JEE Main 2020 - 8 Jan (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: The mean and the standard deviation ( $s, d$ ) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by $p$ and then reduced by $q$ , where $p \neq = 0$ and $q \neq = 0$ . If the new mean and new $s . d .$ become half of their original values, then $q$ is equal to :

Q: Why did we reject $p = 0.5$? Is it just because $q$ became zero?

Exactly! The problem explicitly states that $q \neq 0$. When we tested $p = 0.5$, we found that $q$ had to be $0$ to satisfy the mean equation. Since this violates the given constraint, that entire branch of the solution is invalid. Always check your final results against the constraints provided in the question!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the data transformation as a shift and a scale. Remember that the standard deviation is a measure of dispersion, which means it is completely unaffected by shifting the data (the

- q

part) and only changes when you scale it (the

p

part).

Hint 2

Hint 2 (The Arsenal): Recall the fundamental properties of statistics: for a transformation

y = p x - q

, the new mean is

\overset{y}{ˉ} = p \overset{x}{ˉ} - q

, and the new standard deviation is

σ_{y} = ∣ p ∣ σ_{x}

. Do not forget the absolute value in the standard deviation formula!

Hint 3

Hint 3 (The Execution): Start by solving for

p

using the standard deviation equation:

1 = ∣ p ∣ \cdot 2

. This gives two possible values for

p

. Then, substitute each value of

p

into the mean equation

10 = p (20) - q

and check which one satisfies the constraint

q \neq = 0

Visualized Solution

Original Data Parameters

Given observations: $x_{1}, x_{2}, \dots, x_{10}$
Original Mean $\overset{x}{ˉ} = 20$
Original Standard Deviation $σ_{x} = 2$

Linear Transformation

Transformation rule: $y_{i} = p x_{i} - q$
Constraints: $p \neq = 0$ and $q \neq = 0$

New Data Parameters

New Mean $\overset{y}{ˉ} = \frac{20}{2} = 10$
New S.D. $σ_{y} = \frac{2}{2} = 1$

Property of Standard Deviation

S.D. is independent of change of origin.
S.D. depends on change of scale: $σ_{y} = ∣ p ∣ σ_{x}$

Solving for Scale Factor $p$

$1 = ∣ p ∣ \cdot 2$
$∣ p ∣ = \frac{1}{2}$
Therefore, $p = \frac{1}{2}$ or $p = - \frac{1}{2}$

Property of Mean

Mean is affected by both change of scale and origin.
Formula: $\overset{y}{ˉ} = p \overset{x}{ˉ} - q$

Evaluating Case 1: $p = \frac{1}{2}$

Substitute $p = \frac{1}{2}$ , $\overset{x}{ˉ} = 20$ , $\overset{y}{ˉ} = 10$
$10 = \frac{1}{2} (20) - q$
$10 = 10 - q ⟹ q = 0$

Checking Constraints

Given constraint: $q \neq = 0$
Since $q = 0$ in Case 1, $p = \frac{1}{2}$ is rejected.

Evaluating Case 2: $p = - \frac{1}{2}$

Substitute $p = - \frac{1}{2}$
$10 = - \frac{1}{2} (20) - q$
$10 = - 10 - q$

Final Calculation for $q$

$q = - 10 - 10$
$q = - 20$
This satisfies $q \neq = 0$ .

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Transformation

We are given a dataset with an initial mean

\overset{x}{ˉ} = 20

and a standard deviation

σ_{x} = 2

. We apply the linear transformation

y = p x - q

to each data point.

The resulting dataset has a new mean

\overset{y}{ˉ} = 10

and a new standard deviation

σ_{y} = 1

. We are tasked with finding the value of

q

, under the constraint that $q eq 0$.

The Mystery of the Spread

The standard deviation measures the dispersion of data. When we apply the transformation

y = p x - q

, the shift

q

does not affect the spread of the data, as it merely translates the entire distribution along the number line.

