Question

5.3.35 i Question Help The average time spent sleeping (in hours) for a group of medical residents at a hospital can be approximated by a normal distribution, as shown in the graph to the right Answer parts (a) and (b) below Sleeping Times of Medical Residents 6.5 hours Click to view page 1 of the table Click to view page 2 of the table Hours (a) What is the shortest time spent sleeping that would still place a resident in the top 5% of sleeping times? Residents who get at least 863 hours of sleep are in the top 5% of sleeping times (Round to two decimal places as needed ) (b) Between what two values does the middle 50% of the sleep times lie? The middle 50% of sleep times lies between hours on the low end and hours on the high end (Round to two decimal places as needed) tion
5.3.31 Question Help : * In a survey of women in a certain country (ages 20-29), the mean height was 66.4 inches with a standard deviation of 2 82 inches. Answer the following questions about the specified normal distribution (a) What height represents the 95th percentile? (b) What height represents the first quartile? Click to view page 1 of the table, Click to view page 2 of the table (e) The height that represents the 95th percentile is inches. (Round to two decimal places as needed) (5.3
Standard Normal Table (Page 1) 04 3.4 0.0002 00003 0,0003 0,0003 0 0003 0.0003 0.0003 0.0003 0.0003 0.0003 3.3 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 0.0006 0.0006 0.0007 0.0007 -3.1 0.0007 0.0007 0.0008 0.0008 0.0008 00008 0.0009 0.0009 00009 0.0010 3.0 0.0010 0.0010 0.001 0.0011 0.0011 00012 0.0012 0.0013 0.0013 0.0013 -2.9 0.0014 0 0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.0019 -2.8 0.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0 0023 0.0024 0.0025 0.0026 033 0 0034 O 0035 2.6 10 0036 0 0037 0 0038 0 0039 0 0040 0 0041 0 0043 0 0044 0 0045 0 0047 -2.5 0.0048 0.0049 0.0051 0.0052 0.0054 0.0055 0 0057 0.0059 0.0060 0.0062 -2.4 0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0.0078 0.0080 0.0082 2.3 00084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 00102 0.0104 0.0107 2.2 0.0110 0.0113 0.0116 00119 0 0122 0 0125 00129 0.0132 0.0136 0.0139 2.1 0.0143 0.0146 0.0150 0.0154 00158 0 0162 0.0166 0.0170 0.0174 0.0179 2.0 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 00228 1.9 0.0233 0.0239 0.0244 0.0250 0 0256 0 0262 0.0268 0.0274 0.0281 0.0287 -1.80.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 00351 0.0359 1.7 0.0367 0.0375 0.0384 0.0392 00401 0.0409 0.0418 0.0427 0.0436 0.0446 -1.6 0.0455 0.0465 0.0475 0.0485 00495 0.0505 00516 0.0526 0.0537 0.0548 -1.5 0.0559 00571 0.0582 0.0594 0.0606 0 0618 0 0630 0.0643 0.0655 0.0668 1.4 0.0681 0.0694 0.0708 0.0721 00735 0.0749 0.0764 0.0778 0.0793 0.0808 1.3 0.0823 0.0838 0.0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.0968 1.2 0.0985 0.1003 0 1020 0.1038 0.1056 0.1075 0.1093 0.1112 0.1131 0.1151 1.1 l0.1170 0.1190 0.1210 0.1230 0.1251 0.1271 0.1292 0.1314 0.1335 0.1357 -3.2 0.0005 0.0005 0.0005 0.0006 0.0006 00006 2.7 0.0026 0 0027 0.0028 0 0029 0.0030 0.0031 0.0032 0.0
