Respuesta :
Answer:
The probability that X is nine or more is 0.405.
Step-by-step explanation:
We have a sample of 20 random selected workers, each with a probability p=0.4 of changing jobs for "slightly higher pay".
We have a random variable X that can be modeled by the binomial distribution, with parameters n=20 (sample size) and p=0.4 (probability of success).
The probability that k workers will change jobs for slightly higher pay can be calcualted as:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{20}{k} 0.4^{k} 0.6^{20-k}\\\\\\[/tex]
We have to calculate the probability that the number of workers in the sample who will change jobs for slightly higher pay is 9 or higher. That is P(X≥9).
[tex]P(x\geq9)=1-P(x<9)=1-\sum_{i=0}^{8}P(x=i)\\\\\\P(x=0) = \dbinom{20}{0} p^{0}(1-p)^{20}=1*1*0=0\\\\\\P(x=1) = \dbinom{20}{1} p^{1}(1-p)^{19}=20*0.4*0.0001=0.0005\\\\\\P(x=2) = \dbinom{20}{2} p^{2}(1-p)^{18}=190*0.16*0.0001=0.0031\\\\\\P(x=3) = \dbinom{20}{3} p^{3}(1-p)^{17}=1140*0.064*0.0002=0.0123\\\\\\P(x=4) = \dbinom{20}{4} p^{4}(1-p)^{16}=4845*0.0256*0.0003=0.035\\\\\\[/tex]
[tex]P(x=5) = \dbinom{20}{5} p^{5}(1-p)^{15}=15504*0.0102*0.0005=0.0746\\\\\\P(x=6) = \dbinom{20}{6} p^{6}(1-p)^{14}=38760*0.0041*0.0008=0.1244\\\\\\P(x=7) = \dbinom{20}{7} p^{7}(1-p)^{13}=77520*0.0016*0.0013=0.1659\\\\\\P(x=8) = \dbinom{20}{8} p^{8}(1-p)^{12}=125970*0.0007*0.0022=0.1797\\\\\\\\P(x\geq9)=1-(0+0.0005+0.0031+0.0123+0.035+0.0746+0.1244+0.1659+0.1797)\\\\P(x\geq9)=1-0.595=0.405[/tex]
