Introduction
Statistics is around us. It would be difficult to go without using statistics.
Without statistics we cannot plan our budgets, pay our taxes, enjoy sports, evaluate classroom performance etc.
Statistics is the art and science of gathering, analyzing and making inferences (predictions) from numerical in formation obtained in an experiment. This numerical information is referred as data.
Statistics  a set of concepts, rules, and procedures that help us to: organize numerical information in the form of tables, graphs, and charts; understand statistical techniques that help to make informed decisions that affect our lives and wellbeing. The use of statistics associated with numbers gathered for governments, has grown significantly and is applied in all walks of life.
An average is a number that is a representation of a group of data. It is commonly referred as "Measures of Central Tendency"
There are five types of measures of central tendency which are commonly used.
1. Arithmetic mean (or) average.
2. Median.
3. Mode.
4. Geometric mean.
5. Harmonic mean.
Arithmetic Mean
The most commonly used measure of central tendency is arithmetic mean.
Arithmetic mean is the average of numbers i.e. the sum of all observations in the data divided by the number of observations
Arithmetic mean=
Thus if x_{1},x_{2},x_{3, }â€¦, x_{n} be n numbers, their arithmetic mean or simply mean, is given by,
x =
n x =
= X_{i}
And n x = x_{i} =
Find the mean of x, (x + 2), (x + 4), (x + 6), (x + 8)
Here n = 5 and
= x + x + 2 + x + 4 + x + 6 + x + 8 = 5x + 20
âˆ´Mean (x) = = = (x + 4)
Find the mean of the first ten prime numbers.
First ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
= 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29
= 129
âˆ´ Mean = = 12.9
(a) If the arithmetic mean of 6, 8, 5, 7, x and 4 is 7 then find x.
(b) If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 11. Find the mean of the first four observations.
(a) = 6 + 8 + 5 + 7 + x + 4 = 30 + x
Here, n = 6 and Mean = 7
âˆ´Mean (x) = = (30 + x)
â‡’ 7 = (30 + x)
â‡’ x = 12
(b) = x + x + 2 + x + 4 + x + 6 + x + 8
= 5x + 20 = 5(x + 4)
Here, n = 5 and mean = 11
Mean (x) = = [5(x + 4)] = (x + 4)
â‡’ 11= x + 4
â‡’ x = 7 (i)
Mean of the first four observations
=
=
= x + 3
â‡’ 7 + 3 = 10 from (i)
Prove that = 0
= (x_{1 } x )+(x_{2 } x)+(x_{3 } x)+â€¦+(x_{n } x)
= (x_{1 }+ x_{2 }+ x_{3 }+â€¦ + x_{n})  nx
= n ^{}  nx
= n x  n x =0
The following is the data showing the exports (in Rs. crores) of selected agricultural commodities from 1966 to 1982. Find the average export (in Rs. crores) per year.
Year AprilMarch 
Export in Rs. Crores 
Year AprilMarch 
Export in Rs. Crores 

196566  334.9  197576  1800.6  
196667  565.3  197677  2000.3  
197071  585.0  197778  1902.6  
197172  751.5  197879  2340.5  
197273  1006.8  198081  2566.7  
197374  1401.5  198182  2899.5  
197475  1685.5 
Here n = 13 and x_{1 }= 334.9, x_{2 }= 565.3,â€¦, x_{13 }= 2899.5
= 334.9 + 565.3 + 585.0 +...+ 2899.5 = 19840.7
âˆ´Mean (x) = Average Export (in Rs. crores)
= = Ã— 19840.7 = 1526.2.
In a school library the following number of books were issued to students on 20 days.
305  500  450  489  397 
296  413  503  437  474 
386  429  453  427  482 
399  325  373  455  365 
Find the number of books issued per day.
Here n = 20 and x_{1 }= 305, x_{2 }= 500, â€¦, x_{20 }= 365
âˆ´ = 305 + 500 + 450 + 489 + 397 + 296 + 413 + 503 + 437 + 474 + 386 + 429 + 453 + 427 + 482 + 399 + 325 + 373 + 455 + 365 = 8358
âˆ´ Mean (x) = = Ã— 8358 = 417.9
âˆ´ Number of books issued per day = 417.
