4

∴ The standard deviation is 4