JEE Main 2020 (8 January Shift 1)

LEVELBoard

Animated Solution for Mathematics - Statistics: 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:

Q: Why did we reject $p = 1/2$?

The problem explicitly states that $q \neq 0$. When we substituted $p = 1/2$ into our mean equation, we found $q = 0$. Since this violates the given constraint, we must discard that case and look for the alternative value of $p$.

Q: Can I just use the variance instead of standard deviation?

Absolutely! If $\sigma' = |p|\sigma$, then $\sigma'^2 = p^2 \sigma^2$. Using variance is often safer because it avoids the absolute value sign initially, but you must remember to check the sign of $p$ at the end to satisfy the mean equation.

Q: Is it possible for $q$ to be positive?

In this specific problem, the algebra forces $q$ to be $-20$. If you had chosen a different transformation, $q$ could be anything, but here, the constraints on the mean and the standard deviation lock $q$ into a single, unique value.

Select Answer:

Visualized Solution

Initial Parameters

Original Mean: $\overset{x}{ˉ} = 20$
Original Standard Deviation: $σ = 2$

The Transformation

Transformation: $x_{i}^{'} = p x_{i} - q$
Constraints: $p \neq = 0, q \neq = 0$

New Distribution

New Mean: $\overset{x}{ˉ}^{'} = 10$
New Standard Deviation: $σ^{'} = 1$

Effect on Mean

Formula: $\overset{x}{ˉ}^{'} = p \overset{x}{ˉ} - q$
Substitute: $10 = 20 p - q$

Effect on Standard Deviation

Formula: $σ^{'} = ∣ p ∣ σ$
Substitute: $1 = ∣ p ∣ \cdot 2$

Solving for $p$

$∣ p ∣ = \frac{1}{2}$
Possible values: $p = \frac{1}{2}$ or $p = - \frac{1}{2}$

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

Substitute $p = \frac{1}{2}$ into $10 = 20 p - q$
$10 = 20 (\frac{1}{2}) - q ⟹ 10 = 10 - q$
$q = 0$

Constraint Check

Given constraint: $q \neq = 0$
Therefore, $p = \frac{1}{2}$ is invalid.

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

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

Calculate Final Value of $q$

$q = - 10 - 10$
$q = - 20$

00:00 / 00:00

The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Setup

In statistics, we often apply linear transformations to a dataset. Given a set of 10 observations with a mean

\overset{x}{ˉ} = 20

and a standard deviation

σ = 2

, we apply the transformation

y_{i} = p x_{i} - q

We need to determine the values of

p

and

q

such that the new mean

\overset{x}{ˉ}^{'} = 10

and the new standard deviation

σ^{'} = 1

, under the constraint that $q eq 0$.

The Geometry of Transformation

The mean acts as the balance point of the data. When we apply the transformation

y = p x - q

, the new mean

\overset{x}{ˉ}^{'}

is given by:

\overset{x}{ˉ}^{'} = p \overset{x}{ˉ} - q

The standard deviation, however, measures the dispersion of the data. Shifting the data by

q

does not change the distance between points, but scaling by

p

affects the spread. Thus, the new standard deviation

σ^{'}

is:

σ^{'} = ∣ p ∣ σ

Solving the Puzzle

We substitute the given values

\overset{x}{ˉ}^{'} = 10

σ^{'} = 1

\overset{x}{ˉ} = 20

, and

σ = 2

into our equations.

For the standard deviation:

1 = ∣ p ∣ \cdot 2 ⟹ ∣ p ∣ = \frac{1}{2}

This yields two possible values for

p

p = \frac{1}{2}

p = - \frac{1}{2}

For the mean:

10 = 20 p - q

Evaluating the Constraints

First, consider the case where

p = \frac{1}{2}

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

10 = 10 - q ⟹ q = 0

Since the problem explicitly states that $q eq 0$, this solution is invalid. We must reject this path.

Now, consider the case where

p = - \frac{1}{2}

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

10 = - 10 - q

q = - 10 - 10 = - 20

Final Result

By navigating the constraints and applying the properties of linear transformations, we find the unique solution:

$p = - \frac{1}{2}$ and $q = - 20$

Common Traps & FAQs

Wait, why does the standard deviation only depend on

p

and not

q

Standard deviation measures the 'spread' of data. If you shift every data point by a constant

q

, the distance between points remains identical. Imagine moving a cluster of points along a number line; the cluster's shape doesn't change, so the standard deviation remains invariant to translation.

Why is it

∣ p ∣

in the standard deviation formula and not just

p

Great catch! Standard deviation is a measure of dispersion and is defined as the square root of variance, which is always positive. If

p

is negative, the data points are reflected across the origin, which flips the distribution but doesn't shrink or expand the distance between points. Thus, we use the absolute value

∣ p ∣

to ensure the spread remains positive.

Why did we reject

p = 1/2

The problem explicitly states that

q \neq = 0

. When we substituted

p = 1/2

into our mean equation, we found

q = 0

. Since this violates the given constraint, we must discard that case and look for the alternative value of

p

Can I just use the variance instead of standard deviation?

Absolutely! If

σ^{'} = ∣ p ∣ σ

, then

σ^{'2} = p^{2} σ^{2}

. Using variance is often safer because it avoids the absolute value sign initially, but you must remember to check the sign of

p

at the end to satisfy the mean equation.

Is it possible for

q

to be positive?

In this specific problem, the algebra forces

q

to be

- 20

. If you had chosen a different transformation,

q

could be anything, but here, the constraints on the mean and the standard deviation lock

q

into a single, unique value.

Select Answer:

Visualized Solution

Initial Parameters

The Transformation

New Distribution

Effect on Mean

Effect on Standard Deviation

Solving for $p$

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

Constraint Check

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

Calculate Final Value of $q$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Geometry of Transformation

Solving the Puzzle

Evaluating the Constraints

Final Result

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 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 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 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 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

Initial Parameters

The Transformation

New Distribution

Effect on Mean

Effect on Standard Deviation

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

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

Constraint Check

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

Calculate Final Value of .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 Setup

The Geometry of Transformation

Solving the Puzzle

Evaluating the Constraints

Final Result

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 $p$

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

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

Calculate Final Value of $q$