Question

brainly A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5.5 years, and standard deviation of 1.2 years. If 24 items are picked at random, 8% of the time their mean life will be less than how many years?

Answer 1

Answer:

$z=-1.405$

And if we solve for a we got

$a=5.5 -1.405*1.2=3.814$

So the value of interest that separates the bottom 8% of data from the top 72% is 69.764.  

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 variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.5,1.2)$  

Where $\mu=5.5$ and $\sigma=1.2$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.92$   (a)

$P(X$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.08 of the area on the left and 0.92 of the area on the right it's z=-1.405. On this case P(Z<-1.405)=0.08 and P(z>-1.405)=0.5

If we use condition (b) from previous we have this:

$P(X$  

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-1.405$

And if we solve for a we got

$a=5.5 -1.405*1.2=3.814$

So the value of height that separates the bottom 8% of data from the top 72% is 69.764.