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

The correct answer is (A) is false, (R) is true.

Key PointsAssertion (A): In mean deviation, everything is considered positive​ 

Mean deviation is a measure of dispersion that calculates the average absolute deviation of the values in a dataset from their mean. In mean deviation, not everything is considered positive. Deviations can be either positive or negative, and they are calculated as the absolute difference between the value and the mean. Therefore, some deviations will be positive and some will be negative.

Reason (R): The sum of the deviations of the value of items in a series from arithmetic mean is zero and from other averages it is nearly zero

The reason (R) is true. The sum of the deviations of the value of items in a series from the arithmetic mean is indeed zero, but not from other averages such as median or mode. However, the statement "from other averages it is nearly zero" is not necessarily true as it depends on the distribution of the data. For example, if the data is heavily skewed, the sum of deviations from the median may not be nearly zero.

In summary, the assertion is incorrect and the reason is true.

Key Points

 F-test:

• The F-test is a statistical test used to compare the variances of two or more groups. It is used to determine if a significant difference exists between the variances of the groups being compared. The F-test is based on the F-distribution, a continuous probability distribution that arises when the ratio of two independent chi-square random variables is divided by their degrees of freedom.
• The F-test is a hypothesis testing procedure, and the nature of the test depends on the hypothesis being tested. The test can be either one-tailed or two-tailed, depending on the directionality of the hypothesis. A one-tailed test is used when the hypothesis is directional, and the researcher wants to test whether a specific group has a higher or lower variance than the other group. A two-tailed test is used when the hypothesis is non-directional, and the researcher wants to test whether there is any significant difference between the variances of the compared groups.
• The F-test is commonly used in the analysis of variance (ANOVA) to determine if there is a significant difference between the means of two or more groups. ANOVA is a statistical method used to analyze the variance between multiple groups and compare them to each other. The F-test determines if there is a significant difference in the variance between the groups being compared. If the F-test is statistically significant, it indicates that the variances between the groups are different, and the means of the groups should be compared.
• In summary, the nature of the F-test depends on the hypothesis being tested. If the hypothesis is directional, the F-test is a one-tailed test, and if the hypothesis is non-directional, the F-test is a two-tailed test. The F-test is commonly used in ANOVA to determine if there is a significant difference in the variance between two or more groups.

Hence option 3 is the correct answer.

Given:

47, 58, 96, 1, 6, 52, 6, 78, 102, 1

Formula use:

Median = \({{n \over 2} value +({n \over 2} + 1) value} \over 2\)n2value+(n2+1)value2" role="presentation" style="display: inline; position: relative;" tabindex="0">

Calculation:

Arranging the scores in ascending order

1, 1, 6, 6, 47, 52, 58, 78, 96, 102

Median = \({{n \over 2} value +({n \over 2} + 1) value} \over 2\)n2value+(n2+1)value2" role="presentation" style="display: inline; position: relative;" tabindex="0">

Here, n = 10

Therefore, 

Median = \({{10 \over 2} value +({10 \over 2} + 1) value} \over 2\)

Median = \({{5^{th}} value +{6^{th}} value} \over 2\)

Median = \({47 + 52} \over 2\)

Median = 49.5

∴ The median is 49.5.

Additional InformationTo calculate the Median of data

If n is odd then,

Median = \({(n + 1) \over 2}value\)(n+1)2value" id="MathJax-Element-45-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">

If n is even then,

Median = \({{n \over 2} value +({n \over 2} + 1) value} \over 2\)

Concept:

To find the mean using the direct method, for grouped data,

\(Mean (̅{x})=\frac{\sum f_{i}x_{i}}{\sum f_{i}} \)

where x_i is the classmark and f_i are the frequencies. 

Classmark is the average of the lower limit and upper limit of any class interval.

Calculation:

We know that,

\(Mean (̅{x})=\frac{\sum f_{i}x_{i}}{\sum f_{i}} \)

⇒ x̅ = 315/25 = 12.60

Hence, the mean is 12.60.

Calculation:

The empirical relation between Mean(M), median(M_d) and Mode(M₀)

⇒ M₀ = 3M_d - 2M

Mean = Total sum of all values in the sample divided by the number of observation and it is denoted by x̅ 

x̅ = ∑f_i/n and I = 1, 2, 3, -------

Median = Median of a distribution locates a central point which divides the distribution into two equal halves

Median = (n + 1)/2

Mode = Mode is that value of the distribution which occur more frequency

Given:

The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:

The mode is the value that appears most frequently in a data set

At the time of finding Median

First, arrange the given data in the ascending order and then find the term

Formula used:

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

Median = {(n + 1)/2}^th term when n is odd 

Median = 1/2[(n/2)^th term + {(n/2) + 1}^th] term when n is even

Range = Maximum value – Minimum value 

Calculation:

Arranging the given data in ascending order 

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19

Here, Most frequent data is 4 so 

Mode = 4

Total terms in the given data, (n) = 15 (It is odd)

Median = {(n + 1)/2}th term when n is odd 

⇒ {(15 + 1)/2}^th term 

⇒ (8)^th term

⇒ 6 

Now, Range = Maximum value – Minimum value 

⇒ 19 – 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3

⇒ (17 + 4 + 6)/3 

⇒ 27/3 = 9

∴ The mean of the Range, Mode and Median is 9

Formula used:

The mean of grouped data is given by,

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

Where, \(u_i \ = \ \frac{X_i\ -\ a}{h}\)

Xi = mean of ith class

fi = frequency corresponding to ith class

Given:

Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

Then,

We know that, mean of grouped data is given by

\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)

= \(\frac{1535}{43}\)

= 35.7

Hence, the mean of the grouped data is 35.7

Concept:

Mean: The mean or average of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set.

