Question

Suppose the curent measurements in a strip of wire follow a normal distribution with a mean
of 10 milliamperes (mA) and a standard deviation of 2 mA.
(a) Find the probability that the measurement on a random strip of wire exceeds 13 mA. (10
pts.)

(b) Find the probability that the measurement on a random strip of wire is between 5.3 mA and
12.8 mA. (10 pts.)

(c) The wire is considered to be defective if its current measurement is more than a certain mA.
What should be this level if 14% of the wires are defective?

Answer 1

We know that:

mean= 10 mA

standard deviation= 2mA

a.

P(wire>13)= 1-P(wire<13)

= 1-P(Z< (13-10)/2)

= 1-P(Z<1.5) [Note: This value can be found from a z distribution table]

= 1 - 0.9332

= 0.0668

b.

P(5.3<wire<12.8)= P(wire<12.8) - P(wire<5.3)

= P(Z<(12.8-10)/2) - P(Z<(5.3-10)/2)

= P(Z<1.3) - P(Z<-2.35) [Note: This value can be found from a z distribution table]

= 0.9032-0.00939

= 0.89381

c.

We know that the wire is defective if it is more than a certain mA. Thus the area under the curve to the left of this mA is 86% or 0.86.

Thus, the z-value of this point is [Note: This value can be found from a z distribution table] is 1.08.

Thus,

=> (x-10)/2 = 1.08

=> x-10= 2.16

=> x= 12.16