Question

Sixty nine percent of U.S college graduates expect to stay at their first employer for three or more years. You randomly select 18 U.S. college graduates and ask them

whether they are expecting to stay at their first for three or more years. Using binomial distribution with continuity, find the probability that the number who expect to stay at

their first employer for three or more years is

a. Exactly 10

b. Less than 7 WITHOUT EXCEL

c.At least 15 WITHOUT EXCEL

Answer 1

Ans:

Use binomial distribution with n=18 and p=0.69

P(x=r)=18C_r*0.69^r*(1-0.69)^18-r

a)

P(x=10)=18C₁₀*0.69^10*(1-0.69)^8=0.0913

b)

P(less than 7)=P(x<7)

=P(x=0)+P(x=1)+.........+P(x=6)

=18C₀*0.69⁰*(1-0.69)¹⁵+18C₁*0.69¹*(1-0.69)¹⁴+............................+18C₆*0.69⁶*(1-0.69)⁹

^{=0+0+0+....................+0.0001+0.0003+0.0016}

^=0.0020

c)

P(at least 15)=P(x>=15)

=P(x=15)+P(x=16)+P(x=17)+P(x=18)

=18C₁₅*0.69¹⁵*(1-0.69)³+18C₁₆*0.69¹⁶*(1-0.69)²+18C₁₇*0.69¹⁷*(1-0.69)¹+18C₁₈*0.69¹⁸

=0.0930+0.0388+0.0102+0.0013

=0.1432