Answer to Question #339454 in Statistics and Probability for Paul

Question

Question 2 [25] Suppose that the latest census indicates that for every 10 young people available to work only 4 are employed. Suppose a random sample of 20 young graduates is selected. Required: a) What is the probability that they are all employed? b) What is the probability that none of them are employed? c) What is the probability that at least four are employed? d) What is the probability that at most fifteen are employed? e) What is the probability that the number of young graduates who are employed is greater than ten but less than fifteen? f) What is the expected number of graduates who are not employed? g) What is the standard deviation for the number of graduates who are not employed?

Answer 1

Let $X=$ the number of people who are employed: $X\sim Bin(n, p).$

Given $n=20, p=0.4, q=1-p=0.6.$

a)

$$P(X=20)=\dbinom{20}{20}(0.4)^{20}(0.6)^{20-20}$$

$$=0.00000001$$

b)

$$P(X=0)=\dbinom{20}{0}(0.4)^{0}(0.6)^{20-0}$$

$$=0.00003656$$

c)

$$P(X\ge 4)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)$$

$$=1-\dbinom{20}{0}(0.4)^{0}(0.6)^{20-0}-\dbinom{20}{1}(0.4)^{1}(0.6)^{20-1}$$

$$-\dbinom{20}{2}(0.4)^{2}(0.6)^{20-2}-\dbinom{20}{3}(0.4)^{3}(0.6)^{20-3}$$

$$=0.98403884$$

d)

$$P(X\le 15)=1-P(X=16)-P(X=17)$$

$$-P(X=18)-P(X=19)-P(X=20)$$

$$=1-\dbinom{20}{16}(0.4)^{16}(0.6)^{20-16}-\dbinom{20}{17}(0.4)^{17}(0.6)^{20-17}$$

$$-\dbinom{20}{18}(0.4)^{18}(0.6)^{20-18}-\dbinom{20}{19}(0.4)^{19}(0.6)^{20-19}$$

$$-\dbinom{20}{20}(0.4)^{0}(0.6)^{20-20}=0.99968297$$

e)

$$P(10<X<15)=P(X=11)+P(X=12)+P(X=13)$$

$$+P(X=13)+P(X=14)+P(X=15)$$

$$=\dbinom{20}{11}(0.4)^{11}(0.6)^{20-11}+\dbinom{20}{12}(0.4)^{12}(0.6)^{20-12}$$

$$+\dbinom{20}{13}(0.4)^{13}(0.6)^{20-13}+\dbinom{20}{14}(0.4)^{14}(0.6)^{20-14}$$

$$=0.12590972$$

f)

$$E(X^C)=nq=20(0.6)=12$$

g)

$$Var(X^C)=\sigma_{X^C}^2=nqp=20(0.6)(0.4)=4.8$$

$$\sigma_{X^C}=\sqrt{4.8}\approx2.1909$$