Final answer:
The question involves calculating the probability distribution for a sample proportion based on a given population proportion using the normal approximation method, where the probabilities are found by determining Z-scores and consulting the standard normal distribution.
Step-by-step explanation:
The scenario described involves using normal distribution to approximate the sampling distribution of the sample proportion. The population proportion is given as 51% (0.51) for Americans having less than $350 in the bank for an emergency. We want to determine the probability that the sample proportion (o) is within 6% of this value in a sample of 73 Americans and the probability that it is greater by 10% or more.
To calculate these probabilities, we use the formula for the standard error of the sample proportion SE = sqrt[(p(1-p))/n] and the Z-score formula Z = (o-p)/SE. Assuming the Central Limit Theorem applies and the sample size is large enough, we use the standard normal distribution to find the probabilities corresponding to the calculated Z-scores.
To calculate the probability of being within 6%, we find the Z-scores corresponding to proportions 0.57 and 0.45. For the probability of being greater than the population proportion by 10% or more (which is 61%), we find the Z-score for 0.61 and use the standard normal distribution to find the required probabilities. We round our final answers to four decimal places as requested.