Suppose that the distribution of net typing rate in words pe

Suppose that the distribution of net typing rate in words per minute (wpm) for experienced typists can be approximated by a normal curve with mean 58 wpm and standard deviation 25 wpm. (Round all answers to four decimal places.)

(a) What is the probability that a randomly selected typist\'s net rate is at most 58 wpm?


What is the probability that a randomly selected typist\'s net rate is less than 58 wpm?


(b) What is the probability that a randomly selected typist\'s net rate is between 8 and 108 wpm?


(c) Suppose that two typists are independently selected. What is the probability that both their typing rates exceed 108 wpm?


(d) Suppose that special training is to be made available to the slowest 20% of the typists. What typing speeds would qualify individuals for this training? (Round the answer to one decimal place.) or less words per minute

Solution

Normal Distribution
Mean ( u ) =58
Standard Deviation ( sd )=25
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 58) = (58-58)/25
= 0/25 = 0
= P ( Z >0) From Standard Normal Table
= 0.5                  
P(X < = 58) = (1 - P(X > 58)
= 1 - 0.5 = 0.5                  

P(X < 58) = (58-58)/25
= 0/25= 0
= P ( Z <0) From Standard Normal Table
= 0.5                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 8) = (8-58)/25
= -50/25 = -2
= P ( Z <-2) From Standard Normal Table
= 0.02275
P(X < 108) = (108-58)/25
= 50/25 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(8 < X < 108) = 0.97725-0.02275 = 0.9545                  
c)
P(X > 108) = (108-58)/25/ Sqrt ( 2 )
= 50/17.678= 2.8284
= P ( Z >2.8284) From Standard Normal Table
= 0.0023                  
d)
P ( Z < x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is -0.842
P( x-u/s.d < x - 58/25 ) = 0.2
That is, ( x - 58/25 ) = -0.84
--> x = -0.84 * 25 + 58 = 36.95 ~ 37