The daily minimum temperature record (in degrees F) of a region is given below for 35 days. Find the mean of the minimum temperature in degrees F.
80, 73, 60, 66, 75, 65, 69, 77, 66, 58, 69, 67, 66, 69, 60, 66, 72, 76, 57, 64, 72, 66, 58, 69, 63, 68,74, 70, 57, 71, 71, 56, 58, 60, 67.
Here n = 35 and x_{1 }= 80 x_{2 }= 73, â€¦, x_{35 }= 67
âˆ´ = 80 + 73 + 60 + 66 + 75 + 65 + 69 + 77 + 66 + 58 + 69 + 67 + 66 + 69 + 60 + 66 + 72 + 76 + 57 + 64 + 72 + 66 + 58 + 69 + 63 + 68 + 74 + 70 + 57 + 71 + 71 + 56 + 58 + 60 + 67 = 2335
âˆ´ Mean Temperature = (x)=
= = 66.71Â°F.
M being the mean of x_{1}, x_{2}, x_{3}, x_{4}, x_{5} and x_{6} find the value of, (x_{1 } M) + (x_{2 } M) + (x_{3 }M) + (x_{4 } M) + (x_{5 } M) + (x_{6 } M)
Here Mean = M =
=> 6M = x_{1 }+ x_{2 }+ x_{3 }+ x_{4 }+ x_{5 }+ x_{6}
Now, (x_{1 } M) + (x_{2 } M) + (x_{3 } M) +(x_{4 } M) + (x_{5 } M) + (x_{6 } M)
â‡’ (x_{1 }+ x_{2 }+ x_{3 }+ x_{4 }+ x_{5 }+ x_{6})  6M
â‡’ 6M  6M = 0.
(a) If x is the mean of the x observations x_{1}, x_{2}, â€¦, x_{n,} prove that the mean of (x_{1 }+ a), (x_{2 }+ a), â€¦, (x_{n }+ a) is x + a.
x =
Now mean of (x_{1}+a),(x_{2}+a),+â€¦,(x_{n}+a)
=
=
= = x +a
If x is the mean of x observations x_{1}, x_{2, }â€¦, x_{n,} prove that the mean of ax_{1}, ax_{2}, â€¦, ax_{n} is ax
The mean of x_{1},x_{2},â€¦,x_{n} observation is given by,
x =
Now mean of ax_{1}, ax_{2},â€¦, ax_{n} observations =
= ^{)}
= a x.
The mean of 5 numbers is 27. If one is excluded, their mean is 25. Find the excluded number.
Total of 5 numbers = mean Ã— number= 27 Ã— 5 = 135
Let the excluded number be x.
Total of remaining 4 numbers = mean Ã— numbers = 25 Ã— 4 = 100
Total of four numbers + x = total of five numbers
âˆ´ 100 + x = 135
â‡’ x = 35.
The mean of 9 numbers is 65. If one number is excluded, their mean is 60. Find the excluded number.
Total of nine numbers =9 Â´ x = 9 Ã— 65 = 585.
Let the excluded number be x
Then total of eight numbers = 585  x
âˆ´Mean of the eight numbers = = 60
â‡’ x = 105.
The mean weight of a class of 40 students is 40 kg. If the weight of the class teacher be included, the mean rises by 2 kg. Find the weight of the teacher.
Mean weight of 40 students = 40 kg.
Total weight of 40 students = 40 Ã— 40 = 1600 kg.
Mean weight of 40 students and the class teacher = 40 + 2 = 42 kg
Total weight of 40 students and class teacher = 42 Ã— 41 = 1722 kg.
Weight of the teacher = 1722  1600 = 122 kg.
The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean.
Mean of 10 numbers = 20
âˆ´Sum of 10 numbers = 10 Ã— 20 = 200
After subtracting 5 from every number, new sum = 200  (10 Ã— 5)
= 150
New mean = = 15.
The sum of the deviations of a set of values x_{1},x_{2},â€¦,x_{n} measures from 50 is (10) and the sum of the deviations of the values from 46 is 70. Find the mean and the value of n.
