Test Index

Statistics Part 2

Section: Mathematics

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Question : 45 of 50

Marks: +1, -0

Let sets

A

B

have 5 elements each. Let mean of the elements in sets

A

B

be 5 and 8 respectively and the variance of the elements in sets

A

B

be 12 and 20 respectively. A new set

C

of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of

C

[11-Apr-2023 shift 1]

36
40
32
38

Solution:

\; \omega A = \{a_1, a_2, a_3, a_4, a_5\}

\; B = \{b_1, b_2, b_3, b_4, b_5\}

\; \frac{\sum\limits_{i=1}^{5} a_i^2}{5} - \left( \frac{\sum\limits_{i=1}^{5} a_i}{5} \right)^2 = 12, \; \frac{\sum\limits_{i=1}^{5} b_i^2}{5} - \left( \frac{\sum\limits_{i=1}^{5} b_i}{5} \right)^2 = 20

\; \Rightarrow \sum\limits_{i=1}^{5} a_i^2=185, \;\; \sum\limits_{i=1}^{5} b_i^2=420

Now,

C = \{C_1, C_2, \ldots, C_{10}\}

\; \therefore \; \text{ Mean of } C, \overline{C} = \frac{ \left(\sum a_i - 15\right) + \left(\sum b_i - 10\right) }{10}

\; \overline{C} = \frac{10+50}{10}=6

\; \therefore \sigma^2 = \frac{ \sum\limits_{i=1}^{10} C_i^2 }{10} = (\overline{C})^2

\; = \frac{ \sum (a_i-3)^2 + \sum (b_i-2)^2 + }{10} - (6)^2

\; = \frac{ \sum a_i^2 + \sum b_i^2 - 6 \sum a_i + 4 \sum b_i + 65 }{10} - 36

\; = \frac{185+420-150+160+65}{10} - 36

\; = 32

\; \therefore \; \text{ Mean } + \text{ Variance } = \overline{C} + \sigma^2 = 6+32=38

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