The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted

Q: Wait, why do we subtract $14^2$ from the sum of squares? Shouldn't we just subtract $14$?

That is a classic trap! Think about the variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. The term $\sum x_i^2$ represents the sum of the *squares* of the observations. If you remove an observation $x_k$, you must remove its contribution to the sum of squares, which is $x_k^2$. Subtracting just $14$ would be mathematically incorrect because you are dealing with squared deviations.

Q: Can I just calculate the new mean and variance by shifting the data or using a shortcut?

While there are shortcut formulas for adding or removing data, they are essentially just algebraic rearrangements of the fundamental definitions. For JEE Advanced, it is safer and more robust to derive the new sum and new sum of squares from first principles. It prevents errors and builds a deeper conceptual understanding.

Q: Why does the variance change so drastically when we remove just one observation?

Variance measures the 'spread' of data. When you remove a data point, you are changing both the mean (the center) and the sum of squares (the dispersion). Since variance depends on the square of the distance of each point from the mean, removing a point that is far from the mean (like $14$ compared to the mean of $8$) will significantly alter the spread, leading to a noticeable change in the variance.

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 $a + 3b - 5$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Anatomy of Data: A Statistical Journey

Statistics is often misunderstood as a dry collection of formulas, but in reality, it is the art of distilling chaos into clarity. When we talk about mean and variance, we are talking about the heartbeat and the breath of a data set.

Today, we are going to dissect a problem that tests your understanding of these fundamental concepts. Imagine you have a set of $7$ observations. We know their collective behavior: a mean of $8$ and a variance of $16$. Our goal is to see what happens when we surgically remove one observation, the value $14$.

### Unlocking the Original State

Before we can remove anything, we must understand what we have. The mean is defined as $\bar{x} = \frac{\sum x_i}{n}$.

With $n = 7$ and $\bar{x} = 8$, the total sum of our observations is:
$$\sum x_i = 7 	imes 8 = 56$$

Next, we look at the variance, $\sigma^2 = 16$. The formula for variance is:
$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

By rearranging this, we get $\sum x_i^2 = n(\sigma^2 + (\bar{x})^2)$. Substituting our values:
$$\sum x_i^2 = 7(16 + 8^2) = 7(16 + 64) = 7(80) = 560$$

We now have the two pillars of our original data: the sum is $56$ and the sum of squares is $560$.

### The Surgical Removal

Now, we remove the value $14$. When you remove a value, you must update both the sum and the sum of squares.

The new sum is:
$$\sum x_i' = 56 - 14 = 42$$

The new sum of squares is:
$$\sum (x_i')^2 = 560 - 14^2 = 560 - 196 = 364$$

Our new count of observations is $n' = 6$. This is the 'new reality' of our data set.

### Calculating the New Parameters

With our updated sums, finding the new mean $a$ is straightforward:
$$a = \frac{\sum x_i'}{n'} = \frac{42}{6} = 7$$

Notice how the mean dropped from $8$ to $7$. This makes sense because we removed a value ($14$) that was significantly higher than the original mean.

Now for the new variance $b$:
$$b = \frac{\sum (x_i')^2}{n'} - a^2$$

Substituting our new values:
$$b = \frac{364}{6} - 7^2 = \frac{182}{3} - 49$$

To subtract these, we find a common denominator:
$$b = \frac{182 - 147}{3} = \frac{35}{3}$$

### The Final Synthesis

We have arrived at the final step. The problem asks for the value of $a + 3b - 5$.

Substituting our values $a = 7$ and $b = \frac{35}{3}$:
$$7 + 3\left(\frac{35}{3}ight) - 5$$

The beauty of this expression lies in the cancellation of the $3$s:
$$7 + 35 - 5 = 37$$

The final result is **$37$**. This journey shows that statistics is not about memorizing formulas, but about understanding how every single data point contributes to the aggregate properties of the set.

Animated Solution for Mathematics - Statistics: 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 ______.

Enter Numerical Value:

Visualized Solution

Calculate Original Sum $\sum x_{i}$

Calculate Original Sum of Squares $\sum x_{i}^{2}$

Update Sum after Omission

Calculate New Mean $a$

Update Sum of Squares

Calculate New Variance $b$

Final Expression Evaluation

The Sigma Insight: Variance and Standard Deviation

The Anatomy of Data

Unlocking the Original State

The Surgical Removal

Calculating the New Parameters

The Final Synthesis

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