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# Assignment No.1 (Course STA301) before or on 01-06-2015, (23:59)P.M

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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Ao G :p

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

AoA! For Idea

hiiiiiiiiiiiii fellows ! I need help in assignment no.1

AOA!  Idea for ST301

Answer (1a):  Mean – Mode = 3 (Mean - Median)

 Class Interval Frequency Class-Limits Class-Boundaries

Answer(2a):  the equations not shows by web

Answer(2b):   To find out who is better as run getter and who more consistent player is, first calculate the Mean and Standard Deviation from the given data with help of  sum of total values/no of total values and the equation not shows by web . Then find the coefficient of variance. From here you will get your desired ANSWER, please.

Arrange Data question (1 b part):

 55 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

frequency number must be equal to raw data number.these number are 31 i think so but zain nasir your frequency total32.

sidra 32 hogin value because 55 is starting value bqi 31 values di hue hain so 32 in total

sidra ap 55.0 ko include kr k count kro.....

yes sis there is total 32 values in our data where starting value is 55.0 , Zain bro is correct :)

Spring 2015_STA301_1 Solved by Zain nasar corrected

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