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

Formula

Mean = ∑x_if_i/n

f = frequency

x = observatons

n = number of observations

Calculation

Mean = 83/14

∴ Mean of this series is 83/14

The arithmetic mean is denotted by X̅ is given by

X̅ = (x1₁ + x₂ + ------ x_n)/n

X̅ = ∑x_i/n

Where as (x₁ + x₂ + ------ x_n) are observations

n = Number of observation

1 - For any distribution the sum of deviation from the arithmetic mean is always zero

2 - If each value of the variate x is increased or decreased by a constant value then the arithmetic mean of the variate so obtained is also increased or decreased by the same constant value

3 - The sum of the square of the deviation of a set of values is minimum when taken about mean.

4 - If the values of the variate are multiplied or divided by a constant value, the arithmetic mean so obtained is same as the initial arithmetic mean multiplied ro divided by the constant value

Concept:

Consecutive even or odd numbers have a difference of 2 between successive terms.

e.g. 2, 4, 6, 8, 10, 12...

4 - 2 = 6 - 4 = 8 - 6 = constant = 2

Calculation:

Given:

All numbers are consecutive even numbers.

Let the numbers be (x - 2), x and (x + 2)

mean = \(\frac{(x - 2) + x + (x + 2)}{3}=\frac{3x}{3} = x\)

Largest number = (x + 2)

Required difference = (x + 2) - x

Required difference = 2

Explanation:

Positively Skewed Distribution

• A positively skewed distribution, also known as right-skewed distribution, is a type of frequency distribution in which most of the data points are concentrated on the left side, with the tail extending to the right. This distribution is characterized by a long tail on the right-hand side. In such distributions, the mean, median, and mode are not equal and typically follow a specific order.

Characteristics: In a positively skewed distribution:

• The mode is the highest peak of the distribution and is located at the leftmost part of the distribution.
• The median is located to the right of the mode but to the left of the mean.
• The mean is located to the right of both the median and the mode, as it is pulled in the direction of the long tail on the right.

(Monday + Tuesday + Wednesday) / 3 = 41

(Monday + Tuesday + Wednesday) = 123

Tuesday + Wednesday = 123 - Monday      ----(1)

(Tuesday + Wednesday + Thursday) / 3= 43

(Tuesday + Wednesday + Thursday) = 129      ----(2)    

We know, Thursday = 1.15 Monday

⇒ 123 - Monday + 1.15 Monday = 129

⇒ Monday = 400C

Equation (2) becomes

⇒ 123 - Monday + Thursday = 129  

⇒ 123 - 40 + Thursday = 129

⇒ Thursday = 460C

Hence, the correct answer is 460C.

CONCEPT:
Direct Method:

If a variable X takes value x1, x2, x3, ….., xn with corresponding frequencies f1, f2, …, fn respectively then the arithmetic mean of these values is given by: \(\bar X = \frac{{\mathop \sum \nolimits_{i = 1}^n \left( {{f_i} ⋅ {x_i}} \right)}}{N}\) where \(N = \;\mathop \sum \limits_{i = 1}^{n\;} {f_i}\)

CALCULATION:

Given: Mean of the data is 40

i.e \(\bar X = 40\)

As we know that, \(\bar X = \frac{{\mathop \sum \nolimits_{i = 1}^n \left( {{f_i} ⋅ {x_i}} \right)}}{N}\)

By using the table given above we get

⇒ \(\bar X = \frac{3960 + 30k}{88 + k} = 40 \)

⇒ 3960 + 30k = 3520 + 40k

⇒ 10k = 440

⇒ k = 44

Hence, option 1 is the correct answer.

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

So \(\mathop \smallint \limits_0^1 \left( {a + bx} \right)dx = 1\) 

\(a + \frac{b}{2} = 1\),  \(2a + b = 2\) 

Given \(E\left[ X \right] = \frac{2}{3} = \mathop \smallint \limits_0^1 X\left[ {a + bx} \right]dx\)

\(\frac{2}{3} = \frac{a}{2} + \frac{b}{3}\) ,   \(3a + 2b = 4\) 

From above equations, we get      \(a = 0,\;b = 2\)

Explanation:

\(\begin{array}{l} f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{x}{4}\left( {4 - {x^2}} \right),}&{0 \le x \le 2}\\ 0&{,otherwise} \end{array}} \right.\\ {\mu _x} = \mathop \smallint \limits_0^2 xf\left( x \right)dx \end{array}\)

∴ Mean \(\left( {{\mu _x}} \right) = \mathop \smallint \limits_0^2 x\frac{x}{4}\left( {4 - {x^2}} \right)dx\) 

\(\begin{array}{l} {{\rm{Mean}}}= \mathop \smallint \limits_0^2 \left( {{x^2} - \frac{{{x^4}}}{4}} \right)dx\\ {{\rm{Mean}}}= \left[ {\frac{{{x^3}}}{3} - \frac{{{x^5}}}{{20}}} \right] = \frac{8}{3} - \frac{{32}}{{20}} = \frac{{16}}{{15}} \end{array}\)

Mean = 1.066

Calculation:

Let a, b, c ...... be the 'n' set of observations.

The mean of this set of observations,  \(\bar{x}\) = \(a + b+ c ....... \over n\)

After dividing each observation by k and then increasing them by 5, a new set of observations are \(a \over k \) + 5, \(b \over k\) + 5, ....... 

