Question

(4) Consider the probability that no less than 87 in a sample of 153 software users will not call technical support. Assume the probability that a given software user will not call technical support is 63%. (o) Set up a sun for tebblit ig sgma nolation and sosu sampling with replacement. You need not evaluate this sum. (4 points) (b) What is the expected number of users who will not call technical support in a random sample of 153 software users? (2 points) (c) Approximate the probability using the normal distribution, showing all relevant z-score calculations. Round your answer to four decimal places. (4 points)

Answer 1

it is a binomial probability distribution,               because there is fixed number of trials,
               only two outcomes are there, success and failure
               trails are independent of each other
and probability is given by              

P(X=x) = C(n,x)*px*(1-p)(n-x)

where  
Sample size , n =    153
Probability of an event of interest, p =   0.63

a)

P(no less than 87) = P(X≥87) = ΣC(n,x)*p^x*(1-p)^(n-x) where x goes from 87 to 153

P(x≤87) =  ΣC(153 ,x)*0.63^x*(1-0.63)^(153-x) where x goes from 87 to 153

b)

expected users = np =153 * 0.63 =    96.39

c)

Binomial to normal approximations using continuity

Sample size , n =    153          
Probability of an event of interest, p =   0.63          
right tailed              
X ≥   87          
              
Mean = np =    96.39          
std dev ,σ=√np(1-p)=   5.9720          
              
P(X ≥   87   ) = P(Xnormal ≥   86.5   )
              
Z=(Xnormal - µ ) / σ =    -1.656          
              
=P(Z ≥   -1.656   ) =    0.9511 (answer)