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

Calculation:

Given,

The sequence is \(8^{2},\,9^{2},\,10^{2},\,\dots,\,15^{2}\)

Number of terms, \(n = 15 - 8 + 1 = 8\)

The sum of the first \(m\) squares is  \( \displaystyle \sum_{k=1}^{m} k^{2} = \frac{m(m+1)(2m+1)}{6}\)

Sum up to \(15^{2}\):

\(S_{15} = \frac{15 \times 16 \times 31}{6} = 1240\)

Sum up to \(7^{2}\):

\(S_{7} = \frac{7 \times 8 \times 15}{6} = 140\)

Sum of the required terms:

\(S = S_{15} - S_{7} = 1240 - 140 = 1100\)

The arithmetic mean is,

\( \displaystyle \text{Mean} = \frac{S}{n} = \frac{1100}{8} = 137.5\)

∴ The arithmetic mean is \(137.5\).

Hence, the correct answer is option 3.

Explanation:

Given:

\(\overline x = \frac{x_1+x_2+x_3+......... x_n}{n}\)

⇒ \(x_1+x_2+X_3+......x_n= n\overline x \)

⇒ Mean when x_n is replaced by k

⇒ New mean = \(\frac{x_1+x_2+x_3+......... x_{n-1}+k}{n}\)

= \(\frac{n\overline x -x_n+k}{n}\)

∴ Option (d) is correct

Explanation:

Mean  = \(\frac{1+ 4+ 9 ... n^2}{n}\)

⇒ 130 = \(\frac{(n(n+1)(2n+1)}{6n}\)

⇒ 780 = 2n² + 3n + 1

⇒ 2n² + 3n – 779 =0

⇒ 2n² + 41n – 38n – 779 = 0

⇒ n(2n + 41) – 19(2n + 41) = 0 

⇒ (2n + 41) (n – 19) = 0

⇒ x = 19 ..... ( n = \(-\frac{41}{2}\) is not possible)

∴ Option (b) is correct.

Calculation

Arithmetic mean x̅ of 4, 7, 8, 9, 10, 12, 13, 17 is

x̅ = \(\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10\)

∑|x_i - x̅| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24

∴ Mean deviation about mean

= \(\text { M.D. }(\bar{x})=\frac{\sum\left|x_{i}-x\right|}{n}=\frac{24}{8}=3\)

Hence option 1 is correct

Solution:

1. Given Data:  

Marks: 15, 20, 25, 25, 30, 35, 40, 50 .  

2. Calculate Mean ( \(\mu \)):  

\(\mu = \frac{{Sum \quad of \quad observations}}{{Number\quad of \quad observations}} \)
  

\(\mu = \frac{15 + 20 + 25 + 25 + 30 + 35 + 40 + 50}{8} \)

\(\mu = \frac{260}{8} = 32.5 \)

For the median:  

Arrange the data in ascending order (already done).  

Since there are 8 observations (even number),

the median is the average of the 4th and 5th values:  

\({Md} = \frac{25 + 30}{2} = 27.5 \)

The mode is the value that occurs most frequently in the dataset.

Here: \( {Mo} = 25 \quad \)(occurs twice, more frequent than others).

 Find \( \mu + {Md} + {Mo} \) =  32.5 + 27.5 + 25 = 85.0

Hence 85.00 is the answer.

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:

• \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

Calculation:

We know that, \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

\(\therefore {\rm{Mean}} = \frac{{800}}{{24}} = 33.3\)

Concept:

Suppose there are ‘n’ observations {\({\rm{\;}}{{\rm{x}}_{\rm{1}}},{{\rm{x}}_{{\rm{2\;}}}},{{\rm{x}}_{{\rm{3\;}}}}, \ldots ,{{\rm{x}}_{{\rm{n\;}}}}\)}

Mean \(\left( {{\rm{\bar x}}} \right) = \frac{{{\rm{\;}}({{\rm{x}}_1} + {{\rm{x}}_{2{\rm{\;}}}} + {{\rm{x}}_{3{\rm{\;}}}} + \ldots + {{\rm{x}}_{{\rm{n\;}}}})}}{{\rm{n}}}\) \( = {\rm{\;}}\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Sum of the first n natural numbers = \(\rm \frac{n(n+1)}{2}\)

Calculation:

To find:  Mean of the first 99 natural numbers

As we know, Sum of first n natural numbers = \(\rm \frac{n(n+1)}{2}\)

Now, Mean = \(\rm \dfrac { \frac{99(99+1)}{2}}{99}\)

= \(\rm \frac{(99+1)}{2}=50\)

Concept:

\(\rm Mean\ of\ 'n'\ observations=\dfrac{Sum\ of\ observations}{n}\).

