Question

The mean and variance of 7 observations are 8 and 16 respectively If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observation, then $a+3 b-5$ is equal to ______

Answer 1

Correct Answer: 37

Given the mean of 7 observations is 8, total sum \( S = 7 \times 8 = 56 \). The variance of these observations is 16, so the sum of squares \( \Sigma \) is calculated as follows: variance \(\times\) number of observations = 16\(\times\)7, hence \(\Sigma - \frac{56^2}{7} = 112\), resulting in \(\Sigma = 560\).
When observation 14 is omitted, sum of 6 observations = \(56 - 14 = 42\). So, the new mean \(a = \frac{42}{6} = 7\).
Sum of squares of 6 observations: \(\Sigma' = 560 - 14^2 = 364\). Thus, new variance \(b = \frac{364}{6} - 7^2 = \frac{364}{6} - 49 = \frac{364 - 294}{6} = \frac{70}{6} = \frac{35}{3}\).
Finally, calculate \(a + 3b - 5\): \(7 + 3 \times \frac{35}{3} - 5 = 7 + 35 - 5 = 37\).
Thus, the value of \(a + 3b - 5\) is 37, which matches the expected range.