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The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution

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प्रश्न

The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution

योग
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उत्तर

In a binomial distribution

Mean np = 5 → (1)

Standard deviation `sqrt("npq")` = 2

Squaring on body sides

npq = 4 → (2)

Equation ÷ Equation

⇒ `"npq"/"np" = 4/5`

∴ q = `4/5`, p = `1 - "q"`

= `1 - 4/5 = (5 - 4)/5`

p = `1/5`

Substitute p = `1/5` in equation (1)

`"n"(1/5)` = 5

n = 5 × 5

⇒ n = 25

∴ The binomial distribution is

P(X = x) = ncxpxqn-x 

i.e P(X = x) = `[(25),(x)] (1/5)^x (4/5)^(25 - x)`

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अध्याय 7: Probability Distributions - Exercise 7.1 [पृष्ठ १५६]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 7 Probability Distributions
Exercise 7.1 | Q 16 | पृष्ठ १५६

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