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Find the mean and variance of the frequency distribution given below:

    \[\begin{tabular}{|l|l|l|l|l|} \hline$x$ & $1 \leq x<3$ & $3 \leq x<5$ & $5 \leq x<7$ & $7 \leq x<10$ \\ \hline$f$ & 6 & 4 & 5 & 1 \\ \hline \end{tabular}\]

CBSE | Class 11 | Maths | NCERT Exemplar | Statistics
Find the mean and variance of the frequency distribution given below:

    \[\begin{tabular}{|l|l|l|l|l|} \hline$x$ & $1 \leq x<3$ & $3 \leq x<5$ & $5 \leq x<7$ & $7 \leq x<10$ \\ \hline$f$ & 6 & 4 & 5 & 1 \\ \hline \end{tabular}\]

Solution:

The frequency distribution is given

We now need to find the mean and variance

Convert the ranges of x to groups, we can re-write the given table as shown below,

    \[\begin{tabular}{|l|l|l|l|l|} \hline$x($ Class $)$ & $f_{i}$ & $x_{i}$ & $f_{i} x_{i}$ & $f_{i} x_{i}^{2}$ \\ \hline $1-3$ & 6 & 2 & 12 & 24 \\ \hline $3-5$ & 4 & 4 & 16 & 64 \\ \hline $5-7$ & 5 & 6 & 30 & 180 \\ \hline $7-10$ & 1 & $8.5$ & $8.5$ & $72.25$ \\ \hline Total & 16 & & $66.5$ & $340.25$ \\ \hline \end{tabular}\]

And we can write the variance as

\sigma^{2}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^{2}}{\mathrm{n}}-\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}\right)^{2}

Substitute the values from the above table,

\sigma^{2}=\frac{340.25}{16}-\left(\frac{66.5}{16}\right)^{2}

Simplify

\sigma^{2}=21.265-(4.16)^{2}

\sigma^{2}=21.265-17.305=3.96

The mean can also be written as

\overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}

Substitute the values from the above table, to get

\bar{x}=\frac{66.5}{16}=4.16

As a result the mean and variance of the given frequency distribution is respectively 4.16 and 3.96.

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