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

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