A what is the probability that a call lasted less than 160 s

A. what is the probability that a call lasted less than 160 seconds

B. what is the probability that a call lated between 160 and 270 seconds

C. what is the probability that a call lasted between 140 and 160 seconds

D. what is the length of a call if only 1% of all calls are shorter?

Solution

Mean ( u ) =220
Standard Deviation ( sd )=30
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X < 160) = (160-220)/30
= -60/30= -2
= P ( Z <-2) From Standard Normal Table
= 0.0228                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 160) = (160-220)/30
= -60/30 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 270) = (270-220)/30
= 50/30 = 1.6667
= P ( Z <1.6667) From Standard Normal Table
= 0.95221
P(160 < X < 270) = 0.95221-0.02275 = 0.9295                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 140) = (140-220)/30
= -80/30 = -2.6667
= P ( Z <-2.6667) From Standard Normal Table
= 0.00383
P(X < 160) = (160-220)/30
= -60/30 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(140 < X < 160) = 0.02275-0.00383 = 0.0189                  

d)
P ( Z < x ) = 0.01
Value of z to the cumulative probability of 0.01 from normal table is -2.326
P( x-u/s.d < x - 220/30 ) = 0.01
That is, ( x - 220/30 ) = -2.33
--> x = -2.33 * 30 + 220 = 150.22