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svlad2 [7]
3 years ago
9

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independ

ent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is . What is the probability that at least one of successive points indicates a problem when in fact the process is operating correctly?
Mathematics
1 answer:
Illusion [34]3 years ago
7 0

Question has missing details

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is 0.05. What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly?

Answer:

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

Step-by-step explanation:

Given

Let P = Probability that a point signals an error incorrectly = 0.05

Let Q = Probability that a point signals an error correctly

P + Q = 1 ---- Make Q the subject of formula

Q = 1 - P where P = 0.05

So, Q = 1 - P becomes

Q = 1 - 0.05

Q= 0.95

Solving for the probability that at least one of 10 successive points indicates a problem when in fact the process is operating.

If two events (P and Q) are independent

Then

P(P n Q) = P(P) * P(Q)

From De Morgan law;

P(P u Q) = 1 - P(P' n Q')

Where P(P u Q) represent the probability that at least one of 10 successive points

P(P' n Q') is calculated as follows;

P(P' n Q') = 0.95^10

P(P' n Q') = 0.59873693923837890625

So,

P(P u Q) = 1 - P(P' n Q') becomes

P(P u Q) = 1 - 0.59873693923837890625

P(P u Q) = 0.40126306076162109375

P(P u Q) = 0.4013 ----- Approximated

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

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