Question #267184

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

1
Expert's answer
2021-11-17T14:41:30-0500

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

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


P(X3)=P(X=3)+P(X=4)+P(X=5)P(X\geq3)=P(X=3)+P(X=4)+P(X=5)

=(53)(0.8)3(0.2)53+(54)(0.8)4(0.2)54=\dbinom{5}{3}(0.8)^3(0.2)^{5-3}+\dbinom{5}{4}(0.8)^4(0.2)^{5-4}

+(55)(0.8)5(0.2)55=0.2048+0.4096+0.32768+\dbinom{5}{5}(0.8)^5(0.2)^{5-5}=0.2048+0.4096+0.32768

=0.94208=0.94208

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

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

A.


P(X=2)=(52)(0.4)2(0.6)52=0.3456P(X=2)=\dbinom{5}{2}(0.4)^2(0.6)^{5-2}=0.3456

B.


P(X3)=1P(X=4)P(X=5)P(X\leq3)=1-P(X=4)-P(X=5)

=1(54)(0.4)4(0.6)54+(55)(0.4)5(0.6)55=1-\dbinom{5}{4}(0.4)^4(0.6)^{5-4}+\dbinom{5}{5}(0.4)^5(0.6)^{5-5}

=10.07680.01024=0.91296=1-0.0768-0.01024=0.91296

C.


P(X2)=1P(X=0)P(X=1)P(X\geq2)=1-P(X=0)-P(X=1)

=1(50)(0.4)0(0.6)50+(51)(0.4)1(0.6)51=1-\dbinom{5}{0}(0.4)^0(0.6)^{5-0}+\dbinom{5}{1}(0.4)^1(0.6)^{5-1}

=10.077760.2592=0.66304=1-0.07776-0.2592=0.66304

D.



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

=(50)(0.4)0(0.6)50+(51)(0.4)1(0.6)51=\dbinom{5}{0}(0.4)^0(0.6)^{5-0}+\dbinom{5}{1}(0.4)^1(0.6)^{5-1}

+(52)(0.4)2(0.6)52+\dbinom{5}{2}(0.4)^2(0.6)^{5-2}

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

E.


P(X=0)=(50)(0.4)0(0.6)50=0.07776P(X=0)=\dbinom{5}{0}(0.4)^0(0.6)^{5-0}=0.07776

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