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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 105-106

Consider the following for the two (02) items that follow:

The sum and the sum of squares of the observations corresponding to length X (in cm) and weight Y (in gm) of 50 tropical tubers are given as

ΣX=200

,ΣY=250,ΣX^{2}=900andΣY^{2}=1400

Question : 105 of 120

Marks: +1, -0

Which one of the following is correct?

Solution:

Concept:

Variance measures the spread of data. Use the formula: variance =

\frac{\sum X^2}{N} - \left(\frac{\sum X}{N}\right)^2

where

N

is the number of observations.

Explanation:

For length

X

N = 50

\sum X = 200

\sum X^2 = 900

Mean of

X

\frac{200}{50} = 4

Variance of

X

\frac{900}{50} - (4)^2 = 18 - 16 = 2

For weight

Y

\sum Y = 250

\sum Y^2 = 1400

Mean of

Y

\frac{250}{50} = 5

Variance of

Y

\frac{1400}{50} - (5)^2 = 28 - 25 = 3

Since

2 < 3

, variance of

X

is less than variance of

Y

Answer:

Option B:

\text{Variance}(X) < \text{Variance}(Y)

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