Find the mean and variance of frequency distribution given below :

$\begin{matrix} x_{i} : & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 f_{i} : & 6 & 4 & 5 & 1 \end{matrix}$

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Solution

$\begin{matrix} x_{i} & Mid value y_{i} & y_{i}^{2} & f_{i} & f_{i} y_{i} & f_{i} y_{i}^{2} 1 - 3 & 2 & 4 & 6 & 12 & 24 3 - 5 & 4 & 16 & 4 & 16 & 64 5 - 7 & 6 & 36 & 5 & 30 & 180 7 - 10 & 8.5 & 72.25 & 1 & 8.5 & 72.25 N = \sum f_{i} & \sum f_{i} y_{i} & \sum f_{i} y_{i}^{2} = 16 & = 66.5 & = 340.25 \end{matrix}$

Therefore,

Mean = $\frac{\sum f_{i} y_{i}}{\sum, f_{i}} = \frac{66.5}{16} = 4.16$

Variance = $(\frac{1}{N} \sum f_{i} y_{i}^{2}) - {(\frac{1}{N} \sum f_{i} y_{i})}_{i}^{2}$

$= \frac{1}{16} \times 340.25 - {(\frac{1}{16} \times 66.5)}^{2}$ = 21.26 -17.22= 4.04

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Find the mean and variance of frequency distribution given below : xi:1≤x<33≤x<55≤x<77≤x<10fi:6451

xiMid valueyiy2ififiyifiy2i1−324612243−5416416645−76365301807−108.572.2518.572.25N=∑fi∑fiyi∑fiy2i=16=66.5=340.25 Therefore, Mean = ∑fiyi∑,fi=66.516=4.16 Variance =(1N∑fiy2i)−(1N∑fiyi)2 =116×340.25−(116×66.5)2 = 21.26 -17.22= 4.04

Find the mean and variance of frequency distribution given below :

$\begin{matrix} x_{i} : & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 f_{i} : & 6 & 4 & 5 & 1 \end{matrix}$