Question 1: If the heights of $5$ persons are $140 \ cm, 150 \ cm, 152 \ cm, 158 \ cm$ and $161 \ cm$ respectively, find the mean height.

$Mean \ \overline{X} =$ $\frac{140 + 150 + 152 + 158 + 161}{5}$ $=$ $\frac{761}{5}$ $= 152.2 \ cm$

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Question 2: Find the mean of $994, 996, 998, 1002$ and $1000$.

$Mean \ \overline{X} =$ $\frac{994 + 996 + 998 + 1002 + 1000}{5}$ $=$ $\frac{4990}{5}$ $=$ $998$

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Question 3: Find the mean of first five natural numbers.

$Mean \ \overline{X} =$ $\frac{1 + 2 + 3 + 4 + 5}{5}$ $=$ $\frac{15}{5}$ $=$ $3$

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Question 4: Find the mean of all factors of $10$.

Factors of $10$ are $1, 2, 5, 10$

Therefore $Mean \ \overline{X} =$ $\frac{1 + 2 + 5 + 10}{4}$ $=$ $\frac{18}{4}$ $=$ $4.5$

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Question 5: Find the mean of first $10$ even natural numbers.

First $10$ even natural numbers: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$

Therefore $Mean \ \overline{X} =$ $\frac{2+4+6+8+10+12+16+18+20}{10}$ $=$ $\frac{110}{10}$ $=$ $11$

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Question 6: Find the mean of $x, x + 2, x + 4, x + 6, x + 8$.

$Mean \ \overline{X} =$ $\frac{x, + x + 2 + x + 4 + x + 6 + x + 8}{5}$ $=$ $\frac{5x+20}{5}$ $=$ $x+4$

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Question 7: Find the mean of first five multiples of $3$.

First five multiples of $3$ are $3, 6, 9, 12, 15$

Therefore $Mean \ \overline{X} =$ $\frac{3 + 6 + 9 + 12 + 15}{5}$ $=$ $\frac{45}{5}$ $=$ $9$

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Question 8: Following are the weights (in kg) of $10$ new born babies in a hospital on a particular day: $3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6$. Find the mean $\overline{X}$.

$Mean \ \overline{X} =$ $\frac{3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6}{10}$ $=$ $\frac{40}{10}$ $=$ $4$

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Question 9: The percentage of marks obtained by students of a class in mathematics are: $64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1$. Find their mean.

$Mean \ \overline{X} =$ $\frac{64+36+47+23+0+19+81+93+72+35+3+1}{12}$ $=$ $\frac{474}{12}$ $=$ $39.5$

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Question 10: The numbers of children in $10$ families of a locality are: $2, 4,3, 4,2,0,3,5,1,1, 5$. Find the mean number of children per family.

$Mean \ \overline{X} =$ $\frac{2+4+3+4+2+0+3+5+1+1+5}{10}$ $=$ $\frac{30}{10}$ $=$ $3$

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Question 11: If $M$ is the mean of $x_1 , x_2, x_3, x_4, x_5$ and $x_6$, prove that $(x_1- M)+(x_2- M)+(x_3 -M)+ (x_4+-M) +(x_5 - M)+(x_6 -M) = 0$.

$M =$ $\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}$

$\Rightarrow 6M = x_1+x_2+x_3+x_4+x_5+x_6$

$\Rightarrow x_1+x_2+x_3+x_4+x_5+x_6 - 6M = 0$

$\Rightarrow (x_1- M)+(x_2- M)+(x_3 -M)+ (x_4+-M) +(x_5 - M)+(x_6 -M) = 0$

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Question 12: Duration of sunshine (in hours) in Amritsar for first $10$ days of August 1997 as reported by the Meteorological Department are given below: $9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9$.

i) Find the mean $\overline{X}$     ii) Verify that $\sum \limits_{i=1}^{10} ( x_i - \overline{X}) = 0$

i) $Mean \ \overline{X} =$ $\frac{9.6+5.2+3.5+1.5+1.6 + 2.4+ 2.6+8.4+10.3+10.9}{10}$ $=$ $\frac{56}{10}$ $=$ $5.6$

ii) $\sum \limits_{i=1}^{10} ( x_i - \overline{X})$

$= (9.6-5.6) + (5.2-5.6) + (3.5-5.6) + (1.5-5.6) + (1.6-5.6) + (2.4 - 5.6) + (2.6-5.6) + (8.4- 5.6) + (10.3 - 5.6) + (10.9-5.6)$

