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Calculating the Probability in Random Samples

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#25. Companies often make flextime scheduling available to help recruit and keep women employees who have children. Other workers sometimes view
these flextime schedules as unfair. An article in USA Today indicates that 25% of male employees state they have to pick u the slack for moms working flextime schedules. Suppose you select a random sample of 100 employees working for companies offering flextime.

a. What is the probability that 23% or fewer will indicate that they pick up the slack for moms working flextime?
b. What is the probability that 20% or fewer will indicate that they pick up the slack for moms working flextime?
c. If a random sample of 600 is taken, how does this change your answers to (a) and (b)

#23. The manager of a paint supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from nationally known manufacturer's specifications state that the standard deviation of the amount of paint is equal to 0.02 gallon. A random sample of 50 cans is selected, and the sample mean amount of paint per 1-gallon can is 0.972 gallon.

a. Conduct a confidence interval estimate for the population mean amount of paint included in a 1-gallon can.

#22. In a recent survey of 156 adults, 48 adults reported that emails are easy to misinterpret, but 84 reported that the telephone
conversations are easy to interpret.
a. Construct a 95% confidence interval to estimate for the population proportion of adults who report the emails are easy to misinterpret.
b. Construct a 95% confidence interval to estimate for the population proportion of adults who report the emails are easy to interpret.
c. Compare the results of (a) and (b).

17. A multiple choice test has five choices, there are 4 multiple choice questions on the test.
-What is the probability that a student will get 4 questions correct?
-What is the probability that a student will get a least 3 questions correct?
- What is the probability that a student will get no questions correct?
What is the probability that a student will get more than 2 questions correct?

#11. A toll free number is available from 9am-9pm for your customers to register complaints about a product purchased from your company. Past history indicates that an average of 0.7 calls are received per minute.

a. What properties must be true about the situation described here in order to use Poisson distribution to calculate probabilities concerning the number of phone calls received in a 1 minute period.

b. What is the probability that during a 1 minute period zero calls will be received c. What is the probability that during a 1 minute period 3 or more calls will be received What is the maximum number of phone calls that will be received in a minute period 99.99% of the time?

#15. A set of data has values that vary from 92.8 to 177.8.

If these values are grouped into nine classes, indicate the class boundaries.

What is the class interval width?
What are the class midpoints?

#12. In a deck of cards there is a total of 52 card, there are 4 suits of each (hearts, diamonds, clubs, spades), with 13 faces each.
1. What is the probability that the first two cards pulled from the deck will be 2 aces?
2. What is the probability that the first cards pulled from the deck will be a 10 and the second a 5 or 6?

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#25. Companies often make flextime scheduling available to help recruit and keep women employees who have children. Other workers sometimes view
these flextime schedules as unfair. An article in USA Today indicates that 25% of male employees state they have to pick u the slack for moms working flextime schedules. Suppose you select a random sample of 100 employees working for companies offering flextime.

a. What is the probability that 23% or fewer will indicate that they pick up the slack for moms working flextime?
b. What is the probability that 20% or fewer will indicate that they pick up the slack for moms working flextime?
c. If a random sample of 600 is taken, how does this change your answers to (a) and (b)
a. z=(0.23-0.25)/sqrt(0.25*0.75/100)=-0.46
P(z<-0.46)= 0.3228

b. z=(0.20-0.25)/sqrt(0.25*0.75/100)=-1.15
P(z<-1.15)= 0.1251

c. when n=600, for a, z=(0.23-0.25)/sqrt(0.25*0.75/600)=-1.13
P(z<-1.13)= 0.1292
Therefore, the probability changes from 0.3228 to 0.1292
When n=600, for b, z=(0.20-0.25)/sqrt(0.25*0.75/600)=-2.83
P(z<-2.83)=0.0023
Therefore, the probability changes from 0.1292 to 0.0023.

#23. The manager of a paint supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from nationally known manufacturer's specifications state that the standard deviation of the amount of paint is equal to 0.02 gallon. A random sample of 50 cans is selected, and the sample mean amount of paint per 1-gallon can is 0.972 gallon.

a. Conduct a confidence interval estimate for the population mean amount of paint included in a 1 gallon can the critical value for 95% confidence interval is 1.96
Margin of error=1.96*0.02/sqrt(50)= 0.0055
Upper limit: ...

Solution Summary

The following posting discusses statistics problems. Concepts discussed include probability and random samples.

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