Standard Normal Table (Page 1) 2.3 0.0084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 0.0102 0.0104 0.0107 2.2 0.0110 0.0113 0.0116 0.0119 0.0122 0 0125 0.0129 0.0132 0.0136 0.0139 2.1 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.0179 2.0 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 00222 0.0228 1.9 0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.0287 1.8 0.0294 0.0301 0.0307 0.0314 0.0322 0 .0329 0.0336 0.0344 0 0351 0.0359 1.70.0367 0.0375 0.0384 0.0392 0.0401 0.0409 0.0418 0.0427 00436 0.0446 1.6 0.0455 0.0465 0.0475 0.0485 0.0495 00505 00516 0.0526 0.0537 0.0548 1.5 0.0559 0.0571 0.0582 00594 0.0606 0.0618 0.0630 0.0643 0.0655 0.0668 1.4 0.0681 0.0694 0.0708 0.0721 0.0735 0.0749 0.0764 0.0778 0.0793 0.0808 1.3 0.0823 0.0838 0 0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.0968 1.2 0.0985 0.1003 0.1020 0.1038 0. 1056 0.1075 0.1093 0.1112 0.1131 0.1151 1.1 0.1170 0.1190 0.1210 0.1230 0.1251 0.1271 0.1292 0.1314 0.1335 0.1357 1.0 0.1379 0.1401 0.1423 0.1446 0.1469 0.1492 0.1515 0.1539 0.1562 0.1587 0.9 0.1611 0.1635 0.1660 0.1685 0.1711 0.173 0.1762 0.1788 0.1814 0.1841 0.8 0.1867 0.1894 0.1922 0.1949 0. 1977 02005 02033 0 2061 0.2090 0.2119 0.7 0.2148 0.2177 0.2206 0.2236 0.2266 0.2296 0.2327 0.2358 0.2389 0.2420 0.6 0.2451 0.2483 0.2514 0.2546 0.2578 0.2611 0.2643 0.2676 0.2709 0.2743 0.5 0.2776 0.2810 0.2843 0.2877 0 2912 0 2946 0 2981 0.3015 0.3050 0.3085 0.4 0.3121 0.3156 0.3192 0.3228 0.3264 0.3300 0.3336 0.3372 0.3409 0.3446 0.3 0.3483 0.3520 0.3557 0.3594 0.3632 0.3669 0.3707 0.3745 0.3783 0.3821 0.2 0.3859 0.3897 0.3936 0.3974 0.4013 0.4052 0.4090 0.4129 0.4168 0.4207 0.1 0.4247 0.4286 0 4325 0.4364 0.4404 0.4443 0.4483 0.4522 0.4562 0.4602 0.0 0.4641 0.4681 0.4721 0.4761 0.4801 0.4840 0.4880 0 4920 0.4960 0 5000 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00
Standard Normal Table (Page 2) Z 0.00 001 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0 6554 06591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 o.7357 0.7389 0 7422 0.7454 0.7486 0.7517 0.7549 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0 8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0 8461 0.8485 0.8508 0 8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 08749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 09332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 09821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 09864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.991
media%2F5bd%2F5bdb948b-b019-472f-877f-b6
0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
5.3.35 i Question Help The average time spent sleeping (in hours) for a group of medical...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • VI HW Score: 23.53%, 4 of 17 pts 25.3.9 Question Help se the standard normal table...

