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.
Answer:
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%