Mode: The mode is the value that appears most frequently in a data set.

Median: The median is a numeric value that separates the higher half of a set from the lower half. 

Relation b/w mean, mode and median:

Mode = 3(Median) - 2(Mean)

Calculation:

Given that,

mean of data = 4 and mode of  data = 10

We know that

Mode = 3(Median) - 2(Mean)

⇒ 10 = 3(median) - 2(4)

⇒ 3(median) = 18

⇒ median = 6

Hence, the median of data will be 6.

Formula Used∶

• σ² = ∑(x_i – x)²/n
• Standard deviation is same when each element is increased by the same constant

Calculation:

Since each data increases by 10,

There will be no change in standard deviation because (x_i – x) remains same.

∴ The standard deviation of 10, 11, 12, 13 _____ 19 will be will be K.

Alternate Method 

Concept:

Median: The median is the middle number in a sorted- ascending or descending list of numbers.

Case 1: If the number of observations (n) is even

\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)

Case 2: If the number of observations (n) is odd

\({\rm{Median\;}} = {\rm{value\;of\;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{\;observation}}\)

Calculation:

Given values 2, 6, 6, 8, 4, 2, 7, 9

Arrange the observations in ascending order:

2, 2, 4, 6, 6, 7, 8, 9

Here, n = 8 = even

As we know, If n is even then,

\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)

= \(\rm \frac{4^{th} \;\text{observation}+5^{th} \;\text{observation}}{2} \)

= \(\frac{6+6}{2} =6\)

Hence Median = 6

Concept:

Standard deviation:

The standard deviation of the observation set \(\rm \{x_i,i=1,2,3,\cdots\}\) is given as follows:

\(\rm \sigma=\sqrt{\dfrac{\sum\left(x_i-\mu\right)^2}{N}}\)

Where \(\rm N=\mbox{size of the observation set}\) and \(\rm \mu=\mbox{mean of the observations}\) .

Calculations:

First, we will calculate the mean of the given observations.

\(\begin{align*} \mu &= \dfrac{-\sqrt6-\sqrt5-\sqrt4-1+1+\sqrt4+\sqrt5+\sqrt6}{8}= 0 \end{align*}\)

Therefore, the numerator inside the square root term of the standard deviation formula will simply be equal to \(\rm (x_i-\mu)^2=x_i^2\) .

Now we observe that \(\rm N=8\) .

Therefore, the standard deviation is given as follows:

\(\begin{align*} \sigma &= \sqrt{\dfrac{\left(-\sqrt6\right)^2+\left(-\sqrt5\right)^2+\left(-\sqrt4\right)^2+\left(-1\right)^2+\left(1\right)^2+\left(\sqrt4\right)^2+\left(\sqrt5\right)^2+\left(\sqrt6\right)^2}{8}}\\ &= \sqrt{\dfrac{32}{8}}\\ &= \sqrt4\\ &= 2 \end{align*}\)

Therefore, the standard deviation of the given observations is 2.

Calculation:

Let the numbers be x_1, x₂, x₃, x₄.

The mean of four numbers x_1, x₂, x₃, x₄ = 37

The sum of four numbers x_1, x₂, x₃, x₄ = 37 × 4 = 148.

The mean of the smallest three numbers x_1, x₂, x₃ = 34

The sum of the smallest three numbers x_1, x₂, x₃ = 34 × 3 = 102.

∴ The value of the largest number x₄ = 148 – 102 = 46.

The Range (Difference between largest and smallest value) x₄ – x₁ = 15.

∴ Smallest number x₁ = 46 – 15 = 31.

Now,

The sum of x₂, x₃ = Total sum – (sum of smallest and largest number).

⇒ 148 – (46 + 31)

⇒ 148 – 77

⇒ 71

Now,

Concept:

Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

Variance = (Standard Deviation)²

Calculation:

Given coefficient of variation = 45% = 0.45

And mean = 100

As Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

0.45 = \(\rm\text{Standard Deviation} \over100\)

Standard Deviation = 100 × 0.45

SD = 45

∴ Variance = 45² = 2025

\(\bar x\left( {mean} \right)\; = \;\frac{{\sum fx}}{n}\)

⇒ n = total frequency

\(\sum fx = Sum\;of\;the\;product\;of\;mid - interval\;values\;and\;their\;corresponding\;frequency\;\;\)

Mid value of 10 – 12 = (10 + 12)/2 = 11

Mid value of 12 – 14 = (12 + 14 )/2 = 13

Mid value of 14 – 16 = (14 + 16 )/2 = 15

Mid value of 16 – 18 = (16 + 18 )/2 = 17

Mid value of 18 – 20 = (18 + 20 )/2 = 19

\(\Rightarrow \;Mean\; = \;\frac{{11\; \times \;6\; + \;13\; \times \;8\; + \;15\; \times \;5\; + \;17\; \times \;7\; + \;19\; \times \;4}}{{6\; + \;8\; + \;5\; + \;7\; + \;4}} = \;\frac{{440}}{{30}}\)

⇒ Mean = 14.67

∴The mean marks of the given data are 14.67

Concept:

Mode is the value that occurs most often in the data set of values.

Calculation:

Given data values are 3, 8, 6, 7, 1, 6, 10, 6, 7, 2k + 5, 9, 7, and 13

In the above data set, values 6, and 7 have occurred more times i.e., 3 times

But given that mode is 7.

So, 7 should occur more times than 6.

Hence the variable 2k + 5 must be 7

⇒ 2k + 5 = 7

⇒ 2k = 2