Question 827862
The study showed that the length of the calls has a mean of 3.2 minutes, a standard deviation of 0.50 minutes and it follows the normal probability distribution.
For each of these problems:
1st: Find the z-value of the figures
2nd: Find the probability that z satisfies the condition of the problem:
I'll do part "a":

(a) What portion of the calls was between 3.2 and 4 minutes?
z(3.2) = (3.2-3.2)/0.5 = 0
z(4) = (4-3.2)/0.5 = 0.8/0.5 = 1.6
P(3.2< x <4) = P(0< z < 1.6) = normalcdf(0,1.6) = 0.4452
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Cheers,
Stan H.====================== 
 
(b) What portion of the calls was more than 4 minutes? 
(c) What portion of the calls was between 4 and 4.5 minutes? 
(d) What portion of the calls was between 3 and 4.5 minutes?