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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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Replies to This Discussion

Thanks Man (Zain Nasar) I Love You

IDM ko Uninstall kar den Maxthon Browser sa PFD fast dow Hote hi

 SAQIB BS.IT 6 me too bro .....

Thnx bro so nice of u 

 Papa's Doll yes accurate hai apka answer!


Freinds frequency distribution table mein jo class interval ka column hai us mein pehle value 55.0 se 59.9 tak hogi ??? anyone knows about it ??

me Also confuse :(

sir class interval 5 hy tho 55.0-60.0 ho ga mere khayal sy.. ye aap ne jo bataya hy na ye class boundries mai ho ga.. it will ne 54.5-59.5.

im not saying i am right i am just suggesting.. ;)

nope bro u r wrong 

class interval 5 hy it's mean k hmry pass jo classes bny unky bech 5 farq  ho jesy



and so on 

to jab hum inko dekhty hy to hamry pass classes k bech 5 ka frq hy jo interval hy hamry pass 

60.0 - 55.0 = 5

64.9 - 59.9 = 5 


u r right.. i was wrong . i was making the class limits perfect (difference was perfectly 5) so then we cant make class boundaries. and class boundaries are made in that situation when we have difference in the class limits even if the difference is a tiny one..

yes ab aye na line pe 


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