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

a)

As 58 is the mean, by symmetry, it is 0.5. [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    8      
x2 = upper bound =    108      
u = mean =    58      
          
s = standard deviation =    25      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.954499736   [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    108      
u = mean =    58      
          
s = standard deviation =    25      
          
Thus,          
          
z = (x - u) / s =    2      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2   ) =    0.022750132

Thus,

P( two typists exceed 108 wpm) = 0.022750132^2 = 0.000517569 [ANSWER]

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d)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.2      
          
Then, using table or technology,          
          
z =    -0.841621234      
          
As x = u + z * s,          
          
where          
          
u = mean =    58      
z = the critical z score =    -0.841621234      
s = standard deviation =    25      
          
Then          
          
x = critical value =    36.95946916   [ANSWER]