Question

The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4521, 0.3970, 0.1307, 0.0191, and 0.0010, respectively. Find the standard deviation for the probability distribution

Answer 1

a

Explanation:

Answer 2

The standard deviation of given probability distribution is 0.767.

Explanation:

We are given the following information in the question:

X: 0 1 2 3 4

P(x): 0.4521 0.3970 0.1307 0.0191 0.0010

Formula:

`\text{Mean} = \sum X.P(x)\\= 0(0.4521) + 1(0.3970) + 2(0.1307) + 3(0.0191) + 4(0.0010)\\= 0.7197`

`\mu = 0.7197`

Formula:

`\text{Variance} = \sum (X-\mu)^2E(x)\\= (0-0.7197)^2(0.4521)+(1-0.7197)^2(0.3970)+(2-0.7197)^2(0.1307)+(3-0.7197)^2(0.0191)+(4-0.7197)^2(0.0010) \\=0.587`

`\text{Standard Deviation} = \sqrt{\text{Variance}}\\= √(0.587) = 0.767`

The standard deviation of given probability distribution is 0.767.