Question 1047107: Ten candidates apply for a job and are separately interviewed. Based on past experience, any candidate has a 65% chance of passing to a second interview. What is the probability that: [17]
At least eight candidates pass to a second interview
More than two but less than six candidates pass to a second interview
At most three candidates do not pass to a second interview
If 540 candidates applied for the job and were interviewed, what is the variance and the expected number of candidates that will pass to a second interview?
Found 2 solutions by ewatrrr, Timnewman:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! 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 ≤ 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)
Answer by Timnewman(323)  (Show Source):
You can put this solution on YOUR website! xy=14
x+y=9
solution
xy=14---(1)
x+y=9---(2)
from equ 2,
y=9-x ---(3)
put 3 in 1,
x(9-x)=14
9x-x^2=14
x^2-9x+14=0
factorize the above,
(x-2)(x-7)=0
x=2 or x=7
put x=2 in equ 3
y=9-3
y=7.
The value for x and y is 2 and 7 .
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