$= 4 -0.4 - 2.1 - 4.1 - 4 -3.2-3.0+2.8+4.7+ 5.3$

$= 0$

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Question 13: Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant $k$ to each term, (ii) subtracting a constant $k$ from each them, (iii) multiplying each term by a constant $k$ and (iv) dividing each term by a non-zero constant $k$.

i) We have $\overline{X} = \sum \limits_{i=1}^{n} x_i$

Let $\overline{X_1}$ be the mean of observations $x_1+k, x_2+k, x_3+k , ..., x_n+k$. Then

$\overline{X_1} =$ $\frac{(x_1+k) + (x_2+k) + (x_3+k) + ... + (x_n+k)}{n}$

$\Rightarrow \overline{X_1} =$ $\frac{(x_1+x_2+x_2+...+x_n)+ (kn) }{n}$

$\Rightarrow \overline{X_1} =$ $\frac{1}{n}$ $(x_1+x_2+x_2+...+x_n) + k$

$\Rightarrow \overline{X_1} =$ $\frac{1}{n}$ $\sum \limits_{i=1}^{n} x_i + k = \overline{X} + k$

ii) We have $\overline{X} = \sum \limits_{i=1}^{n} x_i$

Let $\overline{X_1}$ be the mean of observations $x_1-k, x_2-k, x_3-k , ..., x_n-k$. Then

$\overline{X_1} =$ $\frac{(x_1-k) + (x_2-k) + (x_3-k) + ... + (x_n-k)}{n}$

$\Rightarrow \overline{X_1} =$ $\frac{(x_1+x_2+x_2+...+x_n)- (kn) }{n}$

$\Rightarrow \overline{X_1} =$ $\frac{1}{n}$ $(x_1+x_2+x_2+...+x_n) - k$

$\Rightarrow \overline{X_1} =$ $\frac{1}{n}$ $\sum \limits_{i=1}^{n} x_i - k = \overline{X} - k$

iii) We have $\overline{X} = \sum \limits_{i=1}^{n} x_i$

Let $\overline{X_1}$ be the mean of observations $k x_1, k x_2, k x_3 , ..., k x_n$. Then

$\overline{X_1} =$ $\frac{(kx_1) + (kx_2) + (kx_3) + ... + (kx_n)}{n}$

$\Rightarrow \overline{X_1} =$ $\frac{k(x_1+x_2+x_2+...+x_n)}{n}$

$\Rightarrow \overline{X_1} =$ $\frac{k}{n}$ $(x_1+x_2+x_2+...+x_n)$

$\Rightarrow \overline{X_1} =$ $\frac{k}{n}$ $\sum \limits_{i=1}^{n} x_i = k\overline{X}$

iv) We have $\overline{X} = \sum \limits_{i=1}^{n} x_i$

Let $\overline{X_1}$ be the mean of observations $\frac{x_1}{k}, \frac{x_2}{k}, \frac{x_3}{k} , ..., \frac{x_n}{k}$. Then

$\overline{X_1} =$ $\frac{(\frac{x_1}{k}) + (\frac{x_2}{k}) + (\frac{x_3}{k}) + ... + (\frac{x_n}{k})}{n}$

$\Rightarrow \overline{X_1} =$ $\frac{\frac{1}{k}(x_1+x_2+x_2+...+x_n)}{n}$

$\Rightarrow \overline{X_1} =$ $\frac{1}{kn}$ $(x_1+x_2+x_2+...+x_n)$

$\Rightarrow \overline{X_1} =$ $\frac{1}{kn}$ $\sum \limits_{i=1}^{n} x_i =$ $\frac{\overline{X}}{k}$

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Question 14: The mean of marks scored by $100$ students was found to be $40$. Later on it was discovered that a score of $53$ was misread as $83$. Find the correct mean.

