Question

For the following distribution, calculate mean using all suitable methods:
Size of item:
1−4
4−9
9−16
16−27
Frequency:
6
12
26
20

Answer 1

DIRECT METHOD:

In this method find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by xi. For each class interval we have the frequency fi corresponding to the class mark xi.

Class marks = ( lower limit + upper limit)/2

Then find the product of fi, & xi for each class interval. Find Σ fi & Σ fixi.

Use the formula :  

MEAN = Σfixi/ Σfi

[‘Σ’ Sigma means ‘summation’ ]

FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT  

From the table : Σfixi = 848 ,Σfi = 64

MEAN = Σfixi/ Σfi

Mean = 848/64

Mean = 13.25

Hence, the mean is 13.25 .

ASSUMED MEAN METHOD :  

In this method, first of all, one among xi 's is chosen as the assumed mean denoted  by ‘A’. After that the difference ‘di’ between ‘A’ and each of the xi's i.e di = xi - A is calculated .  

MEAN =  A +  Σfidi / Σfi

[‘Σ’ Sigma means ‘summation’ ]

★★ We may take Assumed mean 'A’ to be that xi which lies in the middle of x1 ,x2 …..xn.

FREQUENCY DISTRIBUTION TABLE IS IN THE ATTACHMENT  

From the table : Σfidi = 48 , Σfi = 64

Let the assumed mean, A = 12.5

MEAN =  A +  Σfidi / Σfi

MEAN =  12.5  + (48 / 64)

= 12.5  + ¾

= 12.5 + 0.75

= 13.25

MEAN = 13.25

Hence, the Mean is 13.25 .  

Here, STEP DEVIATION METHOD cannot be used to find the mean of the distribution as the width of the class intervals are not equal. Here 'h' is not fixed.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge \bf{Answer :}$

Solution =>

(By direct method :)

Table is in the First attachment

Mean = (sum/N) + A

= 848 / 64

${ \underline{ \sf {\huge{= 13.25}}}}$

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Let's see by assumed mean method :

Table is in second attachment :

Let the assumed mean ( A ) = 65

Mean = A + ( sum / N )

= 6.5 + 6.75

${ \underline{ \sf {\huge{= 13.25}}}}$

Hence,

Mean of the given distribution is 13.25.

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