The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 2 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?

To solve for the probability proportion, we make use of the z statistic. The procedure to do is to calculate for the z value and using the standard probability tables, we can look up for the p value. The formula for z score is:

z =(x – μ) / (σ / sqrt(n))

where,

x = sample score = 11

μ = sample mean= 10

σ = standard deviation = 2

n = sample size

Calculating for the z and p value when n = 5:

z =(11 – 10) / (2 / sqrt(5))

z = 1.12

Using the tables, p(5) = 0.8686

Calculating for the z and p value when n = 6:

z =(11 – 10) / (2 / sqrt(6))

z = 1.22

Using the tables, p(6) = 0.8888

If both days should be occuring, therefore the total probability that each day is at most 11 min is:

p total = p(5) * p(6)

p total = 0.8686 * 0.8888

p total = 0.772

5050

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