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  1. At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variables are normally distributed.

a. If a proofreader in the company is randomly selected, find the probability that his or her age will be between 36 and 37.5 years

Attempt (unverified): Probability distribution of right-hand side of the mean on the bell curve:

35% of 1 Standard Deviation = 1.295 years

36.2+1.295 = 37.495 (approx. 37.5 years)

Therefore, 35% of 34% (half of one standard deviation on the right) = 11.9%

We do the same thing for the left-hand side and end up with: 6.25% of 34% (half of 1s to the left) = 2.125% [gives us 35.97 years =~ 36 years]

Therefore, P(age between 36 and 37.5 years) = 11.9 + 2.125 = 14.025% (final answer)

b. If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders in the sample will be between 36 years and 37.5 years.

P(age between 36 and 37.5) = 14.025% [from A]

n = 15



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Before proceding with answers, do you know how to standardize into standard normal distribution and read these probabilities form the standard normal table? Because that is what part a) looks like to me. Part b is √n-law where the given standard deviation needs to be divided by √15 before standardization. – imranfat Jun 22 '13 at 3:25
I didn't use any table for part a) and just used trial and error to get the closest possible values, and I have no clue how to proceed with b), so after we do 3.7/sqrt(15) what do we do next? – user81167 Jun 22 '13 at 3:29
For part b with the new SD, it goes the same as for a. But you need to standardize to get a z-score. Does that sound familiar? – imranfat Jun 22 '13 at 3:30
unless you're referring to finding the z-score value from the standard deviation we just calculated, where we would then use the distribution table? – user81167 Jun 22 '13 at 3:31
got it!!! Thank you, I shall update my answer soon. – user81167 Jun 22 '13 at 3:33

I do not understand the first calculation. We want the probability that the age is between $\frac{0.2}{3.7}$ standard deviation units below the mean and $\frac{1.3}{3.7}$ standard deviation units above the mean.

Using the table of the standard normal, we find that the probability that the age is $\frac{1.3}{3.7}\approx 0.351$ standard units above the mean or less is approximately $0.637$.

The probability that we are below the mean by $\frac{0.2}{3.7}$ or more standard deviation units is about $0.48$. So our required probability is about $0.637-0.48$, some distance from your number.

For the next problem, if we take $15$ ages and average them, the resulting random variable has normal distribution with mean $36.2$ and standard deviation $\frac{3.7}{\sqrt{15}}\approx 0.955$. Now one needs to essentially repeat the first calculation, but with $0.955$ replacing $3.7$. The probabilities will change dramatically.

Added: Let $X$ be normally distributed with mean $\mu$ and standard deviation $\sigma$. Then $$\Pr(X\le a)=\Pr\left(Z\le \frac{a-\mu}{\sigma}\right),$$ where $Z$ is standard normal.

For reasonable positive values of $z$, $\Pr(Z\le z)$ can be looked up approximately in a table of the standard normal.

For negative $z$, we have to work a bit. If $z$ is negative, then $\Pr(Z\le z)=1-\Pr(Z\le |z|)$.

Tables of the standard normal are available on the web. Typically they are a single page, you can download one and print it. Such a table is still often found at the back of introductory statistics books.

But nowadays the reasonable thing is to let software do the work for you. There are many pieces of software, including spreadsheet programs, that have normal distribution calculations as a built in feature.

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