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# I. a. What percentage of the area of the standard normal distribution

I.
a. What percentage of the area of the standard normal distribution is between z = -2.00 and z = +2.00? How do you know this?

b. Find the following:
P(z > 2.43)

II.
1. A set of 50 data values has a mean of 25 and a variance of 36.
I. Find the standard score (z) for a data value = 31.
II. Find the probability of a data value > 31.

2. Find the area under the standard normal curve:
I. to the right of z = -1.36
II. to the left of z = -1.36

3. Assume that the population of heights of male college students is approximately normally distributed with mean µ of 69 inches and standard deviation σ of 3.75 inches. Show all work.
(A) Find the proportion of male college students whose height is greater than 70 inches.
(B) Find the proportion of male college students whose height is no more than 70 inches.

4. The diameters of apples in a certain orchard are normally distributed with a mean of 5.55 inches and a standard deviation of 0.65 inches. Show all work.
(A) What percentage of the apples in this orchard have diameters less than 5.8 inches?
(B) What percentage of the apples in this orchard are larger than 6.3 inches?

5. Find the normal approximation for the binomial probability that x = 4, where n = 12 and p = 0.5. Compare this probability to the value of P(x=4) found in Table 2 of Appendix B in your textbook.

6. A set of data is normally distributed with a mean of 100 and standard deviation of 15.
•What would be the standard score for a score of 90?
•What percentage of scores is between 100 and 90?
•What would be the percentile rank for a score

III.

1. If the random variable z is the standard normal score and a > 0, is it true that P(z a)? Why or why not?

2. Given a binomial distribution with n = 39 and p = 0.87, would the normal distribution provide a reasonable approximation?

3. Find the area under the standard normal curve for the following:
(A) P(z > 0.22)
(B) P(-1.26 < z < 0)
(C) P(-2.00 < z < 0.89)

4. Find the value of z such that approximately 48.46% of the distribution lies between it and the mean.

5. Assume that the average annual salary for a worker in the United States is \$27,500 and that the annual salaries for Americans are normally distributed with a standard deviation equal to \$6,250. Find the following:

(A)What percentage of Americans earn below \$18,000?
(B)What percentage of Americans earn above \$40,000?

6. X has a normal distribution with a mean of 80.0 and a standard deviation of 4.0. Find the following probabilities:
(A) P(x < 78.0)
(B) P(75.0 < x 82.0)

(A) Find the binomial probability P(x = 5), where n = 12 and p = 0.30.
(B) Set up, without solving, the binomial probability P(x is at most 5) using probability notation.
(C) How would you find the normal approximation to the binomial probability P(x = 5) in part A? Please show how you would calculate µ and σ in the formula for the normal approximation to the binomial, and show the final formula you would use without going through all the calculations.
Welcome and Thanks for using Just Answer.

I can assist you with your questions. But there are far too many. Can I split and post these in 3 parts for 3 separate accepts?

Thanks.
Customer: replied 5 years ago.
I don't mind but how much will it cost me

Perrion Roberts
For 3 accepts, it will cost you three times on what you offered (Total \$90). Please let me know. so I can start on the work.

Thanks.
Customer: replied 5 years ago.
I can't afford that the last expert that helped me did them as a whole I am sorrow I just can't afford that.

Thank you,

Perrion Roberts
Ok, If you feel that is too much, I can post all answers at once, and you can add a \$30 tip after accepting the answers. That way, it will cost you only \$60.

Customer: replied 5 years ago.
ok I was trying to up load Table 2 in question 5 of II.

yes thank you,

Perrion Roberts
No, I have the table in my computer, so I can use it.

I will get down to work on this. By when is this due?

Customer: replied 5 years ago.
It is due on Monday

Thank you,

Perrion Roberts
Sure, I will post the answers much before that.

Thanks.
Customer: replied 5 years ago.
Thank you very much

Yes, you are welcome.

Hi

Here are all the solutions

Let me know if you need any clarifications on my answers. I will be happy to provide them.

Thanks.

Customer: replied 5 years ago.

This is what the my teacher said about the first set of questions can you fix it and also explain the answers to the other questions. Should I except the answers now or wait for you to finish I will accept now if you like.

Thanks Perrion

Okay. You have made a great start on these problems. #1 is correct. A short-cut estimate for this can be derived from the Empirical rule to give 95%. The Empirical rule is handy for checking that answers are reasonable. For #2, the answer is correct. However, for full credit, remember to include a couple lines to explain your reasoning and the key steps or ideas used to derive the answer. For instance, since the table gives areas to the left of a given z value, how did you find the answer?

Hi, sorry my network was down and therefore I was late to reply.

Ok, here are the two answers again ...

(1) Since the z- scores for 95% confidence are ± 1.96 (which is close to 2.00), we can say that the area enclosed between z = -2.00 and z = 2.00 is about 95%

(2) μ = 25, σ = √36 = 6
(I) z = (x - μ)/σ = (31 - 25)/6 = 1
(II) P(x > 31) = 1 - P(x ≤ 31) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587

Let me know if you need any clarifications on my answers. I will be happy to provide them.

Thanks.

Hi [Customer Name],

I'm just following up with you to see how everything is going. Did my answer help?

