Question 633583: At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variable is normally distributed. Please put you answer in decimal form.
a. If a proofreader from the company is randomly selected, find the probability that his or her age will be between 36 and 37.5 years.
Answer by ikeoddy(3)   (Show Source):
You can put this solution on YOUR website! Problem Statement: Given mean, mu = 36.2 and standard deviation, sigma = 3.7, find Pr(36 <= proofreader's age <= 37.5).
Solution:
Let X be the random variable that represents the age of a proofreader. X is normally distributed.
Available to us are tables of probability for a "standard" normal distribution. Our task will be to "standardize" the values of X, so that we can use the table of probability.
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Standardizing a normal random variable
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To standardize a random variable X, we subtract its mean, mu and divide the result by its standard deviation, sigma.
If Z represents the standard random variable, then
Z = (X - mu)/sigma
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Let x1 = 36 and x2 = 37.5
z1 = (x1 - mu)/sigma = (36 - 36.2)/3.7 = -0.05405 ~= -0.05 to 2 decimal places
z2 = (x2 - mu)/sigma = (37.5 - 36.2)/3.7 = 0.3514 ~= 0.35 to 2 decimal places
Pr(36 <= X <= 37.5) = Pr(-0.05 <= Z <= 0.35)
The standard normal probability tables in http://lilt.ilstu.edu/dasacke/eco148/ztable.htm gives the probability that the random variable is less than or equal to a given z-score. That is, Pr(Z <= z-score).
To use this table for Pr(-0.05405 <= Z <= 0.3514), we want to rewrite the probability using probabilities of the form Pr(Z <= z-score)
P(-0.05405 <= Z <= 0.3514) = Pr(-0.05 < Z <= 0.35) + P(Z = -0.05).
Note that P(Z = -0.05) = 0, since Z is a continuous random variable, and -0.05 is a single point. In other words, we don't have to keep track of the equal to sign in the inequalities.
Pr(-0.05 < Z <= 0.35) = Pr(Z <= 0.3514) - Pr (Z <= -0.05405)
From the cumulative normal distribution table,
Pr(Z <= 0.35) = 0.6368
Pr (Z <= -0.05) = 0.4801
Thus,
Pr(-0.05 < Z <= 0.35) = Pr(Z <= 0.3514) - Pr (Z <= -0.05405)
= 0.6368 - 0.4801
= 0.1567
~= 0.16
Hence, the probability that the proofreader's age is between 36 and 37.5 years is approximately 0.16 (or 16%).
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