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

Concept:

The regression line of Y on X: 

Y =  b1X + b0 

The regression line of X on Y: 

X = a1Y +  a0 

Where b0, a0 are constants and b1, a1 are regression coefficients.

Explanation: 

We have 3x + y - 12 = 0  

⇒ 3x = - y + 12

⇒  x = \(-\frac{y}{3} + 4\)

Also,

x +  2y - 14 = 0

⇒ 2y = - x + 14

⇒ y = \(-\frac{x}{2} + 7\)

The regression line of y on x is  y = \(-\frac{x}{2} + 7\)

∴ Statement 1 is correct.

Since the lines of regression passes through \((\bar x, \bar y)\).

∴ \(3\bar x +\bar y - 12 =0 \)  --- (i)

and \(\bar x + 2 \bar y - 14 = 0\) ---- (ii)

Equation (i) - 3 × (ii), we get

⇒ \(3\bar x +\bar y - 12 - 3(\bar x+ 2\bar y - 14) = 0\)

⇒ \(3\bar x +\bar y - 12 - 3\bar x-6\bar y + 42 = 0\)

⇒ \(-5\bar y + 30 = 0\)

⇒ \(\bar y = 6\)

From equation (i),  \(3\bar x +6 - 12 =0 \)

⇒ \(3\bar x -6 =0 \)

⇒ \(\bar x = 2\)

∴ Statement 2 is correct.

The correct answer is option 1) always equal to 0.

 Key Points

• Deviation is the distance between each data point and the mean of the data set.
• The sum of deviations taken away from the mean is calculated by adding up all the distances between each data point and the mean.
• The sum of deviations taken away from the mean is always equal to 0, regardless of the data set.

Additional Information

•  This property of the sum of deviations from the mean is used to calculate variance and standard deviation, which are important statistical measures.
• The formula for calculating the sum of deviations taken away from the mean is: ∑(Xi - X̄) = 0, where Xi represents each data point and X̄ represents the mean of the data set.
• This property is a consequence of the fact that the mean is the balancing point of the data set, where the sum of deviations above the mean is equal to the sum of deviations below the mean. 

Hence, the sum of deviations taken away from the mean is always equal to 0, regardless of the data set.

The correct answer is 13 

Key Points

•  Calculation - 
  • The formula 3 median = mode + 2 mean is an empirical relationship that may hold true for approximately symmetric distributions.
  • Substituting the value in the formula we get,
  • \(3*(39) = Mode + 2*(37)\\ Mode = 13\)
• Mode is the value that occurs most frequently.
• Median is the middle value when the dataset is arranged in order.
• Mean is the sum of all values divided by the number of values.

Additional Information

•  If the mode, median, and mean are all equal, the distribution is perfectly symmetric.
• If the mean is greater than the median, the distribution is positively skewed.
• If the mean is less than the median, the distribution is negatively skewed.

Hence,by using the formula the correct answer we get, 13 

Concept:

Median: The median is the middle value of the observation.

Calculation:

Ascending Order:

2, 3, 3, 4, 4, 8, 11

The middle value is 4.

So, Median is 4

Formula Used:

Mean proportion = √ab

a = First Term

b = Second term

Calculation:   

Mean Proportion = √4 × 16

Mean Proportion = 8

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:

• Frequency of any class = cumulative frequency of class - cumulative frequency of preceding class

• Cumulative frequency is the sum of all the previous frequencies up to the current point.

Calculation:

We are already having the cumulative frequencies. Let's find out the frequencies:

Hence, the frequency of class intervals 30,000 - 40,000 is 6.