7) Suppose that the amount of time it takes to process insurance claims is normally distributed with a mean of 12 weeks and a variance of 9 weeks. What is the probability that the next claim will be processed within

This question is incomplete, the complete question is;

Suppose that the amount of time it takes to process insurance claims is normally distributed with a mean of 12 weeks and a variance of 9 weeks. What is the probability that the next claim will be processed within; a) 10 weeks, b) 11 weeks, c) 12 weeks.

a) the probability that the next claim will be processed within 10 weeks is 25.14%

b) the probability that the next claim will be processed within 11 weeks is 37.07%

c) the probability that the next claim will be processed within 12 weeks is 50.00%

Step-by-step explanation:

Given the data in the question;

mean 12 weeks

Variance = 9 weeks

Standard deviation SD = √Variance = √ 9 = 3 weeks.

Since,  the amount of time it takes to process insurance claims is normally distributed.

z = ( x - mean ) / SD

where x is weeks of next claim.

a) 10 weeks

z = ( x - mean ) / SD

we substitute

z = ( 10 - 12 ) / 3

z = -2/3

z = -0.67

{ from standard normal distribution table }

value of z is 0.2514

Therefore, the probability that the next claim will be processed within 10 weeks is 25.14%

b) 11 weeks

z = ( x - mean ) / SD

we substitute

z = ( 11 - 12 ) / 3

z = -1/3

z = -0.33

{ from standard normal distribution table }

value of z is 0.3707

Therefore, the probability that the next claim will be processed within 11 weeks is 37.07%

c) 12 weeks

z = ( x - mean ) / SD

we substitute

z = ( 12 - 12 ) / 3

z = 0/3

z = 0.00

{ from standard normal distribution table }

value of z is 0.5000

Therefore, the probability that the next claim will be processed within 12 weeks is 50.00%

answer: i think it's a

step-by-step explanation: