Question

Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean 0.12 ohm and standard deviation 0.009 ohm.

a) What is the probability that a randomly selected wire from company A's production will meet the specifications?
b) If fourof these wires are used in each computer system and all are selected from company A, what is the probability that all four in a randomly selected system will meet the specifications?

Answer 1

a) 73.30% probability that a randomly selected wire from company A's production will meet the specifications

b) 28.87% probability that all four in a randomly selected system will meet the specifications

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 0.12, \sigma = 0.009`

a) What is the probability that a randomly selected wire from company A's production will meet the specifications?

This is the pvalue of when X = 0.13 subtracted by the pvalue of Z when X = 0.11. So

X = 13

`Z = (X - \mu)/(\sigma)`

`Z = (0.13 - 0.12)/(0.009)`

`Z = 1.11`

`Z = 1.11` has a pvalue of 0.8665

X = 11

`Z = (X - \mu)/(\sigma)`

`Z = (0.11 - 0.12)/(0.009)`

`Z = -1.11`

`Z = -1.11` has a pvalue of 0.1335

0.8665 - 0.1335 = 0.7330

73.30% probability that a randomly selected wire from company A's production will meet the specifications

b) If four of these wires are used in each computer system and all are selected from company A, what is the probability that all four in a randomly selected system will meet the specifications?

For each of them, 0.7330 probability

(0.7330)^4 = 0.2887

28.87% probability that all four in a randomly selected system will meet the specifications

Answer 2

a)
`P(0.11<X<0.13)=P((0.11-\mu)/(\sigma)<(X-\mu)/(\sigma)<(0.13-\mu)/(\sigma))=P((0.11-0.12)/(0.009)<Z<(0.13-0.12)/(0.009))=P(-1.11<z<1.11)`

And we can find this probability with this difference and with the normal standard table or excel:

`P(-1.11<z<1.11)=P(z<1.11)-P(z<-1.11)=0.8665-0.1335= 0.733`

b)
`P(0.11 < \bar X < 0.13)`

And we can use the z score defined by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And using the limits we got:

`z = (0.11-0.12)/((0.009)/(√(4)))= -2.22`

`z = (0.13-0.12)/((0.009)/(√(4)))= 2.22`

And we want to find this probability:

`P(-2.22< Z<2.22) = P(Z<2.22) -P(Z<-2.22) = 0.9868 -0.0132= 0.9736`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

`X \sim N(0.12,0.009)`

Where
`\mu=0.12` and
`\sigma=0.009`

We are interested on this probability

`P(0.11<X<0.13)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(0.11<X<0.13)=P((0.11-\mu)/(\sigma)<(X-\mu)/(\sigma)<(0.13-\mu)/(\sigma))=P((0.11-0.12)/(0.009)<Z<(0.13-0.12)/(0.009))=P(-1.11<z<1.11)`

And we can find this probability with this difference and with the normal standard table or excel:

`P(-1.11<z<1.11)=P(z<1.11)-P(z<-1.11)=0.8665-0.1335= 0.733`

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And we want this probability:

`P(0.11 < \bar X < 0.13)`

And we can use the z score defined by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And using the limits we got:

`z = (0.11-0.12)/((0.009)/(√(4)))= -2.22`

`z = (0.13-0.12)/((0.009)/(√(4)))= 2.22`

And we want to find this probability:

`P(-2.22< Z<2.22) = P(Z<2.22) -P(Z<-2.22) = 0.9868 -0.0132= 0.9736`