Let me know,
Chirag
Customer: replied 5 years ago.
Hi guru2009,

I am doing well I have not recieved a feed back for the teacher again I hope everything is ok. Your answers did help I understand a little bit better.

Thanks

Perrion Roberts
Yes, thanks for feedback. I had left a follow-up message because I had not heard from you. No other issues.

Thanks.
Customer: replied 5 years ago.
No not right now.

Thanks

Perrion Roberts
Ok, welcome.
Customer: replied 5 years ago.
I gave you a bonus

I need to post more questions please

Yes, thanks for it.

Customer: replied 5 years ago.
I.

a. The maximum error of estimate is a function of three factors: the level of confidence, sample size and standard deviation. T-F, and why or why not?

b. Explain the difference between a point estimate and an interval estimate.

II.

1. Consider a population with µ 63.7 and σ = 4.31.

(A) Calculate the z-score for x = 62.8 from a sample of size 47.

(B) Could this z-score be used in calculating probabilities using Table 3 in Appendix B of the text? Why or why not?

2. Given a level of confidence of 99% and a population standard deviation of 8, answer the following:
(A) What other information is necessary to find the sample size (n)?
(B) Find the Maximum Error of Estimate (E) if n = 98. Show all work.

3. A sample of 102 golfers showed that their average score on a particular golf course was 88.55 with a standard deviation of 4.27.
Answer each of the following (show all work
and state the final answer to at least two decimal places.):
(A) Find the 95% confidence interval of the mean score for all 102 golfers.
(B) Find the 95% confidence interval of the mean score for all golfers if this is a sample of 135 golfers instead of a sample of 102.
(C) Which confidence interval is smaller and why?

4. Assume that the population of heights of male college students is approximately normally distributed with mean m of 68.14 inches and standard deviation s of 5.38 inches. A random sample of 66 heights is obtained. Show all work.

(A) Find the mean and standard error of the x distribution
(B) Find P(x > 67.25) (THAT IS A LINE OVER BOTH X'S)

5. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.72 inches and a standard deviation of 0.57 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger than 5.66 inches?
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.66 inches?

6. A researcher is interested in estimating the noise levels in decibels at area urban hospitals. She wants to be 99% confident that her estimate is correct. If the standard deviation is 4.92, how large a sample is needed to get the desired information to be accurate within 0.71 decibels? Show all work.

III.

1. Consider a normal population with µ = 25 and σ = 8.0.
(A) Calculate the standard score for a value of 22.

(B) Calculate the standard score for a randomly selected sample of 30 with x = 22. (THERE SHOULD BE A LINE OVER THE X)

(C) Explain why the standard scores of 22 are different between A and B above.

2. Assume that a sample is drawn and z(α/2) = 1.65 and σ = 25. Answer the following questions:

(A)If the Maximum Error of Estimate is 0.05 for this sample, what would be the sample size?

(B)Given that the sample Size is 400 with this same z(α/2) and σ, what would be the Maximum Error of Estimate?

(C)What happens to the Maximum Error of Estimate as the sample size gets larger?

(D)What effect does the answer to C above have to the size of the confidence interval?

3. Assume that the mean score on a certain aptitude test across the nation is 100, and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is between 101.5 and 98.5.

4. By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 3.63 seconds.
Answer each of the following (show all work):
(A) How many measurements should be made in order to be 95% certain that the maximum error of estimation will not exceed 2.0 seconds?

(B) What sample size is required for a maximum error of 0.5 seconds?

5. A 95% confidence interval estimate for a population mean was computed to be (36.2, 48.8). Determine the mean of the sample, which was used to determine the interval estimate (show all work).

 6. A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 180 was taken, and the mean amount spent was \$219.76. Assuming a standard deviation equal to \$41.77, find the 99% confidence interval for m, the mean for all such families (show all work).

 7. A confidence interval estimate for the population mean is given to be (35.85, 44.80). If the standard deviation is 12.298 and the sample size is 41, answer each of the following (show all work): (A) Determine the maximum error of the estimate, E. (B) Determine the confidence level used for the given confidence interval.

IV.

a. What is the relationship between α and Type I Error?

b. What decision is reached when the p-value is greater than α?

Can you see 6 and 7 in III. ?

It is not clear how many questions are there totally. And none of the images show up.

Please UPLOAD the file to a free file sharing site such as www.mediafire.com or www.sendspace.com, SHARE the file, and POST THE LINK here.

Thanks.
Customer: replied 5 years ago.
ok
Customer: replied 5 years ago.
http://www.mediafire.com/view/?kfkdbsxz1d7s8ci

or
http://www.mediafire.com/file/kfkdbsxz1d7s8ci/Week_5_Satistic_1-3.docx
There are totally 15 questions with multiple parts in each. I can do all for a \$60 bonus (showing all working steps). Please let me know so that I can start the work.

Thanks.
Customer: replied 5 years ago.
yes that will be ok with me Please soon as possible
Thank you so much

Perrion Roberts
I will post the solutions in about 5 to 6 hours maximum. Is that ok?
Customer: replied 5 years ago.
yes Perfect

Thanks
You are welcome.

Hi

Here are the solutions

I hope that helps. Let me know if you need any clarifications on my answers. I will be happy to provide them.

Thanks.