Find the mean and variance for the following frequency distributions.
$$
\begin{array}{|l|c|c|c|c|c|}
\hline \text { Classes } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\
\hline \text { Frequencies } & 5 & 8 & 15 & 16 & 6 \\
\hline
\end{array}
$$

Answer

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Intro Stats / AP Statistics

NCERT Class 11 - Math

Chapter 15

Statistics

Section 2

Measures of Dispersion

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Video Transcript

hell in this problem we have to find main invariance by using the given data. So here we have given the following table. First of all, we will find the mid values that is X I. So we get five 15, 25 35 and 45. Let assumed mean as equals two 25. So in order to find the mean, first of all, we will find you Y which is equal to X I minus assumed mean Upon 10. So we get EYS minus two minus one 01 two. Now we will find a fi you Y it would be five into -2 is -10 18 to -1 is -8, 15 in 2, 0 is zero 16 into one is 16 and six into 2 is 12. Similarly, we will find similarly, we will find a fire you a square. So we get 20 seat zero 16 and 24. The submission of F i U I is then and The submission of FIUS Square is 68. Now we know that Mean as equals two assumed mean plus sigma, F I U Y. Upon submission of F I into age here edges the height of the class interval by substituting the known will use we get Mean as equals two 25 plus 10 appoints stay into 10 by simplifying it. We get 25 Plus two which is equals to 27. Now we will find variance. We know that variances equals two hide square a bone and square and do and into sigma. F I U I square minus sigma. F I U Y who square by substituting the known will use we get penn square are born 50 square into 15: 68 -10 sq. By simplifying it. We get 100 upon 2500 Into 30 400 0. It would be equals two. Run upon 25 into 3300, Which is equals two, 132. Hence we get me as equals two. Mhm. 27 variances equals two 132.

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