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.

Answer:

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 .

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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 .

Answer:

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 .

Answer:

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} .

Answer:

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.

Answer:

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.

Answer:

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 .

Answer:

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

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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.

Answer:

\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.

Answer:

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

Answer:

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 .

Answer:

Mean \ 68 = \frac{45+52+60+x+69+70+26+81+94}{9}

\Rightarrow 612 = 487 + x

x = 115

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