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 the mean and variance of the remaining 6 observations, then \(a+3b−5\) is equal to

Answer 1

Correct Answer: 37

The mean of the 7 observations is:

\[ \text{Mean} = \frac{x_1 + x_2 + \cdots + x_7}{7} = 8. \]

The total sum of the observations is:

\[ x_1 + x_2 + \cdots + x_7 = 8 \times 7 = 56. \]

When one observation \(14\) is omitted, the sum of the remaining 6 observations is:

\[ x_1 + x_2 + \cdots + x_6 = 56 - 14 = 42. \]

The mean of the remaining 6 observations is:

\[ a = \frac{x_1 + x_2 + \cdots + x_6}{6} = \frac{42}{6} = 7. \]

The variance of the 7 observations is given as \(16\). Using the formula for variance:

\[ \text{Variance} = \frac{\sum_{i=1}^7 x_i^2}{7} - \left(\frac{\sum_{i=1}^7 x_i}{7}\right)^2. \]

Substituting the known values:

\[ \frac{\sum_{i=1}^7 x_i^2}{7} - 8^2 = 16, \]

\[ \frac{\sum_{i=1}^7 x_i^2}{7} - 64 = 16, \]

\[ \frac{\sum_{i=1}^7 x_i^2}{7} = 80. \]

Thus:

\[ \sum_{i=1}^7 x_i^2 = 80 \times 7 = 560. \]

The variance of the remaining 6 observations is given by:

\[ \text{Variance} = \frac{\sum_{i=1}^6 x_i^2}{6} - \left(\frac{\sum_{i=1}^6 x_i}{6}\right)^2. \]

Substitute the known values:

\[ \frac{\sum_{i=1}^6 x_i^2}{6} - 7^2 = b, \]

\[ \frac{\sum_{i=1}^6 x_i^2}{6} - 49 = b, \]

\[ \frac{\sum_{i=1}^6 x_i^2}{6} = b + 49. \]

We also know:

\[ \sum_{i=1}^7 x_i^2 = \sum_{i=1}^6 x_i^2 + 14^2 = 560, \]

\[ \sum_{i=1}^6 x_i^2 = 560 - 196 = 364. \]

Substitute this value into the variance formula:

\[ \frac{364}{6} = b + 49, \]

\[ b = \frac{364}{6} - 49 = \frac{364}{6} - \frac{294}{6} = \frac{70}{6}. \]

We now compute:

\[ a + 3b - 5 = 7 + 3 \times \frac{70}{6} - 5. \]

Simplify \(3 \times \frac{70}{6}\):

\[ 3 \times \frac{70}{6} = \frac{210}{6} = 35. \]

Thus:

\[ a + 3b - 5 = 7 + 35 - 5 = 37. \]

Conclusion

The value of \(a + 3b - 5\) is:

\[ \boxed{37}. \]