Respuesta :
Answer:
a) The probability that the inspection procedure will pass the shipment is 0.918540.
b) The expected number of defectives in this process of inspecting 5 items is 0.5.
c) The probability that you will find 4 defectives in a sample of 5 is 0.0064.
Step-by-step explanation:
Given : A company is interested in evaluating its current inspection procedure on large shipments of identical items. Te procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall.
To find :
a. What is the probability that the inspection procedure will pass the shipment?
b. What is the expected number of defectives in this process of inspecting 5 items?
c. If items are defective at a 20% rate overall, what is the probability that you will find 4 defectives in a sample of 5?
Solution :
Applying binomial distribution,
The probability of success p=10%=0.1
Number of items n=5
a) The probability that the inspection procedure will pass the shipment i.e. [tex]P(X\leq 1)[/tex]
[tex]P(X\leq 1)=P(0\leq X\leq 1)=P(X=0,1)[/tex]
So, [tex]P(X\leq 1)=P(X=0)+P(X=1)[/tex]
[tex]P(X\leq 1)=^5C_0(0.1)^0(1-0.1)^5+^5C_1(0.1)^1(1-0.1)^4[/tex]
[tex]P(X\leq 1)=1(1)(0.9)^5+5(0.1)(0.9)^4[/tex]
[tex]P(X\leq 1)=0.59049+0.32805[/tex]
[tex]P(X\leq 1)=0.918540[/tex]
The probability that the inspection procedure will pass the shipment is 0.918540.
b) The expected number of defectives in this process of inspecting 5 items i.e. E(X)
The mean E(X) is defined as
[tex]E(X)=n\times p[/tex]
[tex]E(X)=5\times 0.1[/tex]
[tex]E(X)=0.5[/tex]
The expected number of defectives in this process of inspecting 5 items is 0.5.
c) The probability that you will find 4 defectives in a sample of 5 i.e P(X=4)
Here, P=20%=0.2
[tex]P(X=4)=^5C_4(0.2)^4(1-0.2)^1[/tex]
[tex]P(X=4)=5(0.2)^4(0.8)^1[/tex]
[tex]P(X=4)=0.0064[/tex]
The probability that you will find 4 defectives in a sample of 5 is 0.0064.
