Step 1 :The problem is asking for the probability that less than 7.1% of a sample of 225 employed adults hold multiple jobs, given that 13% of employed adults in the United States hold multiple jobs. This is a binomial problem, but since the sample size is large, we can use the normal approximation to the binomial distribution.
Step 2 :The mean of the distribution is np, where n is the sample size and p is the probability of success (in this case, holding multiple jobs). So, \(mean = np = 225 * 0.13 = 29.25\)
Step 3 :The standard deviation is \(\sqrt{np(1-p)}\). So, \(std\_dev = \sqrt{225 * 0.13 * (1 - 0.13)} = 5.044551516240071\)
Step 4 :We want to find the probability that the proportion is less than 7.1%, or 0.071. To do this, we can standardize the value we're interested in (0.071) by subtracting the mean and dividing by the standard deviation to get a z-score. So, \(z = \frac{0.071 * 225 - mean}{std\_dev} = -2.6315520730164828\)
Step 5 :Then we can use a z-table or a function like scipy's norm.cdf to find the probability that a standard normal random variable is less than this z-score. So, \(probability = 0.004249792038589428\)
Step 6 :Final Answer: The probability that less than 7.1% of the individuals in this sample hold multiple jobs is \(\boxed{0.0042}\)