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(a) First we convert 10 and 12 to standard `z` scores using `z=(x-bar(x))/s` where `z` is the standard normal score, x is the score, `bar(x)` is the mean, and `s` the standard deviation.
`z_10=(10-10)/4=0` and `z_12=(12-10)/4=.5`
**Note that the `z` score indicates how many standard deviations thescore lies from the mean; it should not be a surprise that `z_10=0` since 10 is the mean. A positive `z` indicates it is abovethe mean, a negative below the mean.
The standard normal table gives the area under the curve below a given `z` score. Thus the area between `z_10` and `z_12` is the area below `z_12` minus the area below `z_10` . Using the table or a graphing calculator we get approximately .6915-.5=.1915
Thus the area between a score of 10 and a score of 12 is approximately .1915
(b) Using the same process we get :
`z_11=(11-10)/4=.25` and `z_14=(14-10)/4=4`
Consulting the tables yields `1.0-.5987~~.4013` **Note that virtually 100% of the distribution will be found less than4 standard deviations from the mean.
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