Calculation:

Here n = 3. Let's say that the third number is x.

∴ \(\rm 16=\dfrac{x+8+10}{3}\)

⇒ x + 18 = 48

⇒ x = 30.

Concept:

\({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

Calculation:

We know that, \({\rm{Mean}} = \frac{{\sum {\rm{xf}}}}{{\sum {\rm{f}}}}\)

\(\therefore {\rm{Mean}} = \frac{{830}}{{20}} = 41.5\)

Concept:
Mean - mode = 3(Mean - median)

Calculation:

Given:

The difference between mean and mode = 48

And median = 12

Mean - mode = 3(Mean - median)

⇒ 48 = 3(mean - 12)

⇒ 48 = 3 mean - 36

⇒ 3 mean = 48 + 36

⇒ 3 mean = 84

⇒ mean = 84/3

⇒ mean = 28

∴ Mean is 28

Concept:

For the observations \(\rm x_1 ,x_2 ,x_3 ,..........,x_n\)

Mean (\(\rm\overline x\)) = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)

Deviation of the observation from the mean = x_i - \(\rm\overline x\)

where i ∈ [1, n]

Calculation:

Mean = \(\rm {73+85+92+105+120\over 5}\)

\(\rm\overline x\) = \(\rm {475\over 5}\) = 95

The sum of the deviation = -22 +(-10) + (-3) + 10 + 25 = 0

Concept:

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

Calculation:

Let the six numbers be a, b, c, d, e, f

So, Mean = \(\rm \frac{a+b+c+d+e+f}{6}=47\)

\(\rm ⇒ a+b+c+d+e+f=282\)          ....(1)

Let, the excluded number be a, 

So mean of remaining five numbers = \(\rm \frac{b+c+d+e+f}{5}=41\)

\(\rm ⇒ b+c+d+e+f=205\)                ....(2)

∴ a + 205 = 282              (from (1) and (2))

⇒ a = 77

Hence, option (1) is correct.

Concept:

\({\rm{Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\), where x_i is observations and n is total number of observation.

Sum of the first n natural numbers: \(\sum {\rm{n}} = 1 + 2 + 3 + \ldots + {\rm{n}} = {\rm{\;}}\frac{{{\rm{n}}\left( {{\rm{n}} + 1} \right)}}{2}\)

Calculation:

Given: Mean of a set of observations x₁, x₂, x₃,..., x₁₀ is 50

Here n = 10,

\({\rm{Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\;}}\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + \ldots + {{\rm{x}}_{10}}}}{{10}} = 50\)

⇒ x₁ + x₂ + x₃ +...+ x₁₀ = 500                      …. (1)

Now find out the mean of new observations x₁ + 5, x₂ + 10, x₃ + 15,..., x₁₀ + 50

\({\rm{\;Mean\;}} = {\rm{\;}}\frac{{\sum {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}} = {\rm{\;}}\frac{{\left( {{{\rm{x}}_1} + 5} \right) + \left( {{{\rm{x}}_2} + 10} \right) + ({{\rm{x}}_3} + 15) + \ldots + ({{\rm{x}}_{10}} + 50)}}{{10}}\)

 \( = {\rm{\;}}\frac{{\left( {{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + \ldots + {{\rm{x}}_{10}}} \right) + \left( {5 + 10 + 15 + \ldots + 50} \right)}}{{10}} \)

From equation (1),

\(= \frac{{500 + \;5\left( {1 + 2 + 3 + \ldots + 10} \right)}}{{10}} \)

\(= 50 + \frac{1}{2} \times \frac{{10 \times 11}}{2} \)

\(= 50 + 27.5 \)

= 77.5

CONCEPT:

The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol  \(\bar x\), read as ‘x bar’.

Mean \(\rm \bar x = \frac{{Sum\;of\;all\;the\;observations}}{{Total\;number\;of\;observations}}\)

CALCULATION:

Given set of number is 8, 5, n, 10, 15, 21.

So \(\rm \bar x = \frac{{Sum\;of\;all\;the\;observations}}{{Total\;number\;of\;observations}} = 11\)

∴ \(11 = \frac{{8\; + \;5\; + \;n\; +\; 10\; + \;15\; + \;21}}{6}\)  ⇒ n = 7