Question

in 1970, 85% of 30 - year - olds in a country earned more than their parents did at age 30 (adjusted for inflation). in 2014, only 45% of 30 - year - olds in the same country earned more than their parents did at age 30. complete parts a to d below. (round to four decimal places as needed.) (b) what is the probability that two randomly selected 30 - year - olds in 1970 earned more than their parents at age 30? the probability is 0.7225. (round to four decimal places as needed.) (c) what is the probability that out of ten randomly selected 30 - year - olds in 1970, at least one did not earn more than their parents at age 30? the probability is 0.8031. (round to four decimal places as needed.) (d) what is the probability that out of ten randomly selected 30 - year - olds in 2014, at least one did not earn more than their parents at age 30? the probability is. (round to four decimal places as needed.)

Explanation:

Step1: Find probability of earning less in 2014

In 2014, the probability that a 30 - year - old earned more than their parents is $p = 0.45$. So the probability that a 30 - year - old did not earn more than their parents is $q=1 - p=1 - 0.45 = 0.55$.

Step2: Use the complement rule

We want to find the probability that out of ten randomly selected 30 - year - olds in 2014, at least one did not earn more than their parents. First, find the probability that all ten earned more than their parents. The probability that all $n = 10$ individuals earned more than their parents is $P(X = 10)=p^{n}=(0.45)^{10}$. Then, using the complement rule $P(X\geq1)=1 - P(X = 0)$. Here, $P(X = 0)=(0.45)^{10}$.

Step3: Calculate the result

$(0.45)^{10}\approx0.000340506$. Then $P(X\geq1)=1-(0.45)^{10}=1 - 0.000340506\approx0.9997$.

Answer:

$0.9997$