Question

The amount of time it takes for a given insurance company to process each insurance claim is normally distributed. To better gauge the processing time, a sample of 16 insurance claims were collected. The sample mean of processing times was 5.4 days and the sample standard deviation was 2.2 days. Based on this sample

a) Find the probability that the standard deviation of the processing time of insurance claims is smaller than 3 days.

b) Find the probability that the average processing time of insurance claims is larger than 6 days.

c) Build a 90% confidence interval for the average processing times of insurance claims through the following steps:

(i) Specify the pivotal statistic and its sampling distribution

(ii) Construct the confidence interval

(Slove both parts in c)!!

Answer 1

Please keep the z tables ready

given that

mean = 5.4 , sd = 2.2 and n = 16

we know that the z score is given as

Z = (x-mean)/(sd/sqrt(n)

putting the values

a) for smaller than 3 days

z = (3 - 5.4)/(2.2/sqrt(16)) = -4.36
P ( Z<?4.36 )=1?P ( Z<4.36 )=1?1=0
The probability that Z<?4.36 is equal to the blue area under the curve.

b) for 6 days

z = (6 - 5.4)/(2.2/sqrt(16)) = 1.09

P ( Z>1.09 )=1?P ( Z<1.09 )=1?0.8621=0.1379

The probability that Z>1.09 is equal to the blue area under the curve.

c)

we know that the confience interval is given as

mean +- z*sd/sqrt(n)

here

z = 1.645 , from the z table for 90%

putting the values and solving for plus and minus sign we get

5.4 +- 1.645*2.2/sqrt(16)

= 4.49 and 6.3

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