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NCERT Exemplar · Q36

Q.Consider the first 10 positive integers. If we multiply each number by −1-1 and then add 1 to each number, the variance of the numbers so obtained is
(A) 8.25
(B) 6.5
(C) 3.87
(D) 2.87

Telangana TsbieMCQ· 1mImportance★★★★★est
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The variance of a dataset is unaffected by adding a constant to each data point, but it is scaled by the square of any multiplier. For the first 10 positive integers, the variance is 102−112=8.25\frac{10^2-1}{12} = 8.25. Multiplying by −1-1 and adding 11 changes the variance by (−1)2=1(-1)^2 = 1, so the new variance is 8.25\boxed{8.25}.

Understanding how variance behaves under linear transformations is a fundamental concept in statistics. Variance measures the spread or dispersion of data points around their mean. A larger variance indicates that data points are more spread out, while a smaller variance means they are clustered closer to the mean.

Consider a set of data points x1,x2,…,xnx_1, x_2, \dots, x_n. The variance, denoted Var(X)\text{Var}(X) or σ2\sigma^2, is defined as the average of the squared differences from the mean xˉ\bar{x}:

Var(X)=1n∑i=1n(xi−xˉ)2\text{Var}(X) = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2

Now, let's see what happens if we transform each data point linearly. Suppose we create a new set of data points yiy_i such that yi=axi+by_i = ax_i + b, where aa and bb are constants.

  1. Effect of adding a constant (bb): If we just add a constant bb to each xix_i, the entire distribution shifts. The mean also shifts by bb, so the new mean yˉ=xˉ+b\bar{y} = \bar{x} + b. However, the differences (yi−yˉ)(y_i - \bar{y}) remain the same:

yi−yˉ=(xi+b)−(xˉ+b)=xi−xˉy_i - \bar{y} = (x_i + b) - (\bar{x} + b) = x_i - \bar{x}

Since the differences from the mean are unchanged, their squares are unchanged, and thus the variance remains the same. Adding a constant does not affect the spread of the data.

2. Effect of multiplying by a constant (aa): If we multiply each xix_i by a constant aa, the spread of the data changes. The new mean yˉ=axˉ+b\bar{y} = a\bar{x} + b. The differences become:

yi−yˉ=(axi+b)−(axˉ+b)=a(xi−xˉ)y_i - \bar{y} = (ax_i + b) - (a\bar{x} + b) = a(x_i - \bar{x})

When we square these differences for the variance calculation, we get:

(yi−yˉ)2=(a(xi−xˉ))2=a2(xi−xˉ)2(y_i - \bar{y})^2 = (a(x_i - \bar{x}))^2 = a^2 (x_i - \bar{x})^2

So, the new variance will be:

Var(Y)=1n∑i=1na2(xi−xˉ)2=a2(1n∑i=1n(xi−xˉ)2)=a2Var(X)\text{Var}(Y) = \frac{1}{n} \sum_{i=1}^n a^2 (x_i - \bar{x})^2 = a^2 \left( \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \right) = a^2 \text{Var}(X)

This means the variance is scaled by the square of the multiplier $a$.

If Y=aX+bY = aX + b is a linear transformation of a random variable XX, then its variance is given by:

Var(Y)=a2Var(X)\text{Var}(Y) = a^2 \text{Var}(X)

Now, let's apply this understanding to the given problem.

  1. Identify the original data set:

    The original numbers are the first 10 positive integers: X={1,2,3,4,5,6,7,8,9,10}X = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. Here, n=10n=10.

  2. Calculate the variance of the original numbers:

    For the first nn natural numbers, the variance has a standard formula.

    ›Proof

    Derivation of Variance for First nn Natural Numbers

    Let X={1,2,…,n}X = \{1, 2, \dots, n\}.

    The sum of the first nn natural numbers is ∑xi=n(n+1)2\sum x_i = \frac{n(n+1)}{2}.

    The mean is xˉ=∑xin=n(n+1)/2n=n+12\bar{x} = \frac{\sum x_i}{n} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}.

    The sum of the squares of the first nn natural numbers is ∑xi2=n(n+1)(2n+1)6\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}.

    The variance can be calculated using the formula Var(X)=∑xi2n−(xˉ)2\text{Var}(X) = \frac{\sum x_i^2}{n} - (\bar{x})^2.

    Substituting the sums:

    Var(X)=n(n+1)(2n+1)6n−(n+12)2\text{Var}(X) = \frac{n(n+1)(2n+1)}{6n} - \left(\frac{n+1}{2}\right)^2

    Var(X)=(n+1)(2n+1)6−(n+1)24\text{Var}(X) = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}

    Factor out (n+1)(n+1): …

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