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

Answer:

$P(8.6$

And using the normal standard distirbution or excel we got:

$P(-1$

Step-by-step 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

$P(8.6$

And the z score formula is given by:

$z=\frac{x-\mu}{\sigma}$

Using this formula we got:

$P(8.6$

And using the normal standard distirbution or excel we got:

$P(-1$

Answer 2

Answer:

P = 0.8185

Step-by-step explanation:

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

$z=\frac{x-m}{s}$

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

$z=\frac{8.6-9}{0.4}=-1\\\\z=\frac{9.8-9}{0.4}=2$

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:

$P(8.6$