The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 marks ea

Q: Wait, why do we use midpoints? Can't we just use the class interval values?

Great question! In grouped data, we don't know the exact marks of each student, only that they fall within a range. The midpoint is the most statistically sound 'representative' value for that entire interval. When we perform calculations on grouped data, we assume all observations in that class are concentrated at the midpoint. It's an approximation, but it's the standard convention in statistics.

Q: Why can't we just adjust the mean and variance directly? Like, subtract the difference?

That is a very common trap! The mean is a linear function, so you *can* adjust it by shifting the sum. However, variance is a non-linear function (it involves squares). If you try to 'adjust' the variance directly, you will miss the cross-terms and the shift in the mean's square. Always go back to the raw sums ($\sum x_i$ and $\sum x_i^2$), correct them, and then compute the new variance from scratch. It's the only way to be 100% accurate.

Q: Why do we need $\sum x_i^2$? Isn't variance just the average of the squared deviations?

You are absolutely right! Variance is defined as the average of the squared deviations from the mean. But the computational formula, $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$, is a mathematical shortcut derived from that definition. It is much easier to work with $\sum x_i^2$ than to recalculate every single deviation from the new mean. Think of $\sum x_i^2$ as the 'energy' of your data set—it's a vital statistic you must maintain.

Question

The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 marks each. The mean and the variance obtained from this distribution are found to be 40 and 49 respectively. It was later discovered that two observations belonging to the class interval (21-30) were included in the class interval (31-40) by mistake. Find the mean and the variance after correcting the error.

Sigma Percentile · Accepted Answer

### The Data Detective: A Masterclass in Statistical Correction

Welcome, future engineers! Today, we are stepping into the shoes of a data scientist. Imagine you have just received a report on the performance of forty students.

The mean is $40$, and the variance is $49$. But wait—a discovery! Two observations were filed into the wrong class interval.

They were placed in $(31-40)$ when they actually belonged to $(21-30)$. This is not just a clerical error; it is a challenge to our understanding of how statistical parameters are built. Let's fix this, step by step.

### Phase 1: The Midpoint Reality

In grouped data, we don't see individual scores; we see intervals. To perform any math, we must collapse these intervals into a single, representative value: the midpoint ($x_i$).

For the interval $(21-30)$, the midpoint is $x_1 = \frac{21+30}{2} = 25.5$. For the interval $(31-40)$, the midpoint is $x_2 = \frac{31+40}{2} = 35.5$.

Our error is simple: we have two extra values of $35.5$ and two missing values of $25.5$. We need to perform a surgical extraction and insertion.

### Phase 2: The Summation Strategy

To correct the mean, we must first find the 'incorrect' sum of all observations. Since the mean $\bar{x} = \frac{\sum x_i}{N}$, we know that $\sum x_i = N 	imes \bar{x}$.

With $N=40$ and $\bar{x}=40$, our incorrect sum is $1600$. Now, let's apply our correction:

$$	ext{Corrected } \sum x_i = 1600 - 2(35.5) + 2(25.5)$$

Calculating this, we get $1600 - 71 + 51 = 1580$. With this corrected sum, the new mean is simply:

$$\bar{x}_{new} = \frac{1580}{40} = 39.5$$

We have successfully shifted the center of our data!

### Phase 3: The Variance Trap

Now, for the part that separates the novices from the masters: the variance. The variance formula is $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$.

To correct this, we need the sum of squares, $\sum x_i^2$. Rearranging our formula, we get $\sum x_i^2 = N(\sigma^2 + \bar{x}^2)$.

Plugging in our old values ($N=40, \sigma^2=49, \bar{x}=40$), we find the incorrect sum of squares:

$$\sum x_i^2 = 40(49 + 40^2) = 40(1649) = 65960$$

Now, we correct this sum of squares by removing the squares of the wrong values and adding the squares of the correct ones:

$$	ext{Corrected } \sum x_i^2 = 65960 - 2(35.5^2) + 2(25.5^2)$$

Using $35.5^2 = 1260.25$ and $25.5^2 = 650.25$, we get $65960 - 2520.5 + 1300.5 = 64740$. Finally, we compute the new variance:

$$\sigma^2_{new} = \frac{64740}{40} - (39.5)^2 = 1618.5 - 1560.25 = 58.25$$

### The Masterstroke

We started with a mean of $40$ and a variance of $49$. After correcting our data, we arrived at a mean of **$39.5$** and a variance of **$58.25$**.

The lesson here is profound: never try to 'patch' the variance. Always deconstruct the problem into its fundamental sums—$\sum x_i$ and $\sum x_i^2$—correct them, and rebuild your statistics from the ground up.

This is how you handle data with precision and confidence. Keep practicing, and keep questioning the numbers!

Visualized Solution

Initial Data and Error Identification

Defining Midpoints $x_{i}$

Formula for Mean $\overset{x}{ˉ}$

Correcting the Sum $\sum x_{i}$

Calculating New Mean $\overset{x}{ˉ}_{n e w}$

Formula for Variance $σ^{2}$

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

Calculating New Variance $σ_{n e w}^{2}$

Final Results Summary

The Sigma Insight: Variance and Standard Deviation

The Data Detective

Phase 1

Phase 2

Phase 3

The Masterstroke

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