Typing speeds among college students are normally distributed with a mean of 60 words, and a standard deviation of 10 words. determine the typing speed that separates the top 6% typists from the rest.
I think I have it...nevermind .94 on z chart =1.55...then multiply 1.55 times error (10) then time 60=75.5?
right.
To determine the typing speed that separates the top 6% typists from the rest, we can use the concept of the standard normal distribution.
Step 1: Find the z-score corresponding to the top 6%.
The top 6% of typists correspond to an area under the normal distribution curve of 0.06. Since the normal distribution is symmetric, we can find the z-score that corresponds to an area of 0.06 in the upper tail.
Using a standard normal distribution table or a calculator, we find that the z-score for an upper tail area of 0.06 is approximately 1.56. This means that the top 6% of typists have a z-score of 1.56 or higher.
Step 2: Convert the z-score back to the original distribution.
To convert the z-score back to the original distribution, we can use the formula:
x = μ + z * σ
Where:
x = Typing speed we're looking for
μ = Mean of the distribution (60 words)
z = Z-score (1.56)
σ = Standard deviation of the distribution (10 words)
Substituting the values into the formula, we get:
x = 60 + 1.56 * 10 = 60 + 15.6 = 75.6
Therefore, the typing speed that separates the top 6% typists from the rest is approximately 75.6 words per minute.