Given: $\frac{x_1 + x_2 + x_ 3 + ... + x_{99}+x_{100}}{100}$ $= 40$

(Let us assume $x_{100}$ was the incorrect entry)

$\Rightarrow x_1 + x_2 + x_ 3 + ... + x_{99}+x_{100} = 4000$

$\Rightarrow x_1 + x_2 + x_ 3 + ... + x_{99} = 3917$

Hence Correct Mean $\overline{X} =$ $\frac{x_1 + x_2 + x_ 3 + ... + x_{99} + 53}{100}$ $=$ $\frac{3917+53}{100}$ $= 39.7$

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Question 15: The traffic police recorded the speed (in km/hr) of $10$ motorists as $47 , 53, 49 , 60, 39, 42, 55, 57, 52, 48$. Later on an error in recording instrument was found. Find the correct overage speed of the motorists if the instrument recorded $5 \ km/hr$ less in each case.

We will use Property 5: If $\overline{X}$ is the mean of $n$ observations, $x_1, x_2, x_3, ..., x_n$ then the mean of the observations $x_1-a, x_2-a, x_3-a, ..., x_n-a$ is $\overline{X}- a$, where $a$ is any real number.

$Mean \ \overline{X} =$ $\frac{47+53+49+60+39+42+55+57+52+48}{10}$ $=$ $\frac{502}{10}$ $= 50.2 \ km/hr$

Hence the correct $Mean \ \overline{X_1} = 50.2 + 5 = 55.2 \ km/hr$

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Question 16: The mean of five numbers is $27$. If one number is excluded, their mean is $25$. Find the excluded number.

Given: $\frac{x_1+x_2+x_3+x_4+x_5}{5}$ $= 27$

$\Rightarrow x_1+x_2+x_3+x_4+x_5 = 135$

Let us assume that the excluded number is $x_5$

$\Rightarrow x_5 = 135 - (x_1+x_2+x_3+x_4)$   … … … … … i)

Also $\frac{x_1+x_2+x_3+x_4}{4}$ $= 25$

$\Rightarrow x_1+x_2+x_3+x_4 = 100$   … … … … … ii)

From i) and ii) we get

$x_5 = 135 - 100 = 35$

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Question 17: The mean weight per student in a group of $7$ students is $55 \ kg$. The individual weights of $6$ of them (in kg) are $52, 54, 55,53, 56$ and $54$. Find the weight of the seventh student.

Given: $\frac{x_1+x_2+x_3+x_4+x_5+x_6+x_7}{7}$ $= 55$

$\Rightarrow \frac{52+54+55+53+56+54+x_7}{7}$ $= 55$

$\Rightarrow x_7 = 385 - 324 = 61 \ kg$

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Question 18: The mean weight of $8$ numbers is $15$. if each number is multiplied by $2$, what will be the new mean?

We will use Property 3: If $\overline{X}$ is the mean of $x_1, x_2, x_2, ..., x_n$ then the mean of $ax_1, ax_2, ax_3, ..., ax_n$ is $a \overline{X}$, where $a$ is any number different from zero. i.e.  is each observation is multiplied by a non zero number $a$, then the mean is also multiplied by $a$.

$\Rightarrow New \ Mean = 15 \times 2 = 30$

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Question 19: The mean of $5$ numbers is $18$. If one number is excluded, their mean is $16$. Find the excluded number.

Given: $\frac{x_1+x_2+x_3+x_4+x_5}{5}$ $= 18$

$\Rightarrow x_1+x_2+x_3+x_4+x_5 = 90$

Let us assume that the excluded number is $x_5$

$\Rightarrow x_5 = 90 - (x_1+x_2+x_3+x_4)$   … … … … … i)

Also $\frac{x_1+x_2+x_3+x_4}{4}$ $= 16$

$\Rightarrow x_1+x_2+x_3+x_4 = 64$   … … … … … ii)

From i) and ii) we get

$x_5 = 90 - 64 = 26$

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Question 20: The mean of $200$ items was $50$. Later on it was discovered that the two items were misread as $92$ and $8$ instead of $192$ and $88$. Find the correct mean.

