Question

The resistance for resistors of a certain type is a random variable Xhaving the normal distribution with mean 9 ohms and standard devia-tion 0.4 ohms. A resistor is acceptable if its resistance is between 8.6and 9.8 ohms.(a) What is the probability that a randomly chosen resistor is accept-able

Answer 1

P = 0.8185

Explanation:

First we need to standardize the values 8.6 and 9.8 ohms using the following equation:

`z=(x-m)/(s)`

Where m is the mean and s is the standard deviation, so 8.6 and 9.8 are equivalent to:

`z=(8.6-9)/(0.4)=-1\\\\z=(9.8-9)/(0.4)=2`

Finally, using the normal distribution table, we can calculated the probability that a resistor has a resistance between 8.6 and 9.8 ohms as:

`P(8.6<z<9.8)=P(-1<z<2)\\P(-1<z<2)=P(z<2)-P(z<-1)\\P(-1<z<2)=0.9772-0.1587\\P(-1<z<2)=0.8185`

Answer 2

`P(8.6<X<9.8)=P((8.6-\mu)/(\sigma)<(X-\mu)/(\sigma)<(9.8-\mu)/(\sigma))=P((8.6-9)/(0.4)<Z<(9.8-9)/(0.4))=P(-1<z<2)`

And using the normal standard distirbution or excel we got:

`P(-1<z<2)=P(z<2)-P(z<-1)= 0.97725-0.158655 = 0.7907`

Explanation:

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

`X \sim N(9,0.4)`

Where
`\mu=9` and
`\sigma=0.4`

We are interested on this probability

`P(8.6<X<9.8)`

And the z score formula is given by:

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

Using this formula we got:

`P(8.6<X<9.8)=P((8.6-\mu)/(\sigma)<(X-\mu)/(\sigma)<(9.8-\mu)/(\sigma))=P((8.6-9)/(0.4)<Z<(9.8-9)/(0.4))=P(-1<z<2)`

And using the normal standard distirbution or excel we got:

`P(-1<z<2)=P(z<2)-P(z<-1)= 0.97725-0.158655 = 0.7907`