Weights of parts are normally distributed with variance about the process mean.

(a) Without measurement error, what is the probability that a part exceeds the specifications?

(b) With measurement error, what is the probability that a part is measured as being beyond specifications? Does this imply it is truly beyond specifications?

(c) What is the probability that a part is measured as being beyond specifications if the true weight of the part is below the upper specification limit?

Step 1 of 3:

Let the random variable are mutually distributed.

Let X denotes the weight of the part with N() and

Let Y denotes the measurement error with N(0, 0.5)

Our goal is:

a). We need to find the probability that part exceeds the specification.

b). We need to find the probability the probability that part is measured beyond specification.

c). We need to find the probability that a part is measured as being beyond specifications if the true weight of the part is 1 below the upper specification limit.

a). Without measurement error, the probability that part exceeds the specification is

P

P

P

P

Using area under the normal curve,

P

P

P

Therefore, the probability that part exceeds the specification is 0.0026.

Step 2 of 3:

b). The weight of the measured part with error is denoted by T.

T=X+Y.

Now we are computing E(T) and V(T).

E(T) = E(X+Y)

E(T) = E(X)+E(Y)

E(T) =

Then V(T) is

V(T) = V(X+Y)

V(T) =

V(T) = 1.5

A linear combination of 2 independent normal random variables is is also normal.

With the measurement error , the probability that part is measured beyond specification is given by

PP

We assume that Z = .

P

P

P

2P

Using areas under the normal curve table,

2(1-0.9929)

2(0.007)

0.0142

P0.0142

Therefore, the probability that part is measured beyond specification is 0.0142.

This does not empty that part is truly beyond specification because the true weight X could be within the specification limit.