Math Help - Probabilities and Standard Deviations

1. Probabilities and Standard Deviations

3. The computer club has 100 members. There are 50 women and 50 men. Thirty of the women are 21 years of age and older, 45 of the men are 21 years of age and older. What is the probability that a member who is 21 or older and selected at random will be a woman?

4.The mean salary for the engineering department of a large company is $100,000. The salaries for the engineering department follow a normal distribution. About 95% of the salaries are between$60,000 and $140,000. What is the standard deviation of the salaries for the engineering department? 2. 2) Can you get cat and red at the same time? Have you got a formula for this type of probability? 4) Have you learned the rules regarding 68,95,99 ? Show us your worth so far, so that we can find your errors. 3. 2. Yes you can get cat and red at the same time. And no I do not know the formula 4. No I have never heard of the rules regarding 68,95,99. 4. Originally Posted by wytiaz ... 4) Have you learned the rules regarding 68,95,99 ? ... wytiaz, that is neat! I like your thinking. 5. Originally Posted by krzyrice 3. The computer club has 100 members. There are 50 women and 50 men. Thirty of the women are 21 years of age and older, 45 of the men are 21 years of age and older. What is the probability that a member who is 21 or older and selected at random will be a woman? 4.The mean salary for the engineering department of a large company is$100,000. The salaries for the engineering department follow a normal distribution. About 95% of the salaries are between $60,000 and$140,000. What is the standard deviation of the salaries for the engineering department?
2. Use the formula $\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$.

3. Pr(woman | over 21) = Pr(woman and over 21)/Pr(over 21).

Pr(woman and over 21) = .....

Pr(over 21) = ....

Therefore ....

4. See this thread for a similar question: http://www.mathhelpforum.com/math-he...deviation.html

6. So for
4). Would the equation be (140,000 - 60,000) = 100000(o)
Answer = 1.25

7. Originally Posted by krzyrice
So for
4). Would the equation be (140,000 - 60,000) = 100000(o)
Answer = 1.25
I hope in hindsight you can see why 1.25 is a ridiculous answer.

Solve $140,000 - 100,000 = 2 \sigma$ for $\sigma$.

8. Can you explain to me where you get the 2 from? Or is that just part of the equation. Thanks so far.

9. Originally Posted by krzyrice
Can you explain to me where you get the 2 from? Or is that just part of the equation. Thanks so far.
Did you read the thread whose link I gave you? The answer is in there.