1. ## Normal distribution

I am trying to answer question related to normal distribution with Z score but I get confused on my answer. The comparison values seems to be higher than mean or SD and I wonder if this example is correct. Question: America Online advertised job for Senior System Analyst and 800 applications applied for the position. The interview results were found to be normally distributed with mean of 40 marks and a standard deviation of 26 points. How many applicants obtained between 160 and 180 points?America Online advertised job for Senior System Analyst and 800 applications applied for the position. The interview results were found to be normally distributed with mean of 40 marks and a standard deviation of 26 points. How many applicants obtained between 160 and 180 points?
My answer (incomplete though): Z = (χ - μ ) / σ
where:
μ = 40 (mean)
σ = 26 (standard deviation)
For χ = 160
Z160 = (160-40)/26
Z160 = 4.62
For χ = 180
Z180 = (180-40)/26
Z180 = 5.38
P(160< χ<180) = P(4.62<Z<5.38) here is were I got lost on getting right values from table

2. ## Re: Normal distribution

There is a nice table of the standard normal distribution at Public Domain Normal Distribution Table, essentially the table you will see in any text book, however it's not really of much use here!

You appear to have calculated the "standard z scores" but "180" is so much higher than the mean of 40, as you say, almost 5 standard distributions above the mean, that it does not show up on the table. The table only goes up to z= 3.09. However, just below it on the same page is a more accurate table of "Far Right Tail Probabilities". That's what you need here. It does not give exactly your z-values but z= 4.6 gives P= 0.000002112 and z= 5.5 gives "1.899 E-8" or P= 0.00000001899. Subtract those to get the probability that the results will be in that interval. Obviously, because the interval is so far from the mean, the probability will be very, very low!

3. ## Re: Normal distribution

Originally Posted by benedec
P(160< χ<180) = P(4.62<Z<5.38) here is were I got lost on getting right values from table
So far so good. The probability P(160<X<180) is equal to the area under the normal curve between values z = 4.62 and z= 5.38. The table you are referring to priovides a value for the area between z=0 and z=x (this area is called the cumulative distribution function), so the area you need to find is equal to the difference in CDFs for z=5.38 and z=4.62. One issue - not sure how your table works, but many don't provide values for such high values of z, so yuo may need another techniques, such as using Excel's CUMDIST function.

4. ## Re: Normal distribution

Thanks HallsofIvy and ebaines. Suppose that I don't calculate the "standard z scores" and this question appears as below. How could be question be solved, probably I use wrong formula.

Question: America Online advertised job for Senior System Analyst and 800 applications applied for the position. The interview results were found to be normally distributed with mean of 40 marks and a standard deviation of 26 points. How many applicants obtained between 160 and 180 points?America Online advertised job for Senior System Analyst and 800 applications applied for the position. The interview results were found to be normally distributed with mean of 40 marks and a standard deviation of 26 points. How many applicants obtained between 160 and 180 points?

5. ## Re: Normal distribution

Unless you are willing to use some numerical method of integrating the normal distribution, you have to use tables of the normal distribution or (equivalently) a table of the "erf" function. And because there are so many different such distributions they only make tables for the standard normal distributions. To use those, you need to calculate the standard scores. I'm sure there are programs that accept the mean and standard distribution as parameters and a and an "x" value and return the probability. But such a program would itself calculate the correct z-score.

6. ## Re: Normal distribution

I think I should use Public Domain Normal Distribution since I don't understand other methods you mention.