Question 1047107
Binomial distribution (EITHER MAKE IT THRU OR NOT)
p(passing to 2nd Interview) = .65 and n = 10
 TI syntax is binomcdf(n, p, largest x-value) for binomial ≤ cumulative probability
P(x >= 8) = find P(x ≥ 8) = 1 – P(x ≤ 7) = 1 - binomcdf(20, .65, 7) = 1 - .006

P(2 < x < 6) = p( 3 <= X &#8804; 5) = binomcdf(20, .65, 5) - binomcdf(20, .65, 3) = .00030- .000006 


P (x <= 3) = binomcdf(20, .65, 3)  = .000006

If 540 candidates applied for the job and were interviewed:
Binomial distribution
mean = np = .65(540)		
variance = npq= np(1-p) = .35(.65*540)