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

Answer:

The standard deviation of given probability distribution is 0.767.

Step-by-step 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:

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

$\text{Standard Deviation} = \sqrt{\text{Variance}}\\= \sqrt{0.587} = 0.767$

The standard deviation of given probability distribution is 0.767.