JEE Main 2022 (28 June Shift 2)

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Animated Solution for Mathematics - Statistics: Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is

Enter Numerical Value:

Visualized Solution

Given Data: $n = 7, \overset{x}{ˉ} = 62$

Number of students $n = 7$
Mean marks $\overset{x}{ˉ} = 62$
Variance $σ^{2} = 20$

Variance Formula: $σ^{2}$

Variance formula: $σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})^{2}$
This measures the average squared deviation from the mean.

Substitute Values

Substitute the given values:
$20 = \frac{1}{7} \sum_{i = 1}^{7} (x_{i} - 62)^{2}$

Total Squared Deviation

Multiply by $7$ to find the total sum of squared deviations:
$\sum_{i = 1}^{7} (x_{i} - 62)^{2} = 7 \times 20 = 140$

Failing Condition: $x_{i} < 50$

A student fails if their marks $x_{i} < 50$ .
The mean is $\overset{x}{ˉ} = 62$ .

Deviation: $x_{i} - 62 < - 12$

Deviation for a failing student:
$x_{i} - 62 < 50 - 62$
$x_{i} - 62 < - 12$

Squared Deviation: $> 144$

Square the inequality (since $x_{i} - 62$ is negative):
$(x_{i} - 62)^{2} > (- 12)^{2}$
$(x_{i} - 62)^{2} > 144$

Contradiction: $144 > 140$

Total sum available: $\sum (x_{i} - 62)^{2} = 140$
Required for $1$ failure: $(x_{i} - 62)^{2} > 144$
Since $144 > 140$ , even one student failing is impossible.
Final Answer: $0$

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

We are given a set of

n = 7

students with a mean score of

\overset{x}{ˉ} = 62

and a variance of

σ^{2} = 20

. A student is defined as having failed if their score

x_{i} < 50

. We aim to find the maximum number of students who could have failed under these constraints.

The Budget of Deviation

The variance is defined by the formula:

σ^{2} = \frac{1}{n} i = 1 \sum n (x_{i} - \overset{x}{ˉ})^{2}

Given

n = 7

\overset{x}{ˉ} = 62

, and

σ^{2} = 20

, we can calculate the total sum of squared deviations (our "deviation budget"):

i = 1 \sum 7 (x_{i} - 62)^{2} = n \times σ^{2} = 7 \times 20 = 140

This value of

140

represents the absolute limit for the sum of squared differences from the mean. No distribution of marks can exceed this total.

The Red Zone

Defining Failure

A student fails if

x_{i} < 50

. We examine the impact of a single failing student on our deviation budget by looking at their deviation from the mean:

(x_{i} - 62) < (50 - 62)

(x_{i} - 62) < - 12

To determine the cost to our budget, we square this deviation. Since we are squaring a value less than

- 12

, the inequality becomes:

(x_{i} - 62)^{2} > (- 12)^{2}

(x_{i} - 62)^{2} > 144

The Clash of Constraints

To have even a single student fail, we would be required to spend at least

144

units of our deviation budget. However, our total available budget is strictly limited to

140

units.

Because

144 > 140

, it is mathematically impossible for any student to score below

50

without violating the given variance constraint. Even if we attempted to make one student fail, the resulting squared deviation would exceed the total allowed budget.

Final Conclusion

Since the cost of a single failure exceeds the total budget, the number of students who can fail is 0. The constraints of the mean and variance ensure that every student must have scored at least

50

Common Traps & FAQs

Wait, why do we reject the negative root here? Or rather, why does the inequality flip?

That is a fantastic observation! When we say a student fails, their marks

x_{i}

are less than

50

. Since the mean

\overset{x}{ˉ}

62

, the deviation

(x_{i} - 62)

must be less than

(50 - 62)

, which is

- 12

. Now, when we square both sides of an inequality like

A < B

where both

A

and

B

are negative, the inequality reverses. For example,

- 15 < - 12

, but

(- 15)^{2} = 225

and

(- 12)^{2} = 144

. Since

225 > 144

, the squared deviation for a failing student must be strictly greater than

144

. It is not about rejecting a root, but about the properties of squaring negative numbers!

Why can't we just use energy conservation or some other physics principle?

I love the interdisciplinary thinking! While this is a statistics problem, you can actually view it through the lens of 'potential energy' or 'cost.' Think of the variance as the total potential energy of the system. Each student's deviation from the mean is like a displacement from an equilibrium position. To have a student fail, you must 'push' them far away from the mean, which costs a massive amount of 'squared deviation energy.' You simply don't have enough 'energy' in your total budget to push even one student that far!

Is it possible for the answer to be more than zero if the variance was higher?

Absolutely! The variance is the key. If the variance were much larger, say

1000

instead of

20

, the total budget of squared deviations would be

7 \times 1000 = 7000

. Since

7000

is much larger than

144

, it would be very easy for one or more students to fail. The variance essentially dictates how 'spread out' the marks are. A low variance means everyone is clustered near the mean, making it impossible for anyone to be far enough away to fail.

Enter Numerical Value:

Visualized Solution

Given Data: $n = 7, \overset{x}{ˉ} = 62$

Variance Formula: $σ^{2}$

Substitute Values

Total Squared Deviation

Failing Condition: $x_{i} < 50$

Deviation: $x_{i} - 62 < - 12$

Squared Deviation: $> 144$

Contradiction: $144 > 140$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Budget of Deviation

The Red Zone

The Clash of Constraints

Final Conclusion

Similar Questions

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

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

If the variance of the following frequency distribution: Class : 10-20, 20-30, 30-40 Frequency : 2, $x$ , 2 is 50, then $x$ is equal to

If the mean of the frequency distribution $Class : Frequency : 0 - 10 2 10 - 20 3 20 - 30 x 30 - 40 5 40 - 50 4$ is 28, then its variance is ________ .

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ , be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, \dots, 5$ is

Enter Numerical Value:

Visualized Solution

Given Data: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=7,xˉ=62

Variance Formula: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Substitute Values

Total Squared Deviation

Failing Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xi​<50

Deviation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xi​−62<−12

Squared Deviation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } >144

Contradiction: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 144>140

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Budget of Deviation

The Red Zone

The Clash of Constraints

Final Conclusion

Similar Questions

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the variance of the following frequency distribution: Class : 10-20, 20-30, 30-40 Frequency : 2, x, 2 is 50, then x is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean of the frequency distribution Class :Frequency :​0−102​10−203​20−30x​30−405​40−504​​ is 28, then its variance is ________ .

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

Given Data: $n = 7, \overset{x}{ˉ} = 62$

Variance Formula: $σ^{2}$

Failing Condition: $x_{i} < 50$

Deviation: $x_{i} - 62 < - 12$

Squared Deviation: $> 144$

Contradiction: $144 > 140$

If the variance of the following frequency distribution: Class : 10-20, 20-30, 30-40 Frequency : 2, $x$ , 2 is 50, then $x$ is equal to

If the mean of the frequency distribution $Class : Frequency : 0 - 10 2 10 - 20 3 20 - 30 x 30 - 40 5 40 - 50 4$ is 28, then its variance is ________ .