Assume that a procedure yields a binomial distribution with n=579n=579 trials and the probability of success for one trial is p=48%p=48%.Find the mean for this binomial distribution.(Round answer to one decimal place.)μ=μ= Find the standard deviation for this distribution.(Round answer to two decimal places.)σ=σ= Use the range rule of thumb to find the minimum usual value μ–2σ and the maximum usual value μ+2σ.Enter answer as an interval using square-brackets only with whole numbers.usual values =

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Asked Nov 9, 2019
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Assume that a procedure yields a binomial distribution with n=579n=579 trials and the probability of success for one trial is p=48%p=48%.

Find the mean for this binomial distribution.
(Round answer to one decimal place.)
μ=μ= 

Find the standard deviation for this distribution.
(Round answer to two decimal places.)
σ=σ= 

Use the range rule of thumb to find the minimum usual value μ–2σ and the maximum usual value μ+2σ.
Enter answer as an interval using square-brackets only with whole numbers.
usual values = 

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Expert Answer

Step 1

From the given information,

Sample size (n) = 579.

Probability of success (p) =0.48

Step 2

a)

The mean for this binomial distribution is-

Since, the sample size is large enough thus, by normal approximation to binomial.

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E(X)= %3Dпхр = 579x0.48 =277.92

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Step 3

b)

 The standard deviation of for this bino...

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p(1-p) n 0.48x (1-0.48) 579 0.02

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