Probability question, please HELP me before tomorrow!!!

[From: ] [author: ] [Date: 11-06-14] [Hit: ]

36 (probability that he/she stayed on the r first full-time job for less than one year)x >= 5 (out of 15 how many)P( x>=5) = P(x=5)+P(x=6)+.....+P(x=15)P(x=k) = 15Ck (0.......

= 1 - P[less than four] = 1 - {p(0) + p(1) +.....+p(4)}
Please find this.

b.
The expected number is E(X) = np = 30*0.36 = 10.8 = 11 (approximately)

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n=15 (number of students chosen)
p=.36 (probability that he/she stayed on the r first full-time job for less than one year)
x >= 5 (out of 15 how many)

P( x>=5) = P(x=5)+P(x=6)+.....+P(x=15)

P(x=k) = 15Ck (0.36)^k (1-0.36)^(15-k) --- binomial distribution

5 0.209347
6 0.196263
7 0.141940
8 0.079841
9 0.034931
10 0.011789
11 0.003014
12 0.000565
13 0.000073
14 0.000006
15 0.000000
Add: .6777

You'd need a statistical calculator.

b)
Expected value = np
= 30(.36) = 10.8 (or 11 students)

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A) ( 15C5 * 335C10 ) + (15C15 * 335C0 ) <--- use graphing calculator to solve
B) 350 * 60% = 210
350:210
30:x X= 18 people that are expected to stay on their full-time job less then 1 year

12

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