+87 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
+6046 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.
+19912 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!