According to the given conditions, we have
(x_{1 } 50) + (x_{2 } 50) + â€¦ + (x_{n } 50) = 10
â‡’ (x_{1 }+ x_{2 }+ â€¦ + x_{n})  50n = 10 (i)
and (x_{1 } 46) + (x_{2 } 46) + â€¦ +(x_{n } 46) = 70
â‡’ (x_{1 }+ x_{2 }+ â€¦ + x_{n})  46n = 70 (ii)
Subtracting eqs (1) from (2) we get,
4n = 80
â‡’ n = 20
Putting n = 20 in eq. (1) we get,
x_{1 }+ x_{2 }+ â€¦ + x_{n } (50 Ã— 20) = 10
=> x_{1 }+ x_{2 }+ â€¦ + x_{n }= 990
âˆ´Mean (x) = = = 49.5.
The sum of the deviations of a set of n values x_{1}, x_{2}, â€¦ x_{n} measured from 60 is 10 and sum of the deviations of the values from 65 is (40). Find the value of n and the mean.
We have (x_{1 } 60) =10
or  = 10
or  60 n = 10 (i)
Also (x_{i } 65) = 40
or  65 n = 40 (ii)
Subtracting eqs (ii) from (i) we get,
5n = 50
â‡’ n = 10
Now from eq (i) we get,
 (60 Ã— 10) = 10
â‡’ = 610
âˆ´ Mean (x) = = Ã— 610 = 61.
(a) The mean of 100 items was 46. Later on it was found that an item 16 was misread 61 and another item 43 was misread as 34. It was also found that the number of items was 90. Find out the correct mean.
The mean of 100 items = 46
âˆ´x = 46 and n = 100
We know that x =
or n x = Î£x_{i}
â‡’ Î£x_{i }= 100 Ã— 46 = 4600
But 16 was misread as 61 and 43 and 34.
Difference in observations = (16  61) + (43  34) = 36
Therefore, actual total Î£x_{i} = 4600  36 = 4564
[Actual Total = Observed Total + actual reading  misread reading]
Correct number of items = 90
âˆ´ Correct mean = = = 50.71
(b) The mean height of 20 students is 155 cm. It was discovered later on that while calculating the mean, the reading 149 cm was wrongly read as 189 cm. Find the correct mean height.
Here n = 20 and wrong mean = 155 cm
âˆ´Î£x_{i} = nx = 20 Ã— 155 = 3100 cm
âˆ´Wrong Î£x_{i} = 3100
Correct Î£x_{i} = Incorrect Î£x_{i} + correct observation  incorrect observation
â‡’ Correct Î£x_{i} = 3100 + 149  189
= 3060
âˆ´ Correct mean = = = 153
(a) The average weights of 36 girls was 52. But it was discovered lateron that 64 kg was misread as 46 kg for computation. Find the correct mean weight.
Here n = 36 and x = 52
Î£x_{i} = n x â‡’ Î£x_{i }= 36 Ã— 52 = 1872
âˆ´Incorrect Î£x_{1} = 1872
Correct Î£x_{1} = 1872 + 64  46 = 1890
âˆ´x = = = 52.5 kg
(b) The mean of 200 items was 50. Later on it was discovered that the observations 192 and 88 were misread as 92 and 8 respectively for computation of mean. Find out the correct mean.
Here incorrect mean (x) = 50 and n = 200
âˆ´Incorrect value of Î£x_{i} = nx = 200 Ã— 50 = 10000
âˆ´Correct Î£x_{i} = Incorrect Î£x_{i} + correct observation  Incorrect observation
âˆ´Correct Î£x_{i} = 10000 + (192 + 88)  (92 + 8) = 10180
âˆ´Correct mean x = = = 50.9.
The average monthly expenditure of a family was Rs. 6050 during the first four months, Rs. 6180 during the next 5 months and Rs. 6500 during the last three months of the year. If the total savings during the year be Rs. 21400, find the average monthly income.
Total expenditure during the year
= Rs. [6050 Ã— 4 + 6180 Ã— 5 + 6500 Ã— 3]
= Rs.[24500 + 30900 + 19500]
= Rs. 74,600
Savings during the year = Rs. 21,400
Total income during the year = Expenditure + Savings
= Rs. 74,600 + Rs. 21,400 = Rs. 96,000
âˆ´ Average monthly income = = Rs. 8,000.