Question

Let \(a, b, c \in \mathbb{N}\) and \(a<b<c\). Let the mean, the mean deviation about the mean and the variance of the 5 observations \(9, 25, a, b, c\) be \(18, 4\) and \(\frac{136}{5}\), respectively. Then \(2a + b - c\) is equal to _______.

Answer 1

Correct Answer: 33

To determine the value of \(2a + b - c\), we utilize the provided statistical data: the mean, mean deviation about the mean, and variance. The mean of the observations \(9, 25, a, b, c\) is given as 18. This yields the equation:

\[\frac{9 + 25 + a + b + c}{5} = 18\]

From this, we derive the sum \(a + b + c\):

\[9 + 25 + a + b + c = 90\]

\[a + b + c = 56\]

The variance is provided as \(\frac{136}{5}\). The variance formula is:

\[\text{Variance} = \frac{\sum (x_i - \mu)^2}{N}\]

Applying this to our data:

\[\frac{(9-18)^2 + (25-18)^2 + (a-18)^2 + (b-18)^2 + (c-18)^2}{5} = \frac{136}{5}\]

Simplifying the equation:

\[81 + 49 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136\]

\[(a-18)^2 + (b-18)^2 + (c-18)^2 = 6\]

We find the values of \(a\), \(b\), and \(c\) by satisfying the following conditions, given that \(a, b, c\) are natural numbers with \(a<b<c\):

\(a+b+c=56\)
\((a-18)^2+(b-18)^2+(c-18)^2=6\)

Analyzing possible combinations for \((a,b,c)\) that meet these criteria leads to the specific values:

\(a=17\), \(b=19\), \(c=20\)

Now, we compute \(2a + b - c\):

\[2(17) + 19 - 20 = 34 + 19 - 20 = 33\]

Verification confirms that 33 is within the specified range. The final result is:

\[\boxed{33}\]