However, scaling the data by

p

directly affects the standard deviation. The relationship is defined by:

σ_{y} = ∣ p ∣ σ_{x}

Substituting the known values, we have:

1 = ∣ p ∣ \cdot 2 ⟹ ∣ p ∣ = 0.5

This yields two possible values for the scaling factor:

p = 0.5

p = - 0.5

The Sensitivity of the Mean

Unlike the standard deviation, the mean is sensitive to both scaling and shifting. The transformation of the mean follows the linear equation:

\overset{y}{ˉ} = p \overset{x}{ˉ} - q

Substituting the given values

\overset{y}{ˉ} = 10

and

\overset{x}{ˉ} = 20

, we obtain the master equation:

10 = 20 p - q

Evaluating the Candidates

We must now test our two potential values for

p

to find the valid

q

that satisfies $q eq 0$.

Case 1: $p = 0.5$

Substituting

p = 0.5

into the mean equation:

10 = 20 (0.5) - q

10 = 10 - q ⟹ q = 0

Since the problem explicitly states that $q eq 0$, we must reject this solution.

Case 2: $p = - 0.5$

Substituting

p = - 0.5

into the mean equation:

10 = 20 (- 0.5) - q

10 = - 10 - q

q = - 10 - 10

q = - 20

This value satisfies all given constraints. Therefore, the final value is $q = - 20$ .

Common Traps & FAQs

Wait, why does the standard deviation ignore the

- q

part? Shouldn't it change if we subtract a constant from every number?

That is a fantastic question! Think of standard deviation as the 'average distance' of data points from their mean. If you subtract

q

from every single data point, you are essentially shifting the entire cluster of data to the left or right on the number line. The distance between the points themselves remains exactly the same! Because the spread hasn't changed, the standard deviation remains unchanged by the shift

q

Why do we get two values for

p

? And why does the absolute value matter?

The standard deviation is always a non-negative value by definition. When we say

σ_{y} = ∣ p ∣ σ_{x}

, the absolute value ensures that even if

p

is negative (which would flip the data distribution), the spread remains positive. Since

∣ p ∣ = 0.5

p

could be

0.5

- 0.5

. We must test both because the problem doesn't specify that

p

is positive.

Why did we reject

p = 0.5

? Is it just because

q

became zero?

Exactly! The problem explicitly states that

q \neq = 0

. When we tested

p = 0.5

, we found that

q

had to be

0

to satisfy the mean equation. Since this violates the given constraint, that entire branch of the solution is invalid. Always check your final results against the constraints provided in the question!

Select Answer:

Visualized Solution

Original Data Parameters

Linear Transformation

New Data Parameters

Property of Standard Deviation

Solving for Scale Factor $p$

Property of Mean

Evaluating Case 1: $p = \frac{1}{2}$

Checking Constraints

Evaluating Case 2: $p = - \frac{1}{2}$

Final Calculation for $q$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Transformation

The Mystery of the Spread

The Sensitivity of the Mean

Evaluating the Candidates

Similar Questions

The mean and standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by $p$ and then reduced by $q$ , where $p \neq = 0$ and $q \neq = 0$ . If the new mean and standard deviation become half of their original values, then $q$ is equal to:

The mean and the standard deviation(s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations $α$ in this data is replaced by $β$ , then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If $α$ and $β$ are the mean and standard deviation respectively for correct data, then $(α, β)$ is:

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation 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 in a series of $2 n$ observations, half of them are equal to $a$ and remaining half are equal to $- a$ . Also by adding a constant $b$ in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of $a^{2} + b^{2}$ is equal to:

Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $x$ . If $\sum_{i = 1}^{10} (x_{i} - p) = 3$ and $\sum_{i = 1}^{10} (x_{i} - p)^{2} = 9$ where $0 \neq = p \in R$ , then the standard deviation of these observations is

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $σ$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 σ^{2}$ is equal to

Select Answer:

Visualized Solution

Original Data Parameters

Linear Transformation

New Data Parameters

Property of Standard Deviation

Solving for Scale Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Property of Mean

Evaluating Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=21​

Checking Constraints

Evaluating Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=−21​

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

The Sigma Insight: Variance and Standard Deviation

Analyzing the Transformation

The Mystery of the Spread

The Sensitivity of the Mean

Evaluating the Candidates

Similar Questions

The mean and the standard deviation(s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation 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

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

Solving for Scale Factor $p$

Evaluating Case 1: $p = \frac{1}{2}$

Evaluating Case 2: $p = - \frac{1}{2}$

Final Calculation for $q$