Question

A machine produces 0, 1 or 2 defective items in a day with probabilities of \( \frac{1}{4}, \frac{1}{2}, \frac{1}{4} \) respectively. Then, the standard deviation of the number of defective items produced by the machine in a day is ............

• A. $\dfrac{1}{2}$
• B. $\dfrac{1}{\sqrt{2}}$
• C. $\dfrac{1}{4}$
• D. $1$

Hint:

To find the standard deviation of a discrete random variable, first calculate the mean and variance. Then, take the square root of the variance to get the standard deviation.

Answer 1

The Correct Option is B

Let $X$ be the number of defective items produced by the machine in a day. The possible values of $X$ are 0, 1, and 2, with probabilities as follows: \[ P(X = 0) = \frac{1}{4}, P(X = 1) = \frac{1}{2}, P(X = 2) = \frac{1}{4} \] 1. Mean ($\mu$) of the distribution:
The mean is given by: \[ \mu = E[X] = \sum_{x} x \cdot P(X = x) \] \[ \mu = 0 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} = 0 + \frac{1}{2} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1 \] 2. Variance ($\sigma^2$) of the distribution:
The variance is given by: \[ \sigma^2 = E[X^2] - \mu^2 \] First, we calculate $E[X^2]$: \[ E[X^2] = \sum_{x} x^2 \cdot P(X = x) \] \[ E[X^2] = 0^2 \cdot \frac{1}{4} + 1^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{1}{4} = 0 + 1 \cdot \frac{1}{2} + 4 \cdot \frac{1}{4} = 0 + \frac{1}{2} + 1 = \frac{3}{2} \] Now, calculate the variance: \[ \sigma^2 = \frac{3}{2} - 1^2 = \frac{3}{2} - 1 = \frac{1}{2} \] 3. Standard deviation ($\sigma$):
The standard deviation is the square root of the variance: \[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Thus, the standard deviation of the number of defective items produced by the machine in a day is $\dfrac{1}{\sqrt{2}}$.