Use a standard normal table to obtain the areas under the no
Solution
a.
It is graph OPTION A, by inspection.
*******************
Thus, the two z scores are
z1 = lower z score = -1.75
z2 = upper z score = 1.34
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.040059157
P(z < z2) = 0.909877328
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.869818171
Thus, those outside this interval is = 0.1302 [ANSWER]
***************************
***************************
b.
It is graph option A, by inspection.
******************************
Thus, the two z scores are
z1 = lower z score = 0.64
z2 = upper z score = 1.69
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.7389137
P(z < z2) = 0.954486023
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.215572322
Thus, those outside this interval is = 0.7844 [ANSWER]
