Nt1310 Unit 4 Lab Report

❶ A quality control engineer is in charge of testing whether or not 95% of the Blu-ray disc players produced by his company conform to specifications. To do this, the engineer randomly selects a batch of 13 Blu-ray players from each day’s production. The day’s production is acceptable provided no more than 1 Blu-ray player fails to meet specifications. Otherwise, the entire day’s production has to be tested.
(a) What is the probability that the engineer incorrectly passes a day’s production as acceptable if only 85% of the day’s Blu-ray players actually conform to specification?
From the given information, we can know both problems are n Bernoulli Trials, where n  13 .
Suppose p is probability that the engineer passes a day’s production
13!
(1  0.85)0 (0.85)130 
(1  0.85)1 (0.85)131
0! (13  0)!
1! (13  1)!

=0 . 1 2+
0 09 . 2 7 7
=4
0.39=
839.83%

Thus, the probability that the engineer incorrectly passes a day’s production as acceptable if only 85% of the day’s Blu-ray players actually conform to specification is 39.83%.
(b) What is the probability that the engineer unnecessarily requires the entire day’s production to be tested if in fact 95% of the Blu-ray players conform to specifications?
Suppose p is the probability that the engineer requires the entire day’s production to be tested, and x is the quantity of defectives, then: p  P( x  2)  1  P( x  1)  1  P( x  0)  P( x  1)  1 

13!
13!
(1  0.95) 0 (0.95)130 
(1  0.95)1 (0.95)131
0!(13  0)!
1!(13  1)!

 1 - 0.5133 - 0.3512  0.1355  13.55%

Thus, the probability that the engineer unnecessarily requires the entire day’s production to be tested if in fact 95% of the Blu-ray players conform to specifications is 13.55%.
❷ The reliability of an electrical fuse is the probability that a fuse, chosen at random from production, will function under its designed conditions. A random sample of 1000 fuses
n  1 i 1

The rejection region is t0  t

2

, n 1

, where t

2

, n 1

 t0.025,12  2.17881 .

Since t0  2.26  t

2

, n 1

 2.17881 ,

we reject the null hypothesis H 0 .That is, there is sufficient evidence to support the claim that the pH meter is not correctly calibrated.
2/7

(c) Find the 95% two-sided confidence interval to estimate the mean. Comment on your result.
The 95% two-sided confidence interval is given by: x  t

s
2

, n 1

n

   x  t

s
2

, n 1

n




6.9977  2.17881

0.0442
13

   6.9977  2.17881

0.0442
13

6.971    7.024

Thus, we can be 95% sure that the mean will be between 6.971 and 7.024. In other words, the probability for the mean to be between 6.971 and 7.024 will be 95%.
❺ A quality control supervisor in a cannery knows that the exact amount each can contains will vary, since there are certain uncontrollable factors that affect the amount of fill. Suppose regulatory agencies specify that the standard deviation of the amount of fill should be less than
0.07 ounce. The quality control supervisor sampled 10 cans and measured the amount of fill