Thank you for visiting At a large publishing company the mean age of proofreaders is 35 6 years and the standard deviation is 3 years Assume the variable is. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The probability that a randomly selected proofreader's age will be between 34 and 38.2 years is approximately 0.510036.
The given problem involves a normally distributed variable, where we are asked to find the probability of a proofreader's age falling between 34 and 38.2 years. We can solve this using the standard normal distribution.
To find the probability, we need to standardize the values using the formula:
[tex]$$Z = \frac{X - \mu}{\sigma}$$[/tex]
Where:
- [tex]\(X\)[/tex] is the value (age) we want to find the probability for (34 and 38.2 in this case)
- [tex]\(\mu\)[/tex] is the mean age of proofreaders (35.6 years)
- [tex]\(\sigma\)[/tex] is the standard deviation (3 years)
Calculating for \(Z\) for both values:
For [tex]\(X = 34\):[/tex]
[tex]$$Z_1 = \frac{34 - 35.6}{3} = -0.5333$$[/tex]
For [tex]\(X = 38.2\):[/tex]
[tex]$$Z_2 = \frac{38.2 - 35.6}{3} = 0.8667$$[/tex]
Now, we can use a standard normal distribution table or a calculator to find the cumulative probability associated with these [tex]\(Z\)[/tex] values. Subtracting the cumulative probability for[tex]\(Z_1\)[/tex] from the cumulative probability for [tex]\(Z_2\)[/tex] will give us the probability that the age falls between 34 and 38.2.
Using a calculator or statistical software, the cumulative probability for [tex](Z_1\ )[/tex] is approximately 0.2969 and for [tex]\(Z_2\)[/tex] is approximately 0.8069. Subtracting [tex]\(0.2969\) from \(0.8069\)[/tex] gives us the final probability of approximately [tex]\(0.510036\).[/tex]
This means there's about a 51.00% chance that a randomly selected proofreader's age will be between 34 and 38.2 years.
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