Nicole missed the lesson on normal distribution and needs to do her homework. Explain to Nicole how to use the mean and standard deviation of a normal distribution to determine the top 5% of the population.
@ganeshie8
@amistre64
@mathstudent55
The starting data value for top 5% of the population can be found by using the Zscore formula \[Z=\dfrac{x-\mu}{\sigma}\] \(\mu, \sigma\) values are given and the \(Z\) value can be found in the Ztable. Look at the z-value corresponding to 95% area, which is 1.65. Therefore the equation to be solved for \(x\) is \[1.65=\dfrac{x-\mu}{\sigma}\]
familiar with z table right ?
And that's all you need? That's pretty simple. And yes I am, my teacher sent out a document of each different percentage on it.
good, so do you see 1.65 for 95% in it ?
yes thats all
wait i'm thinking about the confidence intervals. i'm looking at the z table now but not sure what i'm looking for
see the highlighted area number : `0.9505` http://gyazo.com/7323894fdf4d7217aefd37518f8a6bbb whats the z score corresponding to that area ?
the 1.6?
or 1.65 considering 0.5 at top?
Yes! so the probability for getting a zscore of 1.65 is `0.9505` which is almost same as `95%` therefore, the top 5% data values will have zscores greater than or equal to 1.65
maybe lets work a quick example
okay
how'd you decide to look for the 95% in the question?
so if it was the top 6% you would look at .9406?
that table gives the "area to the left" of a given zscore so if you want the starting zscore of "top 5%", you need to find the zscore that gives the "95%" area
Absolutely! you're correct. For determining the starting zscore of top 6% of population, look at the table and find the zscore that gives the 94% area.
looks you would get zscore = 1.56 for top 6% ?
I have 1 more question if you can help on?
@rational
Was this right??
Join our real-time social learning platform and learn together with your friends!