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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 well-being. 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 x1,x2,x3, …, xn be n numbers, their arithmetic mean or simply mean, is given by,

         x =

       n x =

             = Xi
And n x = xi =

 

Example

Find the mean of x, (x + 2), (x + 4), (x + 6), (x + 8)

Solution

Here n = 5 and

 = x + x + 2 + x + 4 + x + 6 + x + 8 = 5x + 20

Mean (x) =  =  = (x + 4)
 


 

Example

Find the mean of the first ten prime numbers.

Solution

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

 Mean =  = 12.9

 

 

 

Example

(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.

Solution

(a)  = 6 + 8 + 5 + 7 + x + 4 = 30 + x

Here, n = 6 and Mean = 7


Mean (x) =  = (30 + x)
                      ⇒ 7 = (30 + x)

                       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)

       ⇒ 11= x + 4

         x = 7     ----(i)

Mean of the first four observations

                  = 

                  = 

                  = x + 3

                  7 + 3 = 10 from (i) 

 

 

Example

Prove that  = 0

Solution

  = (x1 x )+(x2 x)+(x3 x)+…+(xn x)

                  = (x1 + x2 + x3 +… + xn) - nx

                  = n  - nx

                  = n x - n x =0
 


 

Example

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
April-March
Export
in Rs. Crores
  Year
April-March
Export
in Rs. Crores
1965-66 334.9   1975-76 1800.6
1966-67 565.3   1976-77 2000.3
1970-71  585.0   1977-78 1902.6
1971-72 751.5   1978-79 2340.5
1972-73 1006.8   1980-81  2566.7
1973-74 1401.5   1981-82 2899.5
1974-75 1685.5      

Solution

Here n = 13 and x1 = 334.9, x2 = 565.3,…, x13 = 2899.5

 = 334.9 + 565.3 + 585.0 +...+ 2899.5 = 19840.7


Mean (x) = Average Export (in Rs. crores)

               = 
 =  × 19840.7 = 1526.2. 

 

 

 

Example

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.

Solution

Here n = 20 and x1 = 305, x2 = 500, …, x20 = 365
 = 305 + 500 + 450 + 489 + 397 + 296 + 413 + 503 + 437 + 474 + 386 + 429 + 453 + 427 + 482 + 399 + 325 + 373 + 455 + 365 = 8358
  Mean (x) = = × 8358 = 417.9
 Number of books issued per day = 417.
 


 

Example

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.

Solution

Here n = 35 and x1 = 80 x2 = 73, …, x35 = 67
  = 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

 Mean Temperature = (x)= 

                              =  = 66.71
°F.
 



 

Example

M being the mean of x1, x2, x3, x4, x5 and x6 find the value of, (x1 -  M) + (x2  -  M) + (x3 -M) + (x4  -  M) + (x5 - M) + (x6 - M)

Solution

Here Mean = M = 
=>   6M = x1 + x2 + x3 + x4 + x5 + x6

Now, (x1 - M) + (x2 - M) + (x3 - M) +(x4 - M) + (x5 - M) + (x6 - M)

⇒ (x1 + x2 + x3 + x4 + x5 + x6) - 6M
⇒  6M - 6M = 0.

 


 

Example

(a) If x is the mean of the x observations x1, x2, …, xn, prove that the mean of (x1 + a), (x2 + a),  …, (xn + a) is x + a.

Solution

 x = 

Now mean of (x1+a),(x2+a),+…,(xn+a)





 = x +a

 


 

Example

If x is the mean of x observations x1, x2, …, xn, prove that the mean of ax1, ax2, …, axn is ax

Solution

The mean of x1,x2,…,xn observation is given by,

                                                  x = 

Now mean of ax1, ax2,…, axn observations = 

                                                            = )

                                                            = a x.
 

 

 

Example

The mean of 5 numbers is 27. If one is excluded, their mean is 25. Find the excluded number.

Solution

Total of 5 numbers = mean × number= 27 × 5 = 135
Let the excluded number be x.
Total of remaining 4 numbers = mean
× numbers = 25 × 4 = 100
Total of four numbers + x = total of five numbers
                      
 100 + x = 135

                                ⇒ x = 35. 

