Question

Step 1 of 5: 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. What is the probability that a randomly selected typist's speed is at most 60 wpm? Round your answer to one decimal place.

What is the probability that a randomly selected typist's speed is between 45 and 90 wpm?  Round your answer to four decimal places.

Would you be surprised to find a typist in this population whose speed exceeded 105 wpm? Explain your reason.

Suppose two typists are independently selected. What is the probability that their speed exceed 75 wpm? Round your answer to four decimal places.

What is the probability that a randomly selected typist's speed is less than 60 wpm? Round your answer to one decimal place.

Answer 1

Let X = Typing speed in words per minute

Given

1. = ?

Since, Normal distribution is symmetric around the mean, the probability that an observation is less than the mean would be exact half i.e 50% = 0.5.

From Standard Normal table, that gives the area to the left of the Z score:

2.  

= 0.97725 - 0.15866

= 0.8186 (appx.)

The probability that a randomly selected typist's speed is between 45 and 90 wpm would be 0.8186.

3.

By Empirical rule,  68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

And this is determined using the Z score: A Z score of 1.00, 2.00 and 3.00 implies that the observation lies 1 (µ ± σ), 2 (µ ± 2σ) and 3 (µ ± 3σ) SDs away from the mean.

Hence an observation that falls above Z = 3:

would have the probability of occurrance of atmost 0.15% = 0.0015 which is very low. There is very little chance that an observation > 105 occurs.

Hence, we would be surprised to find a typist in this population whose speed exceeded 105 wpm.

Using Standard Normal table, we may obtain the exact probability:

= 1 - 0.99865

= 0.0013

4. Probability that first person selected types > 75 wpm

= 1 - 0.84134

= 0.1587

Since, the two typists are independently selected, the probability that first person selected types > 75 wpm would be the same as that of first person:

Required probability = (0.15866)(0.15866)

= 0.0252