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