Given: $\frac{x_1 + x_2 + x_ 3 + ... + x_{199}+x_{200}}{200}$ $= 50$

(Let us assume $x_{199} \ and \ x_{200}$ was the incorrect entry)

$\Rightarrow x_1 + x_2 + x_ 3 + ... + x_{199}+x_{200} = 10000$

$\Rightarrow x_1 + x_2 + x_ 3 + ... + x_{198} = 9900$

Hence Correct Mean $\overline{X} =$ $\frac{x_1 + x_2 + x_ 3 + ... + x_{198} + 192 + 88 }{100}$ $=$ $\frac{9900+280}{200}$ $= 50.9$

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Question 21: Find the values of $n$ and $\overline{X}$ in each of the following cases:

i) $\sum \limits_{i=1}^{n} ( x_i - 12) = -10$  and  $\sum \limits_{i=1}^{n} ( x_i - 3) = 62$

iii) $\sum \limits_{i=1}^{n} ( x_i - 10) = 30$  and $\sum \limits_{i=1}^{n} ( x_i - 6) = 150$

i)    $(x_1 + x_2 + ... + x_n) - 12n = -10$     … … … … … i)

$(x_1 + x_2 + ... + x_n) - 3n = 62$     … … … … … ii)

Subtract ii) from i) we get

$-9n = -72 \Rightarrow n = 8$

Substituting in i) we get

$(x_1 + x_2 + ... + x_n) = 12(8) - 10 = 86$

Therefore $\overline{X} =$ $\frac{1}{8}$ $\times 86 = 10.75$

ii)    $(x_1 + x_2 + ... + x_n) - 10n = 30$     … … … … … i)

$(x_1 + x_2 + ... + x_n) - 6n = 150$     … … … … … ii)

Subtract ii) from i) we get

$-4n = -120 \Rightarrow n = 30$

Substituting in i) we get

$(x_1 + x_2 + ... + x_n) = 10(30) + 30 = 330$

Therefore $\overline{X} =$ $\frac{1}{30}$ $\times 330 = 11$

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Question 22: The sums of the deviations of a set of $n$ values $x_1, x_2, ... , x_n$ measured from $15$ and $-3$ are $- 90$ and $54$ respectively. Find the value of $n$ and mean.

$\sum \limits_{i=1}^{n} ( x_i - 15) = -90$

$\Rightarrow \sum \limits_{i=1}^{n} ( x_i ) - 15n = -90$     … … … … … i)

$\sum \limits_{i=1}^{n} ( x_i + 3) = 54$

$\Rightarrow \sum \limits_{i=1}^{n} ( x_i ) +3n = 54$     … … … … … ii)

Subtract ii) from i) we get

$-18n = -144 \Rightarrow n = 8$

Substituting in i) we get

$\sum \limits_{i=1}^{n} ( x_i ) = 15(8) - 90 = 30$

Therefore $\overline{X} =$ $\frac{1}{8}$ $\times 30 = 3.75$

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Question 23: Find the sum of the deviations of the variate values $3, 4, 6,7 , 8, 14$ from their mean.

Property 1: If $\overline{X}$ is the mean of $n$ observations, $x_1, x_2, x_3, ..., x_n$, then $\sum \limits_{i=1}^{n} ( x_i - \overline{X}) = 0$. i.e. the algebraic sum of deviations from mean is zero.

Hence the answer is $0$.

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Question 24: If $\overline{X}$ is the  mean of the ten natural numbers $x_1, x_2, x_3, ... , x_{10}$ show that

$(x_1 - \overline{X}) + (x_2 - \overline{X}) + ... + (x_{10} - \overline{X}) = 0$

We have $\overline{X} =$ $\frac{x_1 + x_2 + ... + x_{10} }{10}$

$\Rightarrow 10 \overline{X} = x_1 + x_2 + ... + x_{10}$

Now rearranging we get

$(x_1 - \overline{X}) + (x_2 - \overline{X}) + ... + (x_{10} - \overline{X}) = 0$

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Question 25: If the mean of the observations $45, 52, 60, x, 69, 70, 26, 81, 94$ is $68,$ find the value of $x$.

$Mean \ 68 =$ $\frac{45+52+60+x+69+70+26+81+94}{9}$
$\Rightarrow 612 = 487 + x$
$x = 115$
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