Question

School is organised slogan competition on “SAVE ENVIRONMENT” for all classes. The following frequency distribution gives the number of slogans submitted by all the
students:

No. of slogans	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
No. of students	2	3	6	7	14	12	4	2

Find the mean, variance and standard deviation. Which value is depicted from this?

Answer 1

Hint: In this question we have to find the mean, variance and standard deviation and we have the standard formula to find these. We have to form a table by given data and then it is easy to find the given question. By the class interval we can find the midpoint and the no. of students are frequency.
Formula used:
1. $ Mean(\overline x ) = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
2. $ Variance({\sigma ^2}) = \dfrac{1}{N}f{({x_i} - \overline x )^2} $
3. standard deviation $ = \sqrt {{\sigma ^2}} $

Complete step-by-step answer:
Now consider the given data and hence form the table

No. of slogans	No. of students ( $ {f_i} $ )	Midpoint( ${x_i}$)	$ {f_i}{x_i} $	$ {({x_i} - \overline x )^2} $	$ f{({x_i} - \overline x )^2} $
11-15	2	13	26	361	722
16-20	3	18	54	196	588
21-25	6	23	138	81	486
26-30	7	28	196	16	112
31-35	14	33	462	1	14
36-40	12	38	456	36	432
41-45	4	43	172	121	484
46-50	2	48	96	256	512
	$ \sum {{f_i}} = 50 $		$ \sum {{f_i}} {x_i} = 1600 $		$ \sum {f{{({x_i} - \overline x )}^2}} = 3350 $

Now we will find the mean for the given data. We have formula for the mean and that is
 $ Mean(\overline x ) = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
Where we have to find the summation of the product of the frequency and the midpoint. Therefore, we have $ Mean(\overline x ) = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
 $ \Rightarrow Mean(\overline x ) = \dfrac{{1600}}{{50}} $
 $ \Rightarrow Mean(\overline x ) = 32 $
So, the correct answer is “ $ Mean= 32 $ ”.

We have obtained the mean and we need to find the standard deviation and variance. We need to form the data in the table.
Now we will find the variance and standard deviation and we have the formula
The variance is given by $ Variance({\sigma ^2}) = \dfrac{1}{N}f{({x_i} - \overline x )^2} $ where $ N = \sum {{f_i}} $ , so we have
 $ \Rightarrow Variance({\sigma ^2}) = \dfrac{1}{{50}}3350 $
 $ \Rightarrow Variance({\sigma ^2}) = 67 $
So, the correct answer is “ $ Variance= 67 $ ”.

And we have to find the standard deviation, it is the square root of variance. Hence, we have
Standard deviation = square root (variance)
 $ \Rightarrow $ standard deviation $ = \sigma $
 $ \Rightarrow $ standard deviation $ = \sqrt {67} $
 $ \Rightarrow $ standard deviation = 8.18
So, the correct answer is “ $ standard \;deviation = 8.18 $ ”.

Note: Candidates must form the table correctly from the class intervals we can find the midpoint and frequency will be given data. By using the formula, we can find the mean. After finding the mean, calculate the $ {({x_i} - \overline x )^2} $ and $ f{({x_i} - \overline x )^2} $ hence we can calculate the variance and standard deviation by applying the formula