In: Statistics and Probability
•An automobile insurer has found that repair claims have a mean of $920 and a standard deviation of $870. Suppose that the next 100 claims can be regarded as a random sample from the long-run claims process.
What is the probability that the average of the 100 claims is larger than $900?
Solution :
Given that,
mean = = 920
standard deviation = = 870
n=100
= =920
= / n = 870/ 100 = 87
P( > 900) = 1 - P( <900 )
= 1 - P[( - ) / < (900-920) /87 ]
= 1 - P(z <-0.23 )
Using z table
= 1 - 0.4090
probability= 0.5910