So, Mean of new set of observations

New mean = \(\frac{{\left( {\frac{a}{k} + 5} \right) + \left( {\frac{b}{k} + 5} \right) + ......}}{n} = \frac{{\left( {\frac{a}{k} + \frac{b}{k} + ....} \right) + \left( {5n} \right)}}{n}\)

Rearranging the above terms, we get, 

New mean = \(\frac{{\left( {\frac{{a + b + .....}}{n}} \right) + 5k}}{k}\)

New mean = \(\dfrac{\bar{x} + 5k}{k}\)

Given that,

\(\begin{array}{*{20}{c}} {{p_x}\left( x \right) = {e^{ - x}}}&{for\;x \ge 0}\\ {\; = 0}&{otherwise} \end{array}\)

\({g_x}\left( x \right) = {e^{\frac{{3x}}{4}}}\)

Expected value of g_x(x)

\(E = \mathop \smallint \limits_0^\infty {p_x}\left( x \right){g_x}\left( x \right)dx\)

\(= \mathop \smallint \limits_0^\infty {e^{ - x}}{e^{\frac{{3x}}{4}}}dx\)

\(= \mathop \smallint \limits_0^\infty {e^{\frac{{ - x}}{4}}}dx\)

\(= \left[ {\frac{{{e^{ - \frac{x}{4}}}}}{{ - \frac{1}{4}}}} \right]_0^\infty \)

= 4

Explanation:

Let E(x) represent mean

Then by the property of mean

E(Cx) = cE(x) where c is constant

And E(x + c) = E(x) + E(c) = E(x) + c

∴ Options P and R always holds true.

Now,

Let Var (x) represent the variance

Then var (x + c) = var (x) + var (c)

 var (x + c) = var (x)    {∵ var (c) = 0}

∴ S.D (x + c) = S.D (x)

Now,

var (cx) =(c²)var (x)

S.D (cx) = (c)S.D (x)

∴ Options Q and S are incorrect.

Concept:

Suppose we have two sets of data containing n1 and n2 observations with means X₁ and X₂

The combined mean is given as:

 \(X = \frac{n_1 ~{X_1} + n_2 ~ {X_2}}{n_1 + n_2}\)  

Given:

n₁ = 13, n₂ = 17, X₁ = 37, X₂ = 43

\(X = \frac{n_1 ~{X_1} + n_2 ~ {X_2}}{n_1 + n_2}= \frac{13 \times 37 + 17\times 43}{13+17}\)

= 40.4

(Monday + Tuesday + Wednesday) / 3 = 41

(Monday + Tuesday + Wednesday) = 123

Tuesday + Wednesday = 123 - Monday      ----(1)

(Tuesday + Wednesday + Thursday) / 3= 43

(Tuesday + Wednesday + Thursday) = 129      ----(2)    

We know, Thursday = 1.15 Monday

⇒ 123 - Monday + 1.15 Monday = 129

⇒ Monday = 400C

Equation (2) becomes

⇒ 123 - Monday + Thursday = 129  

⇒ 123 - 40 + Thursday = 129

⇒ Thursday = 460C

Hence, the correct answer is 460C.

\(\mathop \smallint \limits_{ - \infty }^\infty f\left( x \right)dx = 1\)

So \(\mathop \smallint \limits_0^1 \left( {a + bx} \right)dx = 1\) 

\(a + \frac{b}{2} = 1\),  \(2a + b = 2\) 

Given \(E\left[ X \right] = \frac{2}{3} = \mathop \smallint \limits_0^1 X\left[ {a + bx} \right]dx\)

\(\frac{2}{3} = \frac{a}{2} + \frac{b}{3}\) ,   \(3a + 2b = 4\) 

From above equations, we get      \(a = 0,\;b = 2\)

Formula

Mean = ∑x_if_i/n

f = frequency

x = observatons

n = number of observations

Calculation

Mean = 83/14

∴ Mean of this series is 83/14

The arithmetic mean is denotted by X̅ is given by

X̅ = (x1₁ + x₂ + ------ x_n)/n

X̅ = ∑x_i/n

Where as (x₁ + x₂ + ------ x_n) are observations

n = Number of observation

1 - For any distribution the sum of deviation from the arithmetic mean is always zero

2 - If each value of the variate x is increased or decreased by a constant value then the arithmetic mean of the variate so obtained is also increased or decreased by the same constant value

3 - The sum of the square of the deviation of a set of values is minimum when taken about mean.

4 - If the values of the variate are multiplied or divided by a constant value, the arithmetic mean so obtained is same as the initial arithmetic mean multiplied ro divided by the constant value

Explanation:

\(\begin{array}{l} f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{x}{4}\left( {4 - {x^2}} \right),}&{0 \le x \le 2}\\ 0&{,otherwise} \end{array}} \right.\\ {\mu _x} = \mathop \smallint \limits_0^2 xf\left( x \right)dx \end{array}\)

∴ Mean \(\left( {{\mu _x}} \right) = \mathop \smallint \limits_0^2 x\frac{x}{4}\left( {4 - {x^2}} \right)dx\) 

\(\begin{array}{l} {{\rm{Mean}}}= \mathop \smallint \limits_0^2 \left( {{x^2} - \frac{{{x^4}}}{4}} \right)dx\\ {{\rm{Mean}}}= \left[ {\frac{{{x^3}}}{3} - \frac{{{x^5}}}{{20}}} \right] = \frac{8}{3} - \frac{{32}}{{20}} = \frac{{16}}{{15}} \end{array}\)

Mean = 1.066