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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    how i can type this in word......... mujy pta nhi chal rha can you help me plzzz

Question 1:

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:

Arranged data:

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 – 98.9 84.95 – 89.95 1 90.0 – 94.9 989.95 – 94.95 1 95.0 – 99.9 94.95 – 99.95 1

You have been told to have only 5 classes here you have 9 classes. This is wrong.

the correct way should be as follows:

Step 1:

We identify the smallest and the largest measurement in the data set.

So

Smallest value (X0)= 55

Largest value ( Xm) = 95.5

Step 2:

Range = Xm - X0 = 95.5 – 55 = 40.5

Step 3:

We pick the class interval. As we have already been give the class interval which is 5

Step 4:

We will define now the class width denote by h

h =  =  = 8.1= 8 After rounding it up we get 9

Step 5:

To choose lower limit of the lowest class. As the lowest value in the given data is 55, we will choose a number equal to or slightly less than this.

In this case the lower class limit of the lowest class could be: 50.9

Step 6:

We determine the lower class limits of the successive classes by adding  h = 8 successively. The following table is obtained

 Class Number Lower Class Limit 1 50.9 2 50.9 +  9  = 59.9 3 59.9 + 9 = 68.9 4 68.9 + 9 = 77.9 5 77.9 + 9 = 86.9

Step 7 :

We determine the upper class limit of every class. The following table shows it.

 Class Number Lower Class Limit Upper Class Limit 1 50.9 61.5 2 50.9 +  9  = 59.9 61.5 + 9 = 69.5 3 59.9 + 9 = 68.9 69.5 + 9 = 77.5 4 68.9 + 9 = 77.9 77.5 + 9 = 86.5 5 77.9 + 9 = 86.9 86.5 + 9 = 95.5

So we obtain the following classes:

 Classes 50.9 – 61.5 59.9 – 69.5 68.9 – 77.5 77.9 – 86.5 86.9 – 95.5

Step 8:

Distribute the data into the appropriate callsess and find the frequency of each class. In this case

 Class - Limits Tally Frequency 50.9 – 61.5 ||| 4 59.9 - 69.5 || ||||| 6 68.9 - 77.5 |||| |||| ||| 11 77.9 + 86.9 |||| |||| 8 86.9  - 95.5 || 2 32

For class boundaries we get use the formula

Class Boundary =

Doing so we get the following table.

 Class - Limits Class Boundaries Frequency 50.9 – 61.5 48.2 - 56.2 4 59.9 - 69.5 56.2 - 64.7 6 68.9 - 77.5 64.7 – 82.2 11 77.9 + 86.9 82.2 – 91.2 8 86.9  - 95.5 91.2  - 99.2 2 32
 sta301 assingiment no 1  solution
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Check it Please and must tell me.

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zain ap k question no1 (B) main ap ki 2nd claass boundry correct ni hy but mathed is correct.         enjoy kro ..............

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#sidra thnx btnay k liay typo mistake the menay 59.5 ki jaga 69.5 type krdiya tha now menay correct krdi hai file!

Stand Deviation ka formula MS word main kesy likhen gy???

aap ko course ki website sy mil jae ga word format mae jahan download ki option he page number 9 par.. other wise i add file check it

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Fellows,

for the question 1(b) I have seen totally different way of doing the problem here in the group. for the lower class limits you guys have selected 55 as the starting one and are adding 5 to it. However in the lectures we have been taught to first compute the Range and then the width of the classes and should add the selected lower limit value to the width. so in our example it should be

Min Value = 55

Max Value = 95.5

Range = Max - Min = 95.5 - 55 = 40.5

width = Range/No.ofClasses = 40.5/ 5 = 8.1 = 8

Now the lower class limits should be

55

55 + 8 = 63

63 + 8 = 71

71 + 8 = 79

79 + 8 = 87

I dont think they should be more than 5 as we have been given the number of class interval as 5. Can anybody please let  me know which one is correct. I am still struggling with this question.

thanks

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