Set up the equation for the mean. The mean of the six observations is given as 55. So,
\[ \frac{a + b + 68 + 44 + 48 + 60}{6} = 55. \]
Multiply both sides by 6 to eliminate the denominator:
\[ a + b + 68 + 44 + 48 + 60 = 330. \]
Simplify to get:
\[ a + b = 110 \quad \text{(Equation 1)}. \]
Set up the equation for the variance. The variance of the six observations is given as 194.
Recall that the variance formula for a set of observations \( x_1, x_2, \ldots, x_n \) with mean \( \overline{x} \) is:
\[ \text{Variance} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \overline{x})^2. \]
Here, the mean \( \overline{x} \) is 55. Applying this to our observations:
\[ \frac{(a - 55)^2 + (b - 55)^2 + (68 - 55)^2 + (44 - 55)^2 + (48 - 55)^2 + (60 - 55)^2}{6} = 194. \]
Calculate known terms in the variance expression. Evaluate each squared term involving the known observations:
\[ (68 - 55)^2 = 13^2 = 169, \quad (44 - 55)^2 = (-11)^2 = 121, \quad (48 - 55)^2 = (-7)^2 = 49, \quad (60 - 55)^2 = 5^2 = 25. \]
Substitute these values into the variance equation:
\[ \frac{(a - 55)^2 + (b - 55)^2 + 169 + 121 + 49 + 25}{6} = 194. \]
Simplify:
\[ \frac{(a - 55)^2 + (b - 55)^2 + 364}{6} = 194. \]
Multiply both sides by 6:
\[ (a - 55)^2 + (b - 55)^2 + 364 = 1164. \]
Subtract 364 from both sides:
\[ (a - 55)^2 + (b - 55)^2 = 800 \quad \text{(Equation 2)}. \]
Solve the system of equations. We have the following two equations: 1. \( a + b = 110 \). 2. \( (a - 55)^2 + (b - 55)^2 = 800 \).
From Equation 1, express \( a \) in terms of \( b \):
\[ a = 110 - b. \]
Substitute \( a = 110 - b \) into Equation 2:
\[ (110 - b - 55)^2 + (b - 55)^2 = 800. \]
Simplify each term:
\[ (55 - b)^2 + (b - 55)^2 = 800. \]
Since \( (55 - b)^2 = (b - 55)^2 \), we can write:
\[ 2(b - 55)^2 = 800. \]
\[ (b - 55)^2 = 400. \]
Taking the square root of both sides:
\[ b - 55 = \pm 20. \]
This gives: 1. \( b = 75 \) (if \( b - 55 = 20 \)), 2. \( b = 35 \) (if \( b - 55 = -20 \)).
Since \( a > b \), we choose \( b = 35 \). Substitute \( b = 35 \) into Equation 1:
\[ a + 35 = 110. \] \[ a = 75. \]
Calculate \( a + 3b \)
\[ a + 3b = 75 + 3 \cdot 35 = 75 + 105 = 180. \]
Thus, the answer is: 180.
Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
---|---|---|---|---|---|---|
Frequency | 11 | 8 | 15 | 7 | 10 | 9 |
Variance of the following discrete frequency distribution is:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0-2 & 2-4 & 4-6 & 6-8 & 8-10 \\ \hline \text{Frequency (}f_i\text{)} & 2 & 3 & 5 & 3 & 2 \\ \hline \end{array} \]
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: