Question

Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 9 min and standard deviation 3 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min? (Round your answer to four decimal places.)

Answer 1

To find the probability that the sample average amount of time taken on each day is at most 11 min, you need to calculate the z-score for a sample mean of 11 min. The probability is approximately 0.7724 or 77.24%.

Step-by-step explanation:

To find the probability that the sample average amount of time taken on each day is at most 11 min, we need to calculate the z-score for a sample mean of 11 min. The formula for calculating the z-score is:

z = (x - μ) / (σ / sqrt(n))

Where z is the z-score, x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the given values:

z = (11 - 9) / (3 / sqrt(5)) = 1 / 1.34 = 0.7463

Using a standard normal distribution table or a calculator, we can find the probability associated with a z-score of 0.7463, which is approximately 0.7724 or 77.24%.