Question

# Assume that a procedure yields a binomial distribution with n=826n=826 trials and the probability of success...

Assume that a procedure yields a binomial distribution with n=826n=826 trials and the probability of success for one trial is p=p=1/12.

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=

The range rule of thumb says that the the minimum usual value is μ-2σ and the maximum usual value is μ+2σ.
Use this to find the minimum and maximum usual values.
Enter answer as an interval using square-brackets -- [ and ] -- and round your answers to be whole numbers. There should be a comma in between your interval endpoints, but no spaces. For instance, [1,10].
usual values =

Given,

n = 826, p = 0.08

a)

Mean = np

= 826* 0.08

= 66.1

b)

Standard deviation = Sqrt( np( 1 - p) )

= sqrt( 826* 0.08 * 0.92)

= 7.80

c)

Minimum usual value - 2 = 66.1- 2 * 7.80

= 50.5

Maximum usual value = + 2

= 6.1 +  2 * 7.80

= 81.7

[51,82]

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