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# Algebra 2

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Suppose the scores on a history quiz are normally distributed with a mean of 73 and a standard deviation of 4. Which group describes 16% of the population of history quiz scores ?

Scores below 65

scores below 75

scores above 77

scores above 81

Apr 9, 2019

#1
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quick rule of thumb

16% of the population is a z-score of -1, i.e. 1 sigma less than the mean.

$$\dfrac{x - \mu}{\sigma} = -1 \Rightarrow x = \mu - \sigma$$

So we would have scores less than 73-4 = 69 being 16%

This doesn't match any of the choices so we note that it could be the top 16% in which case the z-score is 1 and we have

scores greater than x = 73+4 = 77 being 16%

this matches the 3rd choice.

Apr 9, 2019
#2
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Look at your Negative z-score table .....look for a number close to 16% (.16)

I find   .99 z score is .1611

.99 x  standard deviation = .99 x 4 ~~ 4           73 - 4 = 69

This question was submitted earlier ....it has an error in it somewhere

If we use  1.6%    look for number    .016 -ish      z -score -2.14

73  - 2.14 x 4 = 64.44     so I would go with the answer    'Scores below 65'     if it really is 1.6%    and not 16%

Or as Rom points out ....maybe it is the opposite end of the spectrum...the TOP 16% (and we were looking at it from the wrong perspective)

100% -16% = 84     look at positive z-score table for   .84 -ish  approx 1.0 standard deviations ABOVE the mean

73 + 1 Standard deviation = 73+4 = 77 --------> choice #3

Thanx Rom!

Apr 9, 2019
edited by ElectricPavlov  Apr 9, 2019