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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \text{i) Find the mean } \overline{X}$

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

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

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

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

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

$\displaystyle = 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \text{Hence Correct Mean } \overline{X_1} = 50.2 + 5 = 55.2 \ km/hr$

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

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

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

Let us assume that the excluded number is $\displaystyle x_5$

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

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

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

From i) and ii) we get

$\displaystyle x_5 = 135 - 100 = 35$

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

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

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

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

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

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

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

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

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

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

Let us assume that the excluded number is $\displaystyle x_5$

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

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

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

From i) and ii) we get

$\displaystyle x_5 = 90 - 64 = 26$

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

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

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

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

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

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

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

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

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

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

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

Subtract ii) from i) we get

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

Substituting in i) we get

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

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

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

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

Subtract ii) from i) we get

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

Substituting in i) we get

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

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

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

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

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

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

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

Subtract ii) from i) we get

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

Substituting in i) we get

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

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

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

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

Hence the answer is $\displaystyle 0$.

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Question 24: If $\displaystyle \overline{X}$ is the mean of the ten natural numbers $\displaystyle x_1, x_2, x_3, ... , x_{10}$ show that $\displaystyle (x_1 - \overline{X}) + (x_2 - \overline{X}) + ... + (x_{10} - \overline{X}) = 0$

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

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

Now rearranging we get

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

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

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