Calculating the Missing Observation Using Variance

This problem asks us to find a missing positive observation in a set of five observations, given the variance of the entire set and the values of the four known observations. We will use the formula for variance and the definition of the mean to solve for the unknown value.

Let the five positive observations be \(y_1, y_2, y_3, y_4, y_5\). We are given that four of these observations are 2, 2, 4, and 5. Let the fifth, unknown observation be \(x\). So, the observations are 2, 2, 4, 5, \(x\).

The number of observations is \(n = 5\).

Step 1: Calculate the Mean

The mean (\(\bar{y}\)) of the observations is the sum of all observations divided by the number of observations:

\(\bar{y} = \frac{\sum y_i}{n}\)

Substituting the given observations:

\(\bar{y} = \frac{2 + 2 + 4 + 5 + x}{5}\)

\(\bar{y} = \frac{13 + x}{5}\)

Step 2: Use the Variance Formula

The variance (\(\sigma^2\)) of a set of observations is given by the formula:

\(\sigma^2 = \frac{\sum (y_i - \bar{y})^2}{n}\)

We are given that the variance is 3.6. So, \(\sigma^2 = 3.6\).

We need to calculate the sum of the squared differences between each observation and the mean:

\(\sum (y_i - \bar{y})^2 = (2 - \bar{y})^2 + (2 - \bar{y})^2 + (4 - \bar{y})^2 + (5 - \bar{y})^2 + (x - \bar{y})^2\)

Substitute the expression for \(\bar{y}\):

\(\sum (y_i - \bar{y})^2 = \left(2 - \frac{13+x}{5}\right)^2 + \left(2 - \frac{13+x}{5}\right)^2 + \left(4 - \frac{13+x}{5}\right)^2 + \left(5 - \frac{13+x}{5}\right)^2 + \left(x - \frac{13+x}{5}\right)^2\)

Simplify the terms inside the parentheses:

• \(2 - \frac{13+x}{5} = \frac{10 - (13+x)}{5} = \frac{10 - 13 - x}{5} = \frac{-3 - x}{5}\)
• \(4 - \frac{13+x}{5} = \frac{20 - (13+x)}{5} = \frac{20 - 13 - x}{5} = \frac{7 - x}{5}\)
• \(5 - \frac{13+x}{5} = \frac{25 - (13+x)}{5} = \frac{25 - 13 - x}{5} = \frac{12 - x}{5}\)
• \(x - \frac{13+x}{5} = \frac{5x - (13+x)}{5} = \frac{5x - 13 - x}{5} = \frac{4x - 13}{5}\)

Now square these terms:

• \(\left(\frac{-3 - x}{5}\right)^2 = \left(\frac{-(3 + x)}{5}\right)^2 = \frac{(3 + x)^2}{25} = \frac{9 + 6x + x^2}{25}\)
• \(\left(\frac{7 - x}{5}\right)^2 = \frac{(7 - x)^2}{25} = \frac{49 - 14x + x^2}{25}\)
• \(\left(\frac{12 - x}{5}\right)^2 = \frac{(12 - x)^2}{25} = \frac{144 - 24x + x^2}{25}\)
• \(\left(\frac{4x - 13}{5}\right)^2 = \frac{(4x - 13)^2}{25} = \frac{16x^2 - 104x + 169}{25}\)

Sum of squared differences:

\(\sum (y_i - \bar{y})^2 = \frac{(9 + 6x + x^2) + (9 + 6x + x^2) + (49 - 14x + x^2) + (144 - 24x + x^2) + (16x^2 - 104x + 169)}{25}\)

Combine like terms in the numerator:

\(\sum (y_i - \bar{y})^2 = \frac{(x^2 + x^2 + x^2 + x^2 + 16x^2) + (6x + 6x - 14x - 24x - 104x) + (9 + 9 + 49 + 144 + 169)}{25}\)

\(\sum (y_i - \bar{y})^2 = \frac{20x^2 - 130x + 380}{25}\)

Now substitute this into the variance formula \(\sigma^2 = \frac{\sum (y_i - \bar{y})^2}{n}\), with \(\sigma^2 = 3.6\) and \(n = 5\):

\(3.6 = \frac{\frac{20x^2 - 130x + 380}{25}}{5}\)

\(3.6 = \frac{20x^2 - 130x + 380}{25 \times 5}\)

\(3.6 = \frac{20x^2 - 130x + 380}{125}\)

Step 3: Solve for the Missing Observation \(x\)

Multiply both sides by 125:

\(3.6 \times 125 = 20x^2 - 130x + 380\)

\(450 = 20x^2 - 130x + 380\)

Rearrange into a quadratic equation by subtracting 450 from both sides:

\(0 = 20x^2 - 130x + 380 - 450\)

\(0 = 20x^2 - 130x - 70\)

Divide the entire equation by 10 to simplify:

\(0 = 2x^2 - 13x - 7\)

We can solve this quadratic equation by factoring. We look for two numbers that multiply to \(2 \times -7 = -14\) and add up to \(-13\). These numbers are -14 and 1.

Rewrite the middle term \(-13x\) as \(-14x + x\):

\(2x^2 - 14x + x - 7 = 0\)

Factor by grouping:

\(2x(x - 7) + 1(x - 7) = 0\)

\((2x + 1)(x - 7) = 0\)

This gives two possible solutions for \(x\):

\(2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}\)

\(x - 7 = 0 \implies x = 7\)

The problem states that the observations are positive. Since \(x = -\frac{1}{2}\) is negative, it is not a valid solution in this context. Therefore, the remaining observation must be \(x = 7\).

Verification

Let's check if the variance is 3.6 when the observations are 2, 2, 4, 5, 7.

Mean = \(\frac{2+2+4+5+7}{5} = \frac{20}{5} = 4\)

Differences from mean: \(2-4=-2\), \(2-4=-2\), \(4-4=0\), \(5-4=1\​, \(7-4=3\)

Squared differences: \((-2)^2=4\), \((-2)^2=4\), \(0^2=0\), \(1^2=1\), \(3^2=9\)

Sum of squared differences = \(4 + 4 + 0 + 1 + 9 = 18\)

Variance = \(\frac{\text{Sum of squared differences}}{n} = \frac{18}{5} = 3.6\)

The calculated variance matches the given variance, confirming that the missing observation is 7.

The sum of squared differences is 18. Variance = \(\frac{18}{5} = 3.6\).

Revision Table: Key Statistical Concepts

Additional Information: Variance vs. Standard Deviation

Variance is a fundamental measure of dispersion in statistics. While variance measures the average squared deviation from the mean, its square root, the standard deviation (\(\sigma\)), is often more intuitive because it is in the same units as the original data.

• Variance (\(\sigma^2\)): Gives a value in squared units of the data. For example, if the data is in meters, the variance is in square meters.
• Standard Deviation (\(\sigma\)): Is the square root of the variance. It provides a measure of spread in the original units of the data, making it easier to interpret the typical distance of data points from the mean. The formula is \(\sigma = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n}}\).

Both variance and standard deviation are crucial for understanding the variability within a dataset. A higher variance or standard deviation indicates that the data points are more spread out from the mean, while a lower value indicates they are clustered closer to the mean.

In this problem, we were given the variance directly. If we were given the standard deviation, we would square it first to get the variance before using the formula.