Question

calculate the mean , median , mode for the following data pertaining to marks in statistics. there are 80 students in a class and the test is of 140 marks .
marks more than : 0,20,40,60,80,100,120

no of students : 80,76,50,28,18,9,5​

Answer 1

Answer:

Mean= 83.5

Median= 96.36

Mode= 103.07

Answer 2

Concept Introduction:-

The Arithmetic Mean referred as the determined "central" values of a group of the numeral that is the mid of the values.

Given Information:-

We have been given that there are $80$ students in a class and the test is of $140$ marks .

To Find:-

We have to find that calculate the mean, median, mode.

Solution:-

According to the problem

$\Rightarrow \text { Mean }=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{4460}{266}=16.77$

$\Rightarrow \mathrm{N}=266$ and $\frac{\mathrm{N}}{2}=133$

$\therefore$ Median class $=20-40$

$l=$ lower limit of the modal class

$h=$ size of the class intervals

$\mathrm{f}=$ frequency of the modal class

$\mathrm{f}_{1}=$ frequency of the class preceding the modal class

$\mathrm{f}_{2}=$ frequency of the class succeed in the modal class.

$\Rightarrow$ Here, $l=20, \mathrm{f}=76, \mathrm{cf}=80, \mathrm{~h}=20$

Median $=l+\left(\frac{\frac{N}{2}-c f}{f}\right) \times h$

$=20+\left(\frac{133-80}{76}\right) \times 20\\=20+13.94\\=33.95$

We know that,

$Mode }=3(\text { median })-2(\text { mean })$

$=(3 \times 33.95)-(2 \times 16.77) \\&=101.84-33.54\\&=68.30$

Final Answer:-

The value of mean, median and mode are $16.77,33.95$ and $68.30$ respectively.

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