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 σ = $246.

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

lower limit	$
upper limit	$

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

lower limit	$
upper limit	$

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 μ = 42 and standard deviation σ = 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)

Answer 1

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

Please don't forget to rate positively if you found this response helpful.
Feel free to comment on the answer if some part is not clear or you would like to be elaborated upon.
Thanks and have a good day!