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Assignment no.1(Lessons 1 – 11)

Question 1:                                                                                                  Marks: 2+5=7

a)      In a moderately skewed distribution mean = 60 and median = 55. Find Mode?

b)  Arrange the data given below in an array and construct a frequency distribution using a class interval of 5. Calculate the class- limits and class-boundaries and the starting value is 55.0

79.4, 71.6 ,  95.5, 73.0, 74.2, 81.8, 90.6, 55.9, 75.2, 81.9, 68.9, 74.2, 80.7, 65.7, 67.6, 82.9, 88.1, 77.8, 69.4, 83.2, 82.7, 73.8, 64.2, 63.9, 58.3, 83.5, 70.8, 72.1, 71.6, 59.4, 77.6.

Question 2:                                                                                         Marks:2+6=8

a)      For a set on n values of  , it is known that  , , find ?

b)     The following are the scores made by two batsmen A and B in series of innings:

A:        28        22        46        85        9          59        175      42        11        92

B:        52        18        4          95        125      12        90        58        7          76

i)                    Who is better as a run getter?

ii)                  Who is more consistent player?

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hey guyx can any 1 plz tell me how to use math type

Assignment no.1

Question 1:                                                                                                  Marks: 2+5=7

a)      In a moderately skewed distribution mean = 60 and median = 55. Find Mode?

Solution

As we know that

Mode  = 3 Median  - 2 Mean

Mode = 3(55) – 2(60)

Mode = 165 – 120

Mode = 45  (Answer)

b)     Arrange the data given below in an array and construct a frequency distribution using a class interval of 5. Calculate the class- limits and class-boundaries and the starting value is 55.0

79.4, 71.6 ,  95.5, 73.0, 74.2, 81.8, 90.6, 55.9, 75.2, 81.9, 68.9, 74.2, 80.7, 65.7, 67.6, 82.9, 88.1, 77.8, 69.4, 83.2, 82.7, 73.8, 64.2, 63.9, 58.3, 83.5, 70.8, 72.1, 71.6, 59.4, 77.6.

Solution

Data after arrangement and starting with 55.0 as per given in question:

55.0, 55.9, 58.3, 59.4, 63.9, 64.2, 65.7, 67.6, 68.9, 69.4, 70.8, 71.6, 71.6, 72.1, 73, 73.8, 74.2, 74.2, 75.2, 77.6, 77.8, 79.4, 80.7, 81.8, 81.9, 82.7, 82.9, 83.2, 83.5, 88.1, 90.6, 95.5

 Class-limits Class-boundaries Frequency 55.0 – 59.9 54.95 – 59.95 4 60.0 – 64.9 69.95 – 64.95 2 65.0 – 69.9 64.95 – 69.95 4 70.0 – 74.9 69.95 – 74.95 8 75.0 – 79.9 74.95 – 79.95 4 80.0 – 84.9 79.95 – 84.95 7 85.0 – 89.9 84.95 – 89.95 1 90.0 – 94.9 989.95 – 94.95 1 95.0 – 99.9 94.95 – 99.95 1

Question 2:                                                                                         Marks:2+6=8

a)      For a set on n values of  , it is known that  ,,find?

Solution:

s =

s =

s =

s =

s =

s =5

b)     The following are the scores made by two batsmen A and B in series of innings:

A:        28        22        46        85        9          59        175      42        11        92

B:        52        18        4          95        125      12        90        58        7          76

i)                    Who is better as a run getter?

ii)                  Who is more consistent player?

Solution:

For – A

 x ∑ 28 784 22 484 46 2116 85 7225 9 81 59 3481 175 30625 42 1764 11 121 92 8464 569 55145

s(A) =

s(A) =

s(A)=

s(A)=

s(A)=

s(A)= 47.72

For – B

 x ∑ 52 2704 18 324 4 16 95 9025 125 15625 12 144 90 8100 58 3364 7 49 76 5776 537 45127

s(B) =

s(B) =

s(B)=

s(B)=

s(B)=

s(B)= 40.36

No. of Matches Played = 10

Average Score of A = 569/10 = 56.9

Average Score of B = 537/10 = 53.7

 Batsmen A B Mean 56.9 53.7 S.D 47.72 40.36 CV = ×100 83.86 75.16

Solution of (i)

As we can see from table that average score of A is greater than B which shows that A is better run getter.

Solution of (ii)

As

coefficient of variation of B-player > coefficient of variation of A-player

Which shows that player B is more consistent.

Assalam O Alaikum bachooo :P

sub ny kitny marks liye assignment main ??? mery waly to show hi ni ho rahy :'(

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