Question

13) Resistances of wires are meet to the specification if it between 0.12 & 0.14Ω. The actual measured of wires resistances have a normal distribution with mean 0.13 & standard deviation of 0.005. find:

a) Probability that the selected wires meet specification.

b) Probability that the resistances of the selected wires will exceed the mean of the

resistances by two standard deviation.

c) What are the quartiles of this distribution?

d) The resistance of 90 % of wires is below what value?

e) If four wires are used in the system. What is the probability that all four meet the specification?

f) If four wires are used in the system. What is the probability that less than three meet the specification?

Answer 1

Let X be the actual measured of wires resistance of any given wire. X has a normal distribution with mean & standard deviation of .

a) Probability that the selected wires meet specification.

a) Resistances of wires are meet to the specification if it between 0.12 & 0.14Ω.

The Probability that the selected wires meet specification is same as the probability that any given wire has resistance between 0.12 & 0.14Ω

ans: The Probability that the selected wires meet specification is 0.9544

b) Probability that the resistances of the selected wires will exceed the mean of the resistances by two standard deviation (which is mean+2sd= 0.13+2*0.005=0.14) is

ans: The probability that the resistances of the selected wires will exceed the mean of the resistances by two standard deviation is 0.0228

c) What are the quartiles of this distribution?

25% of the wires have resistance below the first quartile Q1. This is same as the probability that the resistance of a wire is less than Q1 is 0.25

We need

In terms of z scores, we need

P(Z<z) =0.25

However, we know that the area under the standard normal curve to the left of mean (which is 0) is 0.5. Since the required area is less than 0.5, we can say that the z value is -ve

That is

Using the standard normal tables, we can get for z=0.67, P(Z<0.67)=0.75

Hence

P(Z<-0.67) = 0.25

We need

We can equate the z score of Q1 to -0.67 and get

In a normal distribution the second quartile, the median is equal to the mean.

75% of the wires have resistance below the third quartile Q3. This is same as the probability that the resistance of a wire is less than Q2 is 0.55

We need

In terms of z scores, we need

P(Z<z) =0.75

Using the standard normal tables, we can get for z=0.67,

P(Z<0.67)=0.75

We need

We can equate the z score of Q3 to 0.67 and get

ans: The quartiles are

• The first Quartile is 0.1267Ω
• The second quartile is 0.13Ω
• The third quartile is 0.1334Ω

d) The resistance of 90 % of wires is below what value?

Let the resistance of 90 % of wires be below q. This is same as the probability that the resistance of any wire is below q is 0.90

P(X<q)=0.90

In terms of z scores, we need

P(Z<z) =0.90

Using the standard normal tables, we can get for z=1.28,

P(Z<1.28)=0.90

We need

We can equate the z score of q to 1.28 and get

ans: The resistance of 90 % of wires is below 0.1364Ω

e) If four wires are used in the system. What is the probability that all four meet the specification?

From part a) The Probability that any randomly selected wire meets specification is 0.9544

Let X be the number of wires out of 4 that meet the specification. We can say that X has a Binomial distribution with parameters, number of trials (number of wires in the system) n=4 and success probability (The Probability that any randomly selected wire meets specification) p=0.9544

The probability that X=x wires meet the specification is

the probability that all four meet the specification is

ans: the probability that all four meet the specification is 0.8297

f) If four wires are used in the system. What is the probability that less than three meet the specification?

the probability that less than three meet the specification is

ans: the probability that less than three meet the specification is 0.0117