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Calculating the Central Measure of the Salary - Essay Example

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Summary
This essay "Calculating the Central Measure of the Salary" demonstrates a median that is particularly useful when in distribution there are a few data points that are vastly different. It may happen that a couple of students have got extraordinarily high salaries…
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Calculating the Central Measure of the Salary
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Extract of sample "Calculating the Central Measure of the Salary"

Q.1

Solution:-

x

f

cf

5

2

2

4

3

5

3

6

11

2

4

15

1

1

16

 

Mean = ∑fx

                 ∑f

            =   (5×2)+(4×3)+(3×6)+(2×4)+(1×1)

                           2+3+6+4+1

            =  49    = 3.0625

                16

                                   

Median =    N+1      th

                       2        value

              

               =   17

                     2       th value

                = 8.5 th value

                Hence median =3

Mode= 3

Q.2

 Solution:-

x

f

cf

1

1

1

2

4

5

3

3

8

4

1

9

5

1

10

 

Mean= ∑fx

               ∑f

            = (1×1)+(4×2)+(3 ×3)+ (4× 1)+ (5 ×1)

                    1+4+3+1+1

             = 27

                10

            =2.7

Median =    N+1      th

                       2        value

              

               =   11

                     2      th value     

                = 5.5 th value

               Hence median = 2

Mode=2

 

Q.3

Solution

 

Mean= ∑fx

               ∑f

10 = ∑fx

        6

∑fx=60

 

Let the new score be x

Mean when the new score is added

Mean= ∑fx

               ∑f

11=60+x

          7

77=60+x

 X=17

 

Q.4

Solution:-

 

  1. The mean is the balance point of the distribution such that if you subtract each value in the distribution from the mean and sum up all of these deviation scores, the result will be zero. It is a kind of center of gravity to which each measure in the distribution contributes in proportion to its size
  2. The median is simply the midpoint of the distribution that is there are as many numbers above it as below it.

If the number of data points is odd, then the median is simply the middle number. Therefore the median of 3, 5, 6, 9, 15 is 6.

If the number of data points is even, then the median is the mean of the middle two numbers. Therefore the median of 2, 7, 15, 20 is (7+15)/2 = 11.

This will take the mean of the salaries of the group to very high values, but the median will truly reflect the placement scenario as it is.

Q.6

Solution:-

x

f

cf

10

2

2

9

3

5

8

5

10

7

6

16

6

3

19

5

1

20

 

Mean= ∑fx

               ∑f

            = (10×2)+(9×3)+(8 ×5)+ (7× 6)+ (6 ×3)

                    2+3+5+6+3+1

             = 152

                20

            =7.6

 

 

Median =    N+1      th

                       2        value

              

               =   21

                     2      th value     

                = 10.5 th value

               Hence median = 8

Mode=7

Q.7

Solution:-

 

Number of times per week

f

cf

5 or more

2

2

4

2

4

3

3

7

2

6

13

1

4

17

0

3

20

 

  1. Mode =2 times per week
  2. Median= N+1 th

                           2      value

              

                =   21

                     2      th value     

                = 10.5 th value

               Hence median = 2

.8

Solution

week

Average daily rainfall (mon-Fri)

Average daily rainfall

(sat-sun)

1

1.2

1.5

2

0.6

2.0

3

0.0

1.8

4

1.6

1.5

5

0.8

2.2

6

2.1

2.4

7

0.2

0.8

8

0.9

1.6

9

1.1

1.2

10

1.4

1.7

 

 Mean for rainfall during weekdays =  ∑x

                                                                    n

                                                                 =(1.2)+(0.6)+(0)+ (1.6)+(0.8)+(2.1)+(0.2 )+( 0.9)+( 1.1)+( 1.4)

                                                                                             10                                                         

 

                                                                 = 9.9

                                                                     10

                                                                =0.99

Mean for rainfall during the weekends=∑x

                                                                          n

                                                                     =(1.5)+(2.0)+(1.8)+ (1.5)+(2.2)+(2.4)+(0.8)+( 1.6)+( 1.2)+( 1.7)

                                                                                                 10                                                          

 

                                                                    = 16.7

                                                                        10

                                                                   =1.67

  1. There is more average rainfall on weekends as compared to that on weekdays for a particular week.

 

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