Question

Consider the current in a piece of cable is assumed to have a normal distribution with a mean of 10 mA and a standard deviation of 2 mA

A. What is the probability that the readings measured of a random piece of cable exceeds 13 mA

B. What is the probability that the readings measured of a random piece of cable is between 5.3 mA and 12.8 mA

C. Consider the cable to be not effective if the measured current is more than a certain amount of milliamperes. What will this level be if 14% of the cables are not effective?

Answer 1

Normal distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 10 mA

Standard deviation = 2 mA

a) P(exceeds 13 mA) = P(X > 13)

= 1 - P(X < 13)

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

= 1 - P(Z < 1.5)

= 1 - 0.9332

= 0.0668

b) P(between 5.3 mA and 12.8 mA) = P(X < 12.8) - P(X < 5.3)

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

= P(Z < 1.4) - P(Z < -2.35)

= 0.9192 - 0.0094

= 0.9098

c) Let the level above which a cable is not effective be N

P(X > N) = 0.14

P(X < N) = 1 - 0.14 = 0.86

(N - 10)/2 = 1.08

(Take z value corresponding to 0.86 from standard normal distribution table)

N = 12.16 mA