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In a binomial distribution, mean = 25, v...

In a binomial distribution, mean = 25, variance = 20 then `P(X le 100)=`

A

`sum_(r=0)^(100)""^(125)C_(r )((1)/(5))^(r ) ((4)/(5))^(125-r)`

B

`sum_(r = 1)^(100) ""^(125)C_(r ) ((1)/(2))^(r )((2)/(5))^(125-r)`

C

`sum_(r = 0)^(100)""^(25)C_(r )((1)/(3))^(r )((2)/(5))^(125-r )`

D

`sum_(r =0)^(100) ""^(225)C_(r )((1)/(5))^(r )((3)/(5))^(125+r)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the properties of the binomial distribution. ### Step 1: Understand the given information We are given: - Mean (μ) = 25 - Variance (σ²) = 20 In a binomial distribution, the mean and variance are given by the formulas: - Mean: \( \mu = np \) - Variance: \( \sigma^2 = npq \) Where: - \( n \) = number of trials - \( p \) = probability of success - \( q \) = probability of failure (where \( q = 1 - p \))
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