Question

The lifetime of a certain type of TV tube has a normal distribution with a mean of 61 and a standard deviation of 6 months. What portion of the tubes lasts between 57 and 59 months?

Answer 1

Answer:

$P(57$

$P(-0.67$

$P(-0.67$

Step-by-step 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".  

Solution to the problem

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

$X \sim N(61,6)$  

Where $\mu=61$ and $\sigma=6$

We are interested on this probability

$P(57$

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

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

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

$P(57$

And we can find this probability like this:

$P(-0.67$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-0.67$