    VI HW Score: 23.53%, 4 of 17 pts 25.3.9 Question Help se the standard normal table to find the 2-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is hallway between two entries, use the score halay between the corresponding 2-cores. If convenient, use technology to find the score P20 Click to view page 1 of the title Click to view. 2 of the...

  • Suppose a random sample of size 15 is taken from a normal population with mean 25...

    Suppose a random sample of size 15 is taken from a normal population with mean 25 and standard deviation 5; and a second, independent random sample of size 21 is taken from a normal population with mean 10 and standard deviation 4. Find the probability P(X- X2 2 17). (Use Appendix Table 3. Give your answer to four decimal places.) p(7. - 12 2 17) = 1.282 Incorrect Table In Standard Normal Distribution Cumulative Probabilities Let 2 be a standard...

  • z   .09   .08   .07   .06   .05   .04   .03   .02   .01   .00   z -3.4   0.0002   0.0003   0.0003  ...

    z   .09   .08   .07   .06   .05   .04   .03   .02   .01   .00   z -3.4   0.0002   0.0003   0.0003   0.0003   0.0003   0.0003   0.0003   0.0003   0.0003   0.0003   -3.4 -3.3   0.0003   0.0004   0.0004   0.0004   0.0004   0.0004   0.0004   0.0005   0.0005   0.0005   -3.3 -3.2   0.0005   0.0005   0.0005   0.0006   0.0006   0.0006   0.0006   0.0006   0.0007   0.0007   -3.2 -3.1   0.0007   0.0007   0.0008   0.0008   0.0008   0.0008   0.0009   0.0009   0.0009   0.0010   -3.1 -3.0   0.0010   0.0010   0.0011   0.0011   0.0011   0.0012   0.0012   0.0013   0.0013   0.0013   -3.0 -2.9   0.0014   0.0014   0.0015   0.0015   0.0016   0.0016   0.0017  ...

  • Find the indicated z-scores shown in the graph Click to view page 2 of the table...

    Find the indicated z-scores shown in the graph Click to view page 2 of the table 0.0212 0.0212 The z-scores are (Use a comma to separate answers as needed. Round to two decimal places as needed) We were unable to transcribe this imagei Standard Normal Table (Page 1) z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 -3.4 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 -3.3 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 00004 0.0005 0.0005...

  • Problem 7 Question Help * David Polston prints up T-shirts to be sold at local concerts....

    Problem 7 Question Help * David Polston prints up T-shirts to be sold at local concerts. The T-shirts sell for $24.20 each but cost David only $8.90 each. However, because the T-shirts have concert-specific information on them, David can sell a leftover shirt for only $1.00. Suppose the demand for shirts can be approximated with a normal distribution and the mean demand is 380 shirts, with a standard deviation of 65. Click the icon to view the normal probability table...

  • A random sample of 20 chocolate energy bars of a certain brand has, on average, 210...

    A random sample of 20 chocolate energy bars of a certain brand has, on average, 210 calories per bar, with a standard deviation of 15 calories. Construct a 99% confidence interval for the true mean calorie content of this brand of energy bar. Assume that the distribution of the calorie content is approximately normal. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. Click here...

  • A survey of 2317 adults in a certain large country aged 18 and older conducted by...

    A survey of 2317 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 401 have donated blood in the past two years. Complete parts (a) through (c) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated...

  • A television sports commentator wants to estimate the proportion of citizens who follow professional football." Complete...

    A television sports commentator wants to estimate the proportion of citizens who follow professional football." Complete parts (a) through (c). Click here to view the standard normal distribution table (page 1) Click here to view the standard normal distribution table (page 2). (a) What sample size should be obtained if he wants to be within 4 percentage points with 95% confidence if he uses an estimate of 54% obtained from a poll? The sample size is (Round up to the...

  • please help me answer this, thank you! In an advertising campaign, a snack company claimed that...

    please help me answer this, thank you! In an advertising campaign, a snack company claimed that every 18-ounce bag of its cookies contained at least 1000 chocolate chips. Two statisticians attempted to verify the claim. The accompanying data represent the number of chips in an 18 ounce bag of the company's cookies based on their study. Complete parts through (e) Click here to view the chocolate chip data table Click here to view the standard normal distribution table (page 1)....

  • A random sample of 16 undergraduate students receiving student loans was obtained, and the amounts of...

    A random sample of 16 undergraduate students receiving student loans was obtained, and the amounts of their loans for the school year were recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. 5,800 2.000 6,200 1,500 6,800 8,400 5,100 3,000 1,800 7,500 3,300 2,500 1,900 5,600 4,500 7,200 Click here to view the table of critical values Click here to view page 1 of the standard normal...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT