Question

A. The personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed, with mean μ = 38 minutes and standard deviation σ = 4 minutes. (Round your answers to four decimal places.)

(i) What is the probability that a first interview will last 40 minutes or longer?____________________

(ii) Ten first interviews are usually scheduled per day. What is the probability that the average length of time for the ten interviews will be 40 minutes or longer? _________________

B. Anystate Auto Insurance Company took a random sample of
**352** insurance claims paid out during a 1-year
period. The average claim paid was **$1510**. Assume
**σ = $246**.

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

lower limit $ ___________ upper limit $ ______________

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

lower limit $ ________________ upper limit $ ________________

**C.** Three experiments investigating the relation
between need for cognitive closure and persuasion were performed.
Part of the study involved administering a "need for closure scale"
to a group of students enrolled in an introductory psychology
course. The "need for closure scale" has scores ranging from
**101 to 201**. For the **80 students**
in the highest quartile of the distribution, the mean score was
**x = 177.90.** Assume a population standard deviation
of **σ = 7.89**. These students were all classified as
high on their need for closure. Assume that the 80 students
represent a random sample of all students who are classified as
high on their need for closure. Find a 95% confidence interval for
the population mean score μ on the "need for closure scale" for all
students with a high need for closure. (Round your answers to two
decimal places.)

lower limit _______________

upper limit ______________

Thank you!

Answer #1

A.

i) P(N(38,4)>40)=0.3085375

ii) Define Xi=length of a first interview for the ith person.

Then (X1+X2+..+X10)/10~N(38,SD=4/sqrt(10))

Therefore, P((X1+X2+..+X10)/10>40)=P(N(38,SD= 1.264911)>40)=0.05692315

B. Since population SD is known, we use Z statistic to find CI.

i) lower limit=mean-z(.05)*sigma/sqrt(n)

upper limit=mean+z(.05)*sigma/sqrt(n)

Now z(.05)=1.6448, n=352, then 90% CI is ( 1488.433, 1531.567)

lower limit=S1488.433, upper limit=$11531.567

ii)

lower limit=mean-z(.005)*sigma/sqrt(n)

upper limit=mean+z(.005)*sigma/sqrt(n)

Now z(.005)=2.5758, n=352,

lower limit=$1476.226, Upper limit=$1543.774

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Three experiments investigating the relation between need for
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