Question

Tutorial/e Cu 6. The iletime of a lighibulb in a certain applicotion is aoro mally distributed wath menn u 1400 bours nnd slandard deviation σ 200 hours. a. What is the probability that a lightbulb willtast more than 1800 bours? b. lFind the 10 percentile of the lifetines e. A particular lightbulb lasts 1645 hours. What percentile is its lifetime on? d. What is the probability that the lifetime of a lightbulb is between 1 350 and 1550 hours? 7. A fiber spinning process cuently produces a fiber w hose strength is normally distributed with a niean of 75 N/mi. The mnimum acceptable strength is 65 N a. len percent of the fiher produced by the current method fails to meet the mininmum specification. What is the standard deviation of fiber strengths in the current process? fiber will fail to meet the specrfication? fiber will fail to meet the specification? b. If the mean remains at 75 N/mi, what must the standard des aation be so that only 12 of the c. If the standard devration is 5 N/mi, to what value must the mean be set so that only 1 of the

Answer 1

6)

a) Given that the probability that a lightbulb will more than 1800 hours is

1800 - 1400 / 200 = 400 / 200

= 2

using the table , Z- score of 2 is 0.9772

then 1 - 0.9772 = 0.0228

b) Let 10th percentile corresponds to a z- score of -1.28 using the chart table

-1.28 = x - 1400 / 200

= (200)(-1.28) = x -1400

= -256 = x - 1400

then x = 1400 - 250 = 1144

so therefore 10th percentile of the lifetime is 1144 hours

c) Given particular light bulb lasts 1645 hours

then given mean = 1400 hours and standard deviation = 200

then 1645 - 1400 / 200 = 245 / 200 = 1.2250

so therefore the lightbulb is in 1st percentile

d) By using the formula X - mean / standard deviation

=   X - 1400 / 200

then 1350 - 1400 / 200 = - 50 / 200 = -0.25

and 1550 - 1400 / 200 = 150 / 200 = 0.75

now by using table values Z - Score of 0.75 is 0.7734

now by using table values Z - Score of -25 is 0.4013

so therefore 0.7734 - 0.4013 = 0.3721

1645 - 1400