find the probability P026
find the probability P(0.26<z<1.33) using the standard normal deviation<1.33)>
find the probability P(0.26<z<1.33) using the standard normal deviation<1.33)>
find the probability P(0.26<z<1.33) using the standard normal deviation<1.33)>
Solution
z1 = lower z score = 0.26
z2 = upper z score = 1.33
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.602568113
P(z < z2) = 0.908240864
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.305672751 [ANSWER]
