We are given a normally distributed set with `bar(x)=500` and `s=100`

(a) The percentage of scores between 300-500:

Convert 300 and 500 to standard `z` scores:

300: `z=(300-500)/100=-2`

`500: z=(500-500)/100=0` (Naturally!)

The percentage of scores between 300 and 500 is the probability of getting a `z` score between -2 and 0

`P(300<x<500)=P(-2<z<0)` From a standard normal table we find the probability that a `z` score is less than -2 to be .0228, and the probability that a score is less than 0 to be .5 (Naturally!)

So `P(-2<z<0)=.5-.0228=.4772` or **approximately 47.7% of the scores will lie between 300 and 500.**

(Using a graphing calculator I got .4772499385)

(b) The percentile score for 300 is the percentage of scores less than 300.

`P(x<300)=P(z<-2)=.0228` as found in (a), **so the percentile for 300 is approximately the 2nd percentile.**

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