Given a binomial distributed random variable with mean 36 an

Given a binomial distributed random variable with mean 36 and standard deviation of 4.8, find n and p. Using the Normal approximation to the Binomial, find the probability that the value of this random variable lies within a 1.5 standard deviation of its mean, that is, within u+ or - 1.5(standard deviation). Find the probability that the value of this random variables lies beyond plue or minus 2.25 standard deviation of its mean.

Solution

a)

As

Mean = n p = 36

s^2 = np(1-p) = 4.8^2

Then

(1-p)= s^2/mean = 4.8^2/36

1-p = 0.64

p = 0.36 [ANSWER]

Thus,

n = mean/p = 36/0.36 = 100 [ANSWER]

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b)

Using the Normal approximation to the Binomial, find the probability that the value of this random variable lies within a 1.5 standard deviation of its mean, that is, within u+ or - 1.5(standard deviation).

This is just like asking the area between z = -1.5 and z = 1.5. Thus,

z1 = lower z score =    -1.5      
z2 = upper z score =     1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866385597   [ANSWER]

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C)

Find the probability that the value of this random variables lies beyond plue or minus 2.25 standard deviation of its mean.

z1 = lower z score =    -2.25      
z2 = upper z score =     2.25      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.012224473      
P(z < z2) =    0.987775527      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.975551055   [ANSWER]