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Probability and Confidence Interval of Normal Distribution

The weekly incomes of a large group of middle managers are normally distributed with a mean of $1,500 and a standard deviation of $150. Use the standard normal distribution to calculate the following:
1. Find the probability that a particular weekly income selected at random is between $750 and $1,500.
2. What is the probability that the income is less than $900?
3. What is the probability that the income is more than $1100?
4. What is the probability that the income is between $1,000 and $1,800? (Be able to draw a graph to illustrate your results)
5. Find the salary that represents the 50th percentile.
6. Find the salary that represents the 90th percentile.
7. 5% of the salaries are below what value?
8. The top 5% of the salaries are above what value?
9. Between what two values will the middle 50% of the data lie?
10. What percent of the managers make more than $1200 a month?
11. Use the empirical rule to determine the following:
A About 68% of the observations lie between what two values?
B About 95% of the observations lie between what two values?
C About 99% of the observations lie between what two values?
12. Use the standard normal distribution to determine the following
A 68% of the observations lie between what two values?
B 95% of the observations lie between what two values?
C 99% of the observations lie between what two values?
13. Discuss the differences in the results for question 12 and question 11

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Let X be the amount of a particular weekly income. We are given that Given mu=1500, sigma=150 and z=(x-mu)/sigma

1. Find the probability that a particular weekly income selected at random is between $750 and $1,500?
P(750<X<1500)=P((750-1500)/150 <(X-mu)/sigma <(1500-1500)/150)=P(-5<Z<0)=0.5 by standard normal table

2. What is the probability that the income is less than $900
P(X<900)=P(Z<(900-1500)/150 )=P(Z<-4)=0

3. What is the probability that he income is more than $1100?
P(X>1100)=P(Z>(1100-1500)/150 )=P(Z>-2.67)=1

4. What is the probability that the income is between $1,000 and $1,800? (Be able to draw a graph to illustrate your results)
P(1000<X<1800)=P((1000-1500)/150 <(X-mu)/sigma <(1800-1500)/150)=P(-3.33<Z<2)=0.9768 by standard normal table

5. Find the salary that represents the 50th percentile.
50th percentile = median, since for normal distribution, mean=median, so 50th percentile = 1500

6. Find the salary that represents the 90th percentile.
We need to find z such that P(Z<z)=0.9. By checking standard normal table,z=1.282. So ...

Solution Summary

The solution gives detailed steps on calculating various probabilities and confidence intervals for some normal distributed random variables. All formula and calcuations are shown and explained.

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