Answer to Question #267184 in Statistics and Probability for Ayomide Adebiyi

Question

1.A survey from teenage research unlimited found that 80% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs

2.PMEA associate market research found that 40% of Nigerians don’t think that having a college education is important to succeed in the business world, if a random sample of five Nigerians are selected, find these probabilities

A.That exactly 2 people will agree with that statement

B.At most 3 people will agree with that statement

C.At least 2 people will agree with that statement

D.Fewer than 3 people will agree with that statement

That none of the people will agree with that statement

Answer 1

1. Let $X=$ the number of teenagers who will have part-time jobs: $X\sim Bin (n, p).$

Given $n=5, p=0.8, q=1-p=0.2$

$$P(X\geq3)=P(X=3)+P(X=4)+P(X=5)$$

$$=\dbinom{5}{3}(0.8)^3(0.2)^{5-3}+\dbinom{5}{4}(0.8)^4(0.2)^{5-4}$$

$$+\dbinom{5}{5}(0.8)^5(0.2)^{5-5}=0.2048+0.4096+0.32768$$

$$=0.94208$$

2. Let $X=$ the number of people who will agree with that statement: $X\sim Bin (n, p).$

Given $n=5, p=0.4, q=1-p=0.6$

A.

$$P(X=2)=\dbinom{5}{2}(0.4)^2(0.6)^{5-2}=0.3456$$

B.

$$P(X\leq3)=1-P(X=4)-P(X=5)$$

$$=1-\dbinom{5}{4}(0.4)^4(0.6)^{5-4}+\dbinom{5}{5}(0.4)^5(0.6)^{5-5}$$

$$=1-0.0768-0.01024=0.91296$$

C.

$$P(X\geq2)=1-P(X=0)-P(X=1)$$

$$=1-\dbinom{5}{0}(0.4)^0(0.6)^{5-0}+\dbinom{5}{1}(0.4)^1(0.6)^{5-1}$$

$$=1-0.07776-0.2592=0.66304$$

D.

$$P(X<3)=P(X=0)+P(X=1)+P(X=2)$$

$$=\dbinom{5}{0}(0.4)^0(0.6)^{5-0}+\dbinom{5}{1}(0.4)^1(0.6)^{5-1}$$

$$+\dbinom{5}{2}(0.4)^2(0.6)^{5-2}$$

$$=0.07776+0.2592+0.3456=0.68256$$

E.

$$P(X=0)=\dbinom{5}{0}(0.4)^0(0.6)^{5-0}=0.07776$$