Question

AMY HAYDEN Question 10 of 13, Step 1 of 1 9/13 Correct The time spent waiting in the line is approximately normally distributed. The mean waiting time is 6 minutes and the variance of the waiting time is 9. Find the probability that a person will wait for more than 9 minutes. Round your answer to four decimal places Answer Tables Keypad If you would like to look up the value in a table, select the table you want to view, then either click the cell at the intersection of the row and column or use the arrow keys to find the appropriate cell in the table and select it using the Space key. Normal Table -- to -- Normal Table -- to

Answer 1

ANSWER:

Given that,

The time spent waiting in the line is approximately normally distributed.The mean waiting time is 6 minutes and the varience of the waiting time is 9.Find the probability that a person will wait for more than 9 minutes.Round your answer to four decimal places.

Mean = $\mu$ = 6

Varience = $\sigma ^{2}$ = 9

Standard deviation =$\sigma$ = sqrt(Varience) = sqrt(9) = 3

We know that,

z = (x-$\mu$) / $\sigma$ = (x-6)/3

P(x>9) = P((x-$\mu$) / $\sigma$ > (9-6)/3)

P(x>9) = P(z > 3/3)

P(x>9) = P(z > 1)

P(x>9) = 1-P(z <1)

P(x>9) = 1-0.84134

P(x>9) = 0.15866

P(x>9) = 0.1587 (Rounded to four decimal places)

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