Given a normal distribution with mean 10 and variance 4. What area of the distribution would lie between a) 10 and 12 units b) 11 and 14 units
(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.