Question: Over a period of 6 months, customers were surveyed and the amount of downtime (in minutes) they had experienced during the previous month was collected. The average downtime was found to be 25.09 min with a variance of 144.
(a) Obtain bounds that will include 95 % of the downtime of all the customers.
This is what I came up with so far using the IQR method. 95% of 25.09 is 23.8355.
$23.8355 = (X*Standard Deviation)+Mean = (X * 12)+25.09$
So X = -.1045416 and using the Z-table, I got the bound to be 0.45837.
I really have no idea how to solve this problem. If you could explain and help me with part a, I'm sure I can do part b alone. Thank you.
(b) Obtain the bound above which 10% of the downtime is included.
If you want to use your tables, first you should normalize $X$,
$$ Z = \frac{X - \mu_X}{\sigma_X} ~~~\Leftrightarrow~~~ X = \sigma_X Z + \mu_X \tag{1} $$
Now you can recast the question as: What is the value of $z$ such that the area under the pdf between $-z$ and $+z$ is 0.95? Since the pdf is symmetric around the mean, this is equivalent to find $z$ such that the area between $0$ and $+z$ is 0.475. By looking at this table you find that
$$ z = 1.96 $$
That is, the area under $-1.96 < z < 1.96$ is 0.95. By using Eq. (1) you get
$$ 1.57\;{\rm min} < x < 48.61\;{\rm min} $$
As for the (b) part, you need to repeat pretty much the same exercise, but now considering the tails instead. Since now the tails enclose 10% of the area, then you are looking for $z$ such that the area under the pdf between $-z$ and $z$ is now 0.9
$$ z = 1.64 $$
which leads to
$$ 5.41\;{\rm min} < x < 44.77\;{\rm min} $$