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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

User Kingcope
by
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2 Answers

6 votes

Answer:

a

Explanation:

User Webaholik
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3 votes

Answer:

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.

User KostyaEsmukov
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