Question

A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then (mean of X)/(standard deviation of X)
is equal to:
(A) 4 (B) 4√3
(C) 3√2 (D) (4√3)/3

Answer 1

Answer: 4/3

Step-by-step explanation:

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by

\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}

in case of discrete data.

Standard Deviation -

In case of discrete frequency distribution

\sigma = \sqrt{\frac{\sum f_{i}x_{i}^{2}}{\sum f_{i}}-\left ( \frac{\sum f_{i}x_{i}}{\sum f_{i}} \right )^{2}}

P(white ball) = \frac{30}{40}=\frac{3}{4}

q=\frac{1}{4}         and    n =  16

mean (X) = np =16\times \frac{3}{4}  

                       = 12

Standard deviation (X) = \sqrt{npq}=\sqrt3

Ans. \frac{12}{\sqrt3}= 4\sqrt3

Option 1)

\frac{4\sqrt3}{3}

Option 2)

4

Option 3)

3\sqrt2

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by

\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}

in case of discrete data.

Standard Deviation -

In case of discrete frequency distribution

\sigma = \sqrt{\frac{\sum f_{i}x_{i}^{2}}{\sum f_{i}}-\left ( \frac{\sum f_{i}x_{i}}{\sum f_{i}} \right )^{2}}

P(white ball) = \frac{30}{40}=\frac{3}{4}

q=\frac{1}{4}         and    n =  16

mean (X) = np =16\times \frac{3}{4}  

                       = 12

Standard deviation (X) = \sqrt{npq}=\sqrt3

Ans. \frac{12}{\sqrt3}= 4\sqrt3

Option 1)

\frac{4\sqrt3}{3}

Option 2)

4

Option 3)

3\sqrt2

Option

4\sqrt3