A certain large shipment comes with a guarantee that it contains no more than 15% defective items. If the proportion of defective items in the shipment is greater than 15%, the shipment may be returned. You draw a random sample of 10 items. Let X be the number of defective items in the sample.
a. If in fact 15% of the items in the shipment are defective (so that the shipment is good, but just barely), what is P(X ≥ 7)?
b. Based on the answer to part (a), if 15% of the items in the shipment are defective, would 7 defectives in a sample of size 10 be an unusually large number?
c. If you found that 7 of the 10 sample items were defective, would this be convincing evidence that the shipment should be returned? Explain.
d. If in fact 15% of the items in the shipment are defective, what is
P(X ≥ 2)?
e. Based on the answer to part (d), if 15% of the items in the shipment are defective, would 2 defectives in a sample of size 10 be an unusually large number?
f. If you found that 2 of the 10 sample items were defective, would this be convincing evidence that the shipment should be returned? Explain.
Solution 19E
Step1 of 7:
Let us consider a random variable X it presents number of defective items in the sample.
Then X follows binomial distribution with parameters “n and p” that is X B(n, p),
The probability mass function of binomial distribution is given by
, x = 0,1,2,...,n.
Where,
n = sample size
= 10
x = random variable
p = probability of success
= 0.15
q = 1 - p (probability of failure)
= 1 - 0.15
= 0.85
Here our goal is:
a).We need to find , when 15% of the items in the shipment are defective.
b).We need to check whether 7 defectives in a sample of size 10 be an unusually large number.
c).We need to check whether this would be convincing evidence that the shipment should be returned or not if yes explain.
d).We need to find , when 15% of the items in the shipment are defective.
e).We need to check whether 2 defectives in a sample of size 10 be an unusually large number.
f).We need to check whether this would be convincing evidence that the shipment should be returned or not if yes explain.
Step2 of 7:
a).
We have n = 10 and p = 0.15.
Here can be obtained from Excel by using the function “=binomdist(X,n,p,false)”
X |
|
7 |
0.0001259 |
8 |
0.000008332 |
9 |
0.0000003267 |
10 |
0.000000005766 |
Total |
0.0001345 |
Therefore, 0.0001345.
Step3 of 7:
b).
From part (a), we have when 15% of the items in the shipment are defective then
0.0001345.
Now take n = 100000 then [n] = [100000
]
= 13.45
13.
Hence, “Yes”,7 defectives in a sample of size 10 be an unusually large number.
Therefore out of 100000 samples have only 7 or more defective items.
Step4 of 7:
c).
“Yes”, because in part (b) we have seen that 7 defectives in a sample of size 10 be an unusually large number for good shipment
Step5 of 7:
d).
We have n = 10 and p = 0.15.
= 1 -
Here can be obtained from Excel by using the function “=binomdist(X,n,p,false)”
X |
|
0 |
0.1968744043 |
1 |
0.3474254194 |
Total |
0.5442998238 |
Now,
= 1 - 0.5442
= 0.4558
Therefore, 0.4558.
Step6 of 7:
e).
From part (d), we have when 15% of the items in the shipment are defective then
0.4558.
Now take n = 100 then [n] = [100
]
= 45.58
45.
Hence, “No”, 2 defectives in a sample of size 10 be an unusually not large number because 45% of samples of size 10, 2 or more items are defective.
Step7 of 7:
f).
“No”, because in part (e) we have seen that 2 defectives in a sample of size 10 be an unusually not large number for good shipment.
