Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboard can be approximated by a normal curve with mean 60 wpm and standard deviation 15 wpm. a) What is the probability that a randomly selected typist's speed is at most 60 wpm? b) What is the probability that a randomly selected typists's speed is less than 60 wpm? c) What is the probability that a randomly selected typist's speed is between 45 and 90 wpm? d) Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? e) Suppose that two typists are independently selected. What is the probability that both their typing speeds exceed 75 wpm? f) Suppose that special training is to be made available to the slowest 20% of the typists. What typing speed would qualify individuals for this training?

a. 0.5 or 50%

b. 0.496 or 49.6%

c. 0.8185 or 81.85%

d. Yes, it would be just 0.0013 or 0.13% of probability to find a typist whose speed exceeds 105 wpm

e. 0.0252 or 2.52%

f. The qualifying speed would be 47.4 wpm.

Step-by-step explanation:

a. Let's find the z-score this way:

μ = 60 σ = 15

z-score = (x - μ) / σ

z-score = (60 - 60) / 15

z-score = 0

Now, let's calculate the p value for z-score = 0, using the z-table:

p (z=0) = 0.5 or 50%

b. z-score = (x - μ) / σ

z-score = (59.9 - 60) / 15

z-score = - 0.01

Now, let's calculate the p value for z-score = -.0.01, using the z-table:

p (z = - 0.01) = 0.496 or 49.6%

If the question were less or equal than 60, a and b would have the same answer. But in this case, the question is "less than 60 wpm".

c.

z-score = (x - μ) / σ

z-score = (45 - 60) / 15

z-score = - 1

Now, let's calculate the p value for z-score = - 1, using the z-table:

p (z = - 1) = 0.1587

z-score = (x - μ) / σ

z-score = (90 - 60) / 15

z-score = 2

Now, let's calculate the p value for z-score = 2, using the z-table:

p (z = 2) = 0.9772

In consequence,

p (-1 ≤ z ≤ 2) = 0.9772 - 0.1587 = 0.8185 or 81.85%

d. z-score = (x - μ) / σ

z-score = (105 - 60) / 15

z-score = 3

Now, let's calculate the p value for z-score = 3, using the z-table:

p (z = 3) = 0.9987

In consequence,

p (z > 3) = 1 - 0.9987 = 0.0013

Yes, it would be just a 0.13% of probability to find a typist whose speed exceeds 105 wpm.

e. z-score = (x - μ) / σ

z-score = (75 - 60) / 15

z-score = 1

Now, let's calculate the p value for z-score = 1, using the z-table:

p (z = 1) = 0.8413

In consequence,

p (z > 1) = 1 - 0.8413 = 0.1587

and if the two typists are independently selected, then

p = 0.1587 * 0.1587

p = 0.0252 or 2.52%

f. p = 0.2, using the z-table, the z-score is - 0.84, then:

z-score = (x - μ) / σ

-0.84 = (x - 60) / 15

-12.6 = x - 60

-x = - 60 + 12.6

-x = - 47.4

x = 47.4

The qualifying speed would be 47.4 wpm.