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Ryan
Ryan, Engineer
Category: Calculus and Above
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Experience:  B.S. in Civil Engineering
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REQUESTING FOR RYAN I will resend you the attachment for

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REQUESTING FOR RYAN

I will resend you the attachment for the STATS questions once it allows me to attach/
Hi,

Thank you for the new request.

Please post the attachment at your convenience.

Thanks,

Ryan
Customer: replied 3 years ago.

Attachment: 2013-09-05_055140_2013-09-04_084854_stats_230___week_3.doc


Let me know if you need anything else. Thanks

Hi,

There is a histogram that goes with problems 13, 14, 15, and 16, but it doesn't show up in this file. There is a line that says "EMBED MtbGraph.Document.15" that I suspect is a placeholder for the actual graph.

Can you take a screenshot of the histogram and post the screenshot image as a separate file?

I can't answer #13 or #14 without the histogram.

Also, since these are multiple choice questions, do you just need the final answers, or do you need the steps taken to get the solutions?

Thanks,

Ryan E
Customer: replied 3 years ago.
Ok I'll ck to see if there's a graph. For the multiple choice, just the answers... No need solutions
Hi,

Ok, sounds great.

I have all of the other answers ready for you. I was just waiting on the histogram for #13 and #14.

If you want or need the other solutions now, just let me know. I can have it posted for you in just a few minutes.

Thanks,

Ryan
Customer: replied 3 years ago.
Sure ill get you the histograms in a few but yes, post the others
Hi,

Ok, no problem. Here are the solutions for the other problems:

StatsSolutions

By the way, it is #14 and #15 that I can't answer yet, not #13 and #14 as I stated earlier.

Thanks,

Ryan
Customer: replied 3 years ago.

Attachment: 2013-09-05_235633_stats_week_3.docx

Attachment: 2013-09-05_235701_stats_week_3.docx


 


Let me know if you can open and read file. I've added three more questions, hope you don't mind answering. I'll be sure to include additional $. Thanks again!

Customer: replied 3 years ago.


Can you resend your Solutions another way, I can not open it from that mediafire from the computer I'm using.

Hi,

Here are the solutions for the first set:


1. E. Both A and C

2. E. Both B and C

3. C. is as quantitative variable

4. B. Percentage of American adults who work full time

5. D. All of the above

6. A. the ratio scale

7. D. Observational studies

8. A. Experimental studies

9. B. the population

10. A. Given the prevalence of identity theft, are you reluctant to provide credit-card information online?

11. D. A scatter diagram

12. A. 27.78%

13. B. 45%

14. C. Right skewed

15. A. Median

16. B. $22,057

17. A. The mean would increase

18. C. the mean

19. B. 30%

20. A. 12.5

21. B. the pie chart

22. C. $56.99

23. A. $31.43

24. E. -1.24

25. C. -0.9265

Conference responses:

1. The 15th percentile corresponds to a left tail area of 0.15. The corresponding z-value obtained from Excel is -1.0364. The IQ corresponding to the 15th percentile is then:

-1.0364 = (x - 100) / 16

x = 100 - (1.0364)(16)

x = 83.42

2. The 75th percentile corresponds to a right tail area of 0.25. The corresponding z-value obtained from Excel is 1.6745. The IQ corresponding to the 75th percentile is then:

1.6745 = (x - 100) / 16

x = 100 + (1.6745)(16)

x = 126.79

I'll post the solutions for the additional problems shortly.

Thanks,

Ryan

Hi again,

Here are the other solutions:

#19a. Prob = 334 / (187 + 334 + 256) = 334 / 777 ≈ 0.4299

#19b. Prob = (411 + 213) / (197 + 411 + 213) = 624 / 821 ≈ 0.7600

#19c. Prob = (334 + 411) / (187 + 197 + 334 + 411 + 256 + 213) = 745 / 1598 ≈ 0.4662

#19d. No, the relative proportions for each category are close to the same for each country:

US / yes: 187 / 777 ≈ 0.2407
Britain / yes: 197 / 821 ≈ 0.2400

US / no: 334 / 777 ≈ 0.4299
Britain / no: 411 / 821 ≈ 0.5006

US / unsure: 256 / 777 ≈ 0.3295
Britain / unsure: 213 / 821 ≈ 0.2594

#25a. P(M or F) = P(M) + P(F) - P(MF) = 0.56 + 0.42 - 0.24 = 0.74

#25b. P(neither M or F) = 1 - P(M or F) = 1 - 0.74 = 0.26


#52a. z = (1000 - 1550) / 300 = -1.8333. P(z < -1.8333) = 0.0334

#52b. From part a, z1 = -1.8333. z2 = (2000 - 1550) / 300 = 1.5
P(1000 < x < 2000) = P(-1.8333 < z < 1.5) = P(z < 1.5) - P(z < -1.8333) = 0.9332 - 0.0334 = 0.8998

#52c. x = µ + zσ = 1550 + 1.645(300) = 2043.5
Since the number of accidents would be a discrete value, this answer should be rounded to 2044.

As always, please feel free to ask if you have questions about any of these solutions.

Thanks,

Ryan
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