Question

The Wall Street Journal reported some interesting statistics on the job market. One statistic is that 40% of all workers say they would change jobs for "slightly higher pay." Suppose 20 workers are randomly selected and asked if they would change jobs for "slightly higher pay." Based on the relevant binomial distribution, answer the following question about X, the number of workers who will change jobs for slightly higher pay: What is the probability that X is nine or more? Round your answer to 3 decimal places.

Answer 1

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:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{20}{k} 0.4^{k} 0.6^{20-k}$

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).

$P(x\geq9)=1-P(x$

$P(x=5) = \dbinom{20}{5} p^{5}(1-p)^{15}=15504*0.0102*0.0005=0.0746\\\\\\P(x=6) = \dbinom{20}{6} p^{6}(1-p)^{14}=38760*0.0041*0.0008=0.1244\\\\\\P(x=7) = \dbinom{20}{7} p^{7}(1-p)^{13}=77520*0.0016*0.0013=0.1659\\\\\\P(x=8) = \dbinom{20}{8} p^{8}(1-p)^{12}=125970*0.0007*0.0022=0.1797\\\\\\\\P(x\geq9)=1-(0+0.0005+0.0031+0.0123+0.035+0.0746+0.1244+0.1659+0.1797)\\\\P(x\geq9)=1-0.595=0.405$