Estadistica I

Estadística I

Problems chapters 5-7
14/12/2011 Surname, name: Final qualification: Each question correctly answered 10 points

1.

The times between failures on a personal computer follow anexponential distribution with a mean of 300,000 hours. What is the probability of: a. A failure in less than 100,00 hours? b. No failure in the next 500,00 hours? c. The next failure occurring between200,00 and 350,00 hours? d. What are the mean and standard deviation of the time between failures? 73. a. b. c. d. 0.2835, found by 1 – e(–1/300,000)(100,000) 0.1889, found by e(–1/300,000)(500,000)0.2020, found by e(–1/300,000)(200,000) – e(–1/300,000)(350,000) Both the mean and standard deviation are 300,000 hours. (LO 8)

2.

The annual sales of romance novels follow the normaldistribution. However, the mean and the standard deviation are unknown. Forty percent of the time sales are more than 470,00 and 10 percent of the time sales are more than 500,00. What are the mean and thestandard deviation?
470  

67.



 0.25

500  



 1.28

 =29,126 and  =462,718 (LO 5)

3.

The Orange County Register, as part of its Sunday health supplement, reported that64 percent of American men over the age of 18 consider nutrition a top priority in their lives. Suppose we select a sample of 60 men. What is the likelihood that: a. 32 or more consider nutritionimportant? b. 44 or more consider nutrition important? c. More than 32 but fewer than 43 consider nutrition important? d. Exactly 44 consider diet important?

1

Estadística I

Problems chapters5-7
14/12/2011

57.

a.

b. c. d.

0.9678, found by:  =60(0.64) = 38.4   13.824  3.72  2 = 60(0.64)(0.36) = 13.824 Then (31.5 – 38.4)/3.72 = – 1.85, for which the area is 0.4678. Then0.5000 + 0.4678 = 0.9678 0.0853, found by (43.5 – 38.4)/3.72 = 1.37, for which the area is 0.4147. Then 0.5000 – 0.4147 = 0.0853 0.8084, found by 0.4441 + 0.3643 0.0348 found by 0.4495 – 0.4147 (LO 7)...