A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for

Q: Why do we have to find the sixth score first? Can't we just ignore it?

That is a great question! In statistics, the standard deviation is a measure of the dispersion of the *entire* dataset. If you ignore the sixth test, you are calculating the standard deviation of only five tests, which would be mathematically incorrect. The problem explicitly states there are six tests, so the sixth score is a necessary component of the final calculation.

Q: Why is the deviation of the sixth score zero?

This is a beautiful property of the mean! The deviation is defined as $d_i = x_i - \bar{x}$. Since we calculated that the sixth score $x_6$ is 48, and the mean $\bar{x}$ is also 48, the deviation $48 - 48$ is naturally zero. This is a lucky break that simplifies our sum of squares significantly!

Q: Is there a faster way to calculate variance without finding deviations?

Yes, you could use the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. However, squaring numbers like 57 or 54 can lead to large, cumbersome values that are prone to arithmetic errors. The deviation method is generally safer and faster for manual calculations in competitive exams like JEE.

Question

A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

Options:

(A) $10/\sqrt{3}$

(B) $100/\sqrt{3}$

(C) $100/3$

(D) $10/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a classroom, looking at a student's performance across six tests. You have the scores for five of them: $45$, $54$, $41$, $57$, and $43$.

The sixth score is a complete mystery, yet you are given one vital piece of information: the mean score across all six tests is exactly $48$. This is the heartbeat of your dataset.

The mean, denoted as $\bar{x}$, is defined as the sum of all observations divided by the total count $n$. Mathematically, we write this as:

$$\bar{x} = \frac{\sum x_i}{n}$$

With $n=6$ and $\bar{x}=48$, we set up our equation:

$$48 = \frac{45 + 54 + 41 + 57 + 43 + x_6}{6}$$

### Unlocking the Sixth Score

Let's break this down. The sum of the five known scores is $240$.

Multiplying both sides of our equation by $6$, we get:

$$288 = 240 + x_6$$

A simple subtraction reveals that $x_6 = 48$. This is a fascinating result, as the sixth score is exactly equal to the mean.

In the world of statistics, this is a gift. It means that when we calculate the deviation of this specific data point from the mean, the result will be zero, saving us precious effort in the next phase.

### The Art of Deviation

Now, we move to the core of the problem: the standard deviation. We use the deviation method, calculating how far each score deviates from the mean: $d_i = x_i - \bar{x}$.

Let's calculate these deviations one by one:

- For $45$: $45 - 48 = -3$
- For $54$: $54 - 48 = 6$
- For $41$: $41 - 48 = -7$
- For $57$: $57 - 48 = 9$
- For $43$: $43 - 48 = -5$
- For $48$: $48 - 48 = 0$

Now, we square each deviation to ensure all values are positive:

$$(-3)^2 = 9, \quad 6^2 = 36, \quad (-7)^2 = 49, \quad 9^2 = 81, \quad (-5)^2 = 25, \quad 0^2 = 0$$

Summing these up, we get:

$$9 + 36 + 49 + 81 + 25 + 0 = 200$$

### The Final Stretch: Variance to Standard Deviation

We are almost at the finish line. The variance $\sigma^2$ is the average of these squared deviations:

$$\sigma^2 = \frac{200}{6} = \frac{100}{3}$$

The standard deviation $\sigma$ is simply the square root of the variance:

$$\sigma = \sqrt{\frac{100}{3}}$$

Taking the square root of the numerator and denominator, we arrive at our final, elegant answer:

$$\sigma = \frac{10}{\sqrt{3}}$$

You have successfully navigated the trap of large numbers and arrived at the solution with precision. Remember, in JEE, the path of least resistance—like using deviations—is often the path to success.

Animated Solution for Mathematics - Statistics: A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

Select Answer:

Visualized Solution

Analyzing the Given Data

Formula for the Mean

Setting Up the Mean Equation

Solving for the $6^{th}$ Score

A Crucial Observation

The Smart Variance Formula

Calculating Deviations

Squaring the Deviations

Summing the Squared Deviations

Calculating the Variance

Final Standard Deviation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

Unlocking the Sixth Score

The Art of Deviation

The Final Stretch

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