A). Anystate Auto Insurance Company took a random sample of 358 insurance claims

Question

A). Anystate Auto Insurance Company took a random sample of 358 insurance claims paid out during a 1-year period. The average claim paid was $1555. Assume = $246.

Find a 0.90 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

Find a 0.99 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

$

B). Find z such that 19% of the area under the standard normal curve lies to the right of z. (Round your answer to two decimal places.)

C). Find z such that 68% of the standard normal curve lies between z and z. (Round your answer to two decimal places.)

D). Given that x is a normal variable with mean = 42 and standard deviation = 6.5, find the following probabilities. (Round your answers to four decimal places.)

(a)  P(x 60)


(b)  P(x 50)


(c)  P(50 x 60)

Explanation / Answer

a) (mean – z* /sqrt(n) , mean + z* /sqrt(n))

for 90 % confidence interval

z = 1.645

(1555 – 1.645 * 246/ sqrt(358)), 1555 + 1.645 * 246/ sqrt(358)))

= (1533.612525 , 1576.3874)

For 99 % , z = 2.576

(1555 – 2.576 * 246/ sqrt(358)), 1555 + 2.576 * 246/ sqrt(358)))

(1521.50812, 1588.49187)

b)

P(Z >z*) =0.19

Z* = 0.878

c)

P(-z < Z< z) = 0.68

Z = 0.994

d)

Z = (Z – 42)/6.5

P(X< 60) =P (Z<2.77)=0.9972

P(X > 50) =P (Z>1.23)=0.1093

P(50 <X<60)

= P ( 1.23<Z<2.77 )=0.1065

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