Question

Let the mean and variance of the frequency distribution be 6 and 6.8 respectively.
x : \( x_1 = 2,\; x_2 = 6,\; x_3 = 8,\; x_4 = 9 \)
f : \( 4,\; 4,\; \alpha,\; \beta \)
If \( x_3 \) is changed from 8 to 7, then the mean for the new data will be:

• A. $\frac{17}{3}$
• B. 5
• C. $\frac{16}{3}$
• D. 4

Hint:

When a single data point in a distribution is changed, it's faster to calculate the new mean by finding the change in the sum ($\Delta \sum f_i x_i$) and adjusting the old sum, rather than recalculating the entire sum from scratch. New Mean = (Old Sum + Change) / Total Frequency.

Answer 1

The Correct Option is A

Total frequency: \[ N = 4 + 4 + \alpha + \beta = 8 + \alpha + \beta \] Sum of observations: \[ \sum fx = 4(2) + 4(6) + \alpha(8) + \beta(9) = 32 + 8\alpha + 9\beta \] Given mean $\bar{x} = 6$, \[ \frac{32 + 8\alpha + 9\beta}{8 + \alpha + \beta} = 6 \] \[ 32 + 8\alpha + 9\beta = 48 + 6\alpha + 6\beta \] \[ 2\alpha + 3\beta = 16 \qquad \text{(1)} \] Now, \[ \sum fx^2 = 4(2^2) + 4(6^2) + \alpha(8^2) + \beta(9^2) = 160 + 64\alpha + 81\beta \] Given variance $\sigma^2 = 6.8$, \[ \frac{\sum fx^2}{N} - \bar{x}^2 = 6.8 \] \[ \frac{160 + 64\alpha + 81\beta}{8 + \alpha + \beta} = 42.8 \] \[ 160 + 64\alpha + 81\beta = 342.4 + 42.8\alpha + 42.8\beta \] \[ 21.2\alpha + 38.2\beta = 182.4 \qquad \text{(2)} \] Solving equations (1) and (2): \[ \alpha = 5, \quad \beta = 2 \] Thus, \[ N = 8 + 5 + 2 = 15 \] \[ \sum fx = 32 + 40 + 18 = 90 \] Change in data: $x_3$ changes from $8$ to $7$ with frequency $5$. Change in total sum: \[ \Delta(\sum fx) = 5(7 - 8) = -5 \] New sum: \[ \sum fx_{\text{new}} = 90 - 5 = 85 \] New mean: \[ \bar{x}_{\text{new}} = \frac{85}{15} = \frac{17}{3} \]