Let the mean and variance of the frequency distribution be 6 and 6.8, respectively.

Question

Let the mean and variance of the frequency distribution be 6 and 6.8, respectively. If $x_3$ is changed from 8 to 7, then the mean for the new data will be

Accepted Answer

Mean $=6$: $\dfrac{8+24+8\alpha+9\beta}{8+\alpha+\beta}=6 \implies 2\alpha+3\beta=16$ ...(i)

Variance $=6.8$: $\dfrac{64+4\alpha+9\beta}{8+\alpha+\beta}=... \implies 64+4\alpha+9\beta = 6.8(8+\alpha+\beta)$

$$640+40\alpha+90\beta = 544+68\alpha+68\beta \implies 28\alpha-22\beta=96 \implies 14\alpha-11\beta=48$$ ...(ii)

Solving (i) and (ii): $\alpha=5, \beta=2$.

If $x_3$ (originally 8, with frequency corresponding to $\alpha$-term) is changed from 8 to 7:
$$\text{New mean} = \frac{8+24+35+18}{15} = \frac{85}{15}=\frac{17}{3}$$

Solutions · 1