Hello everyone, I was hoping someone could shed some light on a question I have.
A normal distribution with a mean of 80 and a standard deviation of 14. Determine the value above which 80% of the values will occur.
Ok, so I used the equation z=x-u/q
x=20
u=mean (80)
q= std dev (14)
so, 20-80/14=-4.29
That part is ok, the trouble is the z table I have only goes up to 3. How would one find a z value for 4.29?? The z value for 40 is .4979 or 49.79%. Can I just divide this by 2? Thanks in advance for any advice.
Steve
Last edited by desmoface; 10-07-2008 at 09:35 AM.
In this situation, you need to first look up the z value in the table that marks the separation between the lower 20% and upper 80% of the normal distribution. Then put that z value into the equation and solve for x, which will be "the value above which 80% of the values will occur."
Hi John, that's what I did, I think LOL. If I find the z for 80, it equals 0; 80-80/14=0. So, I figured I'd just use the z for 20% which is; 20-80/14=-4.29
So, thats fine, but unfortunately my Z chart only goes up to 3; internet searches have not yielded any z charts that go high enough to let me find the z value for 4.29. Thanks again for the reply, I'm still scratchin my head over here LOL.
Is there an equation that will give me the z #?
Steve
You plugged in 20 for x, when the "20" represents a percentage, not a value along the x-axis.....you can use an Excel formula =NORMSINV(probability) which gives you the z score for the area under the curve between negative infinity and z
plugging in .20 for probability gives you z = -0.842
so, -0.842 = (x-80)/14
solve for x --> x = (-0.842*14) + 80
x = 68.21
Now, you still need to understand how to use the standard normal table in your text so that you can enter the correct value in the Excel formula.....
Hi John, I'm sure I'm doing something wrong, but that reply did shed some light on the situation LOL. Thanks again.
Steve
Ok, figured it out with the help of my Professor last night in class. Thanks again, John.
Steve
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