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Permalink Reply by Yasmeen(ADMIN) on November 15, 2011 at 12:29pm

solution by sidra

 

Attachments:

• STA301Assignment2ByInnocentPrince.pdf, 132 KB

Permalink Reply by Yasmeen(ADMIN) on November 16, 2011 at 12:30pm

solution by huma

 

Attachments:

• Fall2011_STA301_21.doc, 70 KB

Permalink Reply by Yasmeen(ADMIN) on November 17, 2011 at 9:26pm

Question 1

 

           Top ten students in MCS class in VU have following marks division in a CS paper.

89, 90, 85, 88, 83, 79, 95, 93, 81, 80

Calculate the fraction of data falling in the limits,  and.

SOLUTION:

 

X	X2
89	7921
90	8100
85	7225
88	7744
83	6889
79	6241
95	9025
93	8649
81	6561
80	6400
∑X=863	∑X2=74755

         

MEAN:

                                              X BAR =         ∑X

                                                                        n

                                                      =          863

                                                                  10

                                                       =         86.3

 

STANDARD DEVIATION:

                                                S.D.       =√[(∑X²/n)-(∑X/n)²]

                                                              =√[(74755/10)-(863/10)²]

                                                              = √[7475.5-7447.7] 

                                                              =√27.81

                                                              =5.3

Now

i)         

       =       X bar +S              ;                X bar - S 

 

                 =       X bar + S              ;          =     X bar - S 

 

                                           =         86.3 + 5.3           ;           =      86.3 -5.3

                                           =            91.6                 ;           =         81

 

ii)       

 

     =        X bar +2S              ;                X bar - 2S 

 

                  =       X bar + 2S              ;          =     X bar - 2S 

                  =         86.3 +2(5.3)          ;          =       86.3 -2(5.3)

                  =         86.3 +10.6             ;          =       86.3 -10.6

                  =          96.9                      ;          =          75.7

 

iii)    

 

                               =        X bar +3S               ;                X bar - 3S

 

                  =       X bar + 3S               ;          =     X bar - 3S

                  =         86.3 +3(5.3)           ;          =       86.3 -3(5.3)

                   =        86.3+15.9                 ;          =    86.3-15.9

                   =           102.2                      ;          =        70.4

                 The fraction of data falling in the limits

  =   91.6     ---------       81,

                              =   96.9     ---------       75.7

And

=   102.2    ---------       70.4      

 

1. For a certain data, Mean = 4, Median = 3 and Standard Deviation = 2.Find Pearson’s coefficient of skewness by using mode and interpret the results.

 

     SOLUTION:

 

                           Mean                                 = 4

                              Median                              = 3                         

                              Standard Deviation          = 2

       

     FORMULA:

                  

                  Coefficient of Skewness    =    [3(mean - Median)]/S.D

         By putting the value into the formula

                   Coefficient of Skewness    =    [3(mean - Median)]/S.D

                                                          =            [3(4-3)]/2

                                                          =              1.5

So,

     

   The Pearson’s Coefficient of Skewness by using mode is 1.5 which is positively skewed.

 

Question 2:

 

First four moments about zero are given below

       Find the coefficient of skewness and kurtosis.

 

SOLUTION:

                 Coefficient of Skewness:

                                               b₁       =   (m₃)²/ (m₂)³

Coefficient of Kurtosis:

                                                b₂     =   (m₄)/ (m₂)²

By putting the values,

                                                b₁      =   (m₃)²/ (m₂)³

                                                        =   (11.6)²/ (8.4)³

                                                        =   134.56/592.704

                                                        =    0.23

 

                                                 b₂     =   (m₄)/ (m₂)²

                                                        =   (308.6)/ (8.4)²

                                                        =    308.6/70.56

                                                        =   4.37

                  So,

                                The Coefficient of Skewness (b₁) is 0.23.

                                The Coefficient of Kurtosis (b₂) is 4.37.        

 

 

                    

1. a.      For a certain data, we have following information

Calculate percentile coefficient of kurtosis and interpret the result.

 

SOLUTION:

 

                               Q₁   = 24.9

                               Q₃    = 84.2

                                P₁₀  = 7.6

                                P₉₀  = 94.7

FORMULA:

                              

                                 K= Q/ (P₉₀-P₁₀)

                                        

                                  Q=1/2(Q₃-Q₁)

                     Q =Semi-inter quartile Range

By putting the values,

                                              Q=1/2(Q₃-Q₁)

                                                 =1/2(84.2-24.9)

                                                 =1/2(59.3)

                                                 = 29.65

  Now

                                              K= Q/ (P₉₀-P₁₀)

                                                  =29.65/ (94.7-7.6)

                                                  =29.65/87.1

                                                  =0.340

RESULT:

           A distribution that is more peaked than normal distribution (0.263) is called Leptokurtic.

        Percentile Coefficient of Kurtosis is 0.340 which is Leptokurtic.

 

Question 3:

Ages and weights of Lahore city are given in the following table

Ages (X)	17.5	18.2	19.5	21	25	26	28	30	35	40	50
Weights (Y)	35	37	39	35	55	49	74	60	80	60	55

Calculate

1. Simple regression line age on weight.
2. Standard error of estimate for this line.
3. By using the line, estimate the value of weights at the age of 55.
4.  Check, is there significant relation between ages and weights?

 

 

 

Permalink Reply by evergreen on December 20, 2011 at 12:36pm

 solution by vu:

Attachments:

• Fall 2011_STA301_2_SOL.doc, 113 KB

Permalink Reply by Nawaz Chouddry on December 30, 2011 at 6:43pm

sta 3013  assighment requried