Sorry...i wrote the previous questions wrong..
The producer of HB pencils has 720 pencils randomly chosen from each day's productions. A defect rate of 15% is acceptable.
a) Assuming that 15% of all manufactured pencils are defective, what is the probability of finding at least 80 defective pencils in this sample?
b) If 120 pencils are found to be defective in today's sample, is it likely that the manufacturing process may need improvements? Explain....
If you can just give me the answer..you don't have to show me how, i Just want to see if i'm right...Thanks..
2 answers
If you tell us your answers, we'll try to check them.
OK, I will try but I am no expert and have not looked at this material for years.
a)80/720 = .111... = 11.1%
What is the probability of finding more than 11.1% defective?
sigma , the standard deviation of p is
sqrt(.15*.85)/sqrt(720)
= sqrt(.1275)/sqrt(720)
=.35355/26.83
sigma =.01318
so we expect about 68% of 720 pencil samples to fall within .15 +/- .01318
now .1111 or 80 out of 720 is
.15 -.1111 or .0389 from mean
which is
.0389/.01318 = 2.95 sigma from mean
that is unlikely but how unlikely?
from my normal distribution table .9984 of samples are left of 2.95 sigma so
1-.9984 = .0016 are out on the tail beyond
and .9984 is the probability of finding at least 80 if the mean is .15*720 = 108
b) now 120/720 = .1667 where the mean was .15 and sigma was .01318
(.1667 -.15)/.01318 = 1.267 sigmas above mean failure rate
about .897 is my table result for fraction of sample results with less than 120 failures
So the probability of getting 120 or more today is 1 - .897 = .103
or about ten percent
That sounds bad until you think about running the plant for 100 days
You can expect to measure 120 or more defects out of your 720 pencil sample about ten days out of 100.
If improving the process is expensive, it is might not be worth it.
a)80/720 = .111... = 11.1%
What is the probability of finding more than 11.1% defective?
sigma , the standard deviation of p is
sqrt(.15*.85)/sqrt(720)
= sqrt(.1275)/sqrt(720)
=.35355/26.83
sigma =.01318
so we expect about 68% of 720 pencil samples to fall within .15 +/- .01318
now .1111 or 80 out of 720 is
.15 -.1111 or .0389 from mean
which is
.0389/.01318 = 2.95 sigma from mean
that is unlikely but how unlikely?
from my normal distribution table .9984 of samples are left of 2.95 sigma so
1-.9984 = .0016 are out on the tail beyond
and .9984 is the probability of finding at least 80 if the mean is .15*720 = 108
b) now 120/720 = .1667 where the mean was .15 and sigma was .01318
(.1667 -.15)/.01318 = 1.267 sigmas above mean failure rate
about .897 is my table result for fraction of sample results with less than 120 failures
So the probability of getting 120 or more today is 1 - .897 = .103
or about ten percent
That sounds bad until you think about running the plant for 100 days
You can expect to measure 120 or more defects out of your 720 pencil sample about ten days out of 100.
If improving the process is expensive, it is might not be worth it.
0 answers