Question

5.In a certain Asian country, the average weight of men is 60 kg and its standard deviation is 8 kg. The average of women is 52 kg and its standard deviation is 5 kg. If we select one man and one woman randomly, (5.1) What is the probability that the sum exceeds 100 kg? (5.2) What is the probability that the womans weight exceeds the mans weight? (53) What is the probability that the womans weight is at least ¾ of the mans weight?

Please use the following distribution table to get the answer!

(b) Table 1. Cumulative distribution function for the standard normal distribution (continued). The table gives the area under the standard normal probability curve between z-andz z Example If z1 2.56, then: P(-00 < z < 2.56) 0.99477 0 z 2 0.00 0.00 0.50000 0.50399 0.50798 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.53586 0.10 0.53983 0.54380 0.54776 0.55172 0.55567 0.55962 0.56356 0.56749 0.57142 0.57535 0.20 0.57926 0.58317 0.58706 0.59095 0.59483 0.59871 0.60257 0.60642 0.61026 0.61409 0.30 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.65173 0.40 0.65542 0.65910 0.66276 0.66640 0.67003 0.67364 0.67724 068082 0.6849 1.68793 0.50 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.72240 0.60 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.75490 0.70 0.75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.78524 0.80 0.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.81327 0.90 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.83891 1.00 0.84134 0.84375 0.84614 0.84849 0.85083 0.85314 0.85543 0.85769 0.85993 086214 1.10 0.86433 0.86650 0.86864 0.87076 0.87286 0.87493 0.87698 0.87900 0.88100 0.88298 1.20 0,88493 0.88686 0.88877 0.89065 0.89251 0.89435 0.89617 0.89796 0,89973 0.90147 1.30 0.90320 0.90490 0.90658 0.90824 0.90988 0.91149 0.91308 0.91466 0.91621 0.91774 1.40 0.91924 0.92073 0.92220 0.92364 0.92507 0.92647 0.92785 0.92922 0.93056 0.93189 1.50 0.93319 0.93448 0.93574 0.93699 0.93822 0.93943 0.94062 0.94179 0.94295 0.94408 1.60 0.94520 0.94630 0.94738 0.94845 0.94950 0.95053 0.95154 0.95254 0.95352 0.95449 1.70 0.95543 0.95637 0.95728 0.95818 0.95907 0.95994 096080 0.96164 0.96246 0.96327 1.80 0.96407 0.96485 0.96562 0.96638 0.96712 0.96784 0.96856 0.96926 0.96995 0.97062 .90 0.97128 0.97193 0.97257 0.97320 0.97381 0.97441 0.97500 0.97558 0.97615 0.97670 2.00 0.97725 0.97778 0.97831 0.97882 0.97932 0.97982 0.98030 0.98077 0.98124 0.98169 2.10 0.98214 0.98257 0.98300 0.98341 0.98382 0.98422 0.98461 0.98500 0.98537 0.98574 2.20 0.98610 0.98645 0.98679 0.98713 0.98745 0.98778 0.98809 0.98840 0.98870 0.98899 2.30 0.98928 0.98956 0.98983 0.99010 0.99036 0.99061 099086 0.99111 090 ЛЯ9158 2.40 0.99180 0.99202 0.99224 0.99245 0.99266 0.99286 0.99305 0.99324 0.99343 0.99361 2.50 0.99379 0.99396 0.994 13 0.99430 0.99446 0.99461 099477 0.93492 D99505 2.60 0.99534 0.99547 0.99560 0.99573 0.99585 0.99598 0.99609 0.99621 D99632 99643 2.70 0.99653 0.99664 0.99674 0.99683 0.99693 0.99702 0.99711 0.99720 0.99728 0.99736 2.80 0.99744 0.99752 0.99760 0.99767 0.99774 0.99781 0.99788 0.99795 0.99801 0.99807 2.90 0.998 13 0.998 19 0.99825 0.99831 0.99836 0.99B41 0.99816 0.99851 D99856 09981 0.04 0.06 0.07 0.08 0.09

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Answer #1

5:

Let X shows the man weight and Y shows the woman weight. Here we have

mu_{X}=60,sigma_{X}=8,mu_{Y}=52,sigma_{Y}=5

1:

Let T= X+Y

Here T will have normal distribution with mean

mu_{T}=mu_{X}+mu_{Y}=60+52=112

and standard deviation

sigma_{T}=sqrt{sigma^{2}_{X}+sigma^{2}_{Y}}=sqrt{64+25}=9.434

The z-score for T = 100 is

T HT 100 - 112 9.434

The required probability is

P(T > 100) = P(z > -1.27) = 1 - P(z <= -1.27) = 1- 0.102 = 0.8980

(2)

Let D =X-Y

Here D will have normal distribution with mean

mu_{D}=mu_{X}-mu_{Y}=60-52=8

and standard deviation

sigma_{D}=sqrt{sigma^{2}_{X}+sigma^{2}_{Y}}=sqrt{64+25}=9.434

The z-score for D = 0 is

0-80.85 -9.434

The required probability is

P(D < 0) = P(z < -0.85) = 0.1977

(3)

Let W =(3X /4)-Y

Here W will have normal distribution with mean

mu_{W}=rac{3mu_{X}}{4}-mu_{Y}=-7

and standard deviation

16

The z-score for W = 0 is

0+0.90 7.810 --

The required probability is

< 0 P(z < 0.90) = 0.8159

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