 


 

Example

The mean of 9 numbers is 65. If one number is excluded, their mean is 60. Find the excluded number.

Solution

Total of nine numbers =9 ´  x = 9 × 65 = 585.
Let the excluded number be x
Then total of eight numbers = 585 - x

Mean of the eight numbers =  = 60

                                        ⇒ x = 105.           


 

Example

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.

Solution

Mean weight of 40 students = 40 kg.
Total weight of 40 students = 40
 × 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
× 41 = 1722 kg.
Weight of the teacher = 1722 - 1600 = 122 kg.


 

Example

The mean of 10 numbers is 20. If 5 is subtracted from every number, what will be the new mean.

Solution

Mean of 10 numbers = 20
Sum of 10 numbers = 10 × 20 = 200
After subtracting 5 from every number, new sum = 200 - (10
 × 5)

                                                                    = 150
New mean =  = 15.

 

 

Example

The sum of the deviations of a set of values x1,x2,…,xn 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.

Solution

According to the given conditions, we have
 (x1 - 50) + (x2 - 50) + … + (xn - 50) = -10

⇒ (x1 + x2  + … + xn) - 50n = -10          (i)
and (x1 - 46) + (x2 - 46) + … +(xn - 46) = 70

⇒ (x1 + x2 + … + xn) - 46n = 70            (ii)
Subtracting eqs (1) from (2) we get,
                   4n = 80

                 ⇒ n = 20
Putting n = 20 in eq. (1) we get,
x1 + x2 + … + xn - (50
× 20) = -10
=>   x1 + x2 + … + xn = 990
Mean (x) =  =  = 49.5.

 

 

Example

The sum of the deviations of a set of n values x1, x2, … xn 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.

Solution

We have (x1 - 60) =10

or  -  = 10

or - 60 n = 10    ----(i)

Also (xi - 65) = -40

or  - 65 n = -40 ----(ii)
Subtracting eqs (ii) from (i) we get,

                5n = 50

               n = 10
Now from eq (i) we get,

          - (60
× 10) = 10

                     = 610
 Mean (x) =  = × 610 = 61.

 


 

Example

(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 mis-read as 34. It was also found that the number of items was 90. Find out the correct mean.

Solution

The mean of 100 items = 46
x = 46 and n = 100
We know that x = 
              or n x = 
Σxi

              Σxi = 100 × 46 = 4600
But 16 was misread as 61 and 43 and 34. 
Difference in observations = (16 - 61) + (43 - 34) = -36
Therefore, actual total
Σxi = 4600 - 36 = 4564
[Actual Total = Observed Total + actual reading - misread reading]
Correct number of items = 90

 Correct mean =  =  = 50.71

 

 

Example

(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.

Solution

Here n = 20 and wrong mean = 155 cm
∴Σxi = nx = 20 × 155 = 3100 cm
Wrong Σxi = 3100
   Correct
Σxi = Incorrect Σxi + correct observation - incorrect observation
⇒ Correct Σxi = 3100 + 149 - 189

                   = 3060
 Correct mean =  =  = 153


 

Example

(a) The average weights of 36 girls was 52. But it was discovered later-on that 64 kg was mis-read as 46 kg for computation. Find the correct mean weight.

Solution

Here n = 36 and x = 52

Σxi = n x ⇒ Σxi = 36 × 52 = 1872
Incorrect Σx1 = 1872
Correct
Σx1 = 1872 + 64 - 46 = 1890
x =  =  = 52.5 kg

 

 

Example

(b) The mean of 200 items was 50. Later on it was discovered that the observations 192 and 88 were mis-read as 92 and 8 respectively for computation of mean. Find out the correct mean.

Solution

Here incorrect mean (x) = 50 and n = 200
Incorrect value of Σxi = nx = 200 × 50 = 10000
Correct Σxi = Incorrect Σxi + correct observation - Incorrect observation
Correct Σxi = 10000 + (192 + 88) - (92 + 8) = 10180
Correct mean x =  =  = 50.9.

 

 

Example

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.

Solution

Total expenditure during the year
            = Rs. [6050
× 4 + 6180 × 5 + 6500 × 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

∴ Average monthly income =  = Rs. 8,000.





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