1 Find the mean for the binomial distribution which has the

1) Find the mean for the binomial distribution which has the following values of n and p. Round answer to nearest tenth: n=37, p=0.2

2) Find the standard deviation for the binomial distribution which has the following values of n and p. Round answer to nearest hundredth: n=22, p= 3/5

3) Use the given values of n and p to find the minimum and maximum usual values: n=97, p=0.24

Solution

1.

u = mean = np =    7.4 [ANSWER]

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2.

s = standard deviation = sqrt(np(1-p)) =    2.297825059 [ANSWER]

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3.

u = mean = np =    23.28
  
s = standard deviation = sqrt(np(1-p)) =    4.206281018

Hence,

u - 2s = 23.28 - 2*4.206281018 = 14.86743796 [ANSWER, MINIMUM USUAL]

u + 2s = 23.28 + 2*4.206281018 = 31.69256204 [ANSWER, MAXIMUM USUAL]