Statistics Question 5
Question 5 - 30 January - Shift 1
The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted a and $b$ are respectively mean and variance of remaining 6 observation, then $a+3 b-5$ is equal to
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Answer: (37)
Solution:
Formula: Arithmetic mean of individual series (ungrouped data), Variance of individual observations (ungrouped data)
$\frac{x_1+X_2+\ldots .+X_7}{7}=8$
$\frac{x_1+x_2+x_3 \ldots .+x_6+14}{7}=8$
$\Rightarrow x_1+x_2+\ldots .+x_6=42$
$\therefore \frac{x_1+x_2 \ldots .+x_6}{6}=\frac{42}{6}=7=a$
$\frac{\Sigma \mathbf{x} _i^{2}}{7}-8^{2}=16$
$\Sigma x i^{2}=560$
$\Rightarrow x_1^{2}+x_2^{2}+\ldots+x_6^{2}=364$
$b=\frac{x_1^{2}+x_2^{2}+\ldots .+x_6^{2}}{6}-7^{2}$
$=\frac{364}{6}-49$
$b=\frac{70}{6}$
$a+3 b-5=7+3 \times \frac{70}{6}-5$
$=37$
