a: top 9% of scores b: scores below the top 9 and above the bottom 64% c: scores below the top 36% and above the bottom 21% d: scores below the top 79% and above the bottom 7% f: bottom 7 % mean 70.4 standard deviation of 7.3 find the oimits for a c

Assuming a normal distribution with

mean = 70.4

standard deviation = 7.3

(a) top 9% of scores

The top .09 scores correspond to the 1 - 0.09 = 91 percentile

the z-score for this value is 1.34 (from table of values)

calculating limit...

1.34 * 7.3 + 70.4 = 80.182

So, the top 9% of scores correspond to any value x such that

x > 80.182

(c) scores below the top 36% and above the bottom 21%

The top .36 scores correspond to the 1 - 0.36 = 64 percentile

the z-score for this value is 0.36 (from table of values)

0.36 * 7.3 + 70.4 = 73.028

The bottom .21 scores correspond to the 21 percentile

the z-score for this value is -0.81 (from table of values)

-0.81 * 7.3 + 70.4 = 64.487

So, the data range corresponds to any value x such that

64.487 < x < 73.028

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