Question

Assume that the duration of human pregnancies can be described by a normal model with mean 269 days and standard deviation 18 days. Answer the following questions a) What percentage of pregnancies should last between 265 and 280 days? % (Round to one decimal place as needed.)

Answer 1

Ans gn In Normal distribution first we calculate the zo score with the below formula Z=x-M offer calculating the Z-score, with the help of Zitable We look for the corresponding probability value. In z=null M= Mean - 269 days a s standard deviation - 18 days. - Random value for which the probability is I calculated. we need to find the percentage of pregnancies that last between 265 and 280 days, For this first find the probability that a pregnancy lasts between 265 and 280 days, For this Case ;) Take = 265 z= ng , x=265, M= 269, + =18 8=265-269 = -4 = -0.22 - – 0.22 18

The from the 2-table below find the probability value corresponding to a Z-value of -0.22, 7 probability value is 0.4129 Let = 0.4129. Next, Case (ii) Take a= 280 solve for Z. z=x-M Zz 280-269 = 11 a 0.61, 18 From the e-table below find the probability value corresponding to a z-value of 0.61, The probability value is 0.7291. Let = 0.7291 Now observe the Normal curve below

Number in the table represents PIZsz) z0 UUUT ，0.000.01 0.020.030.040.050.060.070.080.09 3,6 |.0002 .0002 .0001 .0001 .0001 .0001 .0001 .0001 .0001 =3.5 | .0002 .0002 .0002 .0002 .0002 .0002 .0002 .0002 .0002 .0002 =3.4 | .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002 -3,3 | ,0005 .0005 .0005 .0004 0 004 .0004 .0004 .0004 .0004 .0003 -3.2| ,0007 .0007 .0006 .0006 .0006 0 006 .0006 .0005 .0005 .0005 -31 ,0010 .0009 .000 .000 .000 .000 .000 .000 .0007 .0007 3.0 [.0013 .0013 .0013 .0012 0 012 .0011 .0011 .0011 .0010 .0010 =2.9 | ,0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 =2. ] ,0026 0 025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -27] ,0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 | ,0047 .0045 .0044 ,0043 .0041 .0040 .0039 .0038 .0037 .0036 -2.5 | ,0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 ,0049 .0048 -2.4.0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 -2.3 | .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 -2.2 | ,01390136 .0132 .0129 .0125 .0122 0119 .0116 .0113 .0110 -21 | ,0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 ,0146 .0143 -2.0 | ,0228 .0222 .0217 .0212 ,0207 .0202 .0197 .0192 ,0188 .0183 -1.བྱ| ,0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 ,0239 .0233 -1. ། ,0359 .0351 .0344 .0336 .0329 -0322 .0314 .0307 .0301 .0294 -j| ,0446 .0436 .0427 .041 .0409 .0401 .0392 .0384 .0375 .0367 -1.6 | .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 ,0465 .0455 -1.5 | ,0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559 1.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .070 .0694 .06B1 -1.3.0968 .0951 .0934 .091 .0901 .0685 .0869 .0B53 .0838 .0823 -1.2|.j151 .1131 .j112 .1093 1075 .1056 .j038 .1020 .1003 .0985 -11.1357 .1335 13141292127112511230 .1210 .j190 .1170 -1.0.1587 .1562 .1539 .1515 .1492 .1469 1 446 .1423 .1401 .1379 -0.9 |.1841 .1814 .1788 17621736 17111685 .1660 .1635 .1611 0.s | .2119 2090 2061 2033 2005 1977 1949 1922 1894 .1867 071 2420230923582327 .2296 .2266 .2236 .2206217 .2148 -0,6 | ,2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451 0.5 |.3085 .3050 3 015 .2981 .2946 .2912 .2877 .2843 .2810 .2776 -0.4 | ,3446 34093372 .3336 3300 326432283192.31563121 པ| 382137833745 .3707 .3669 .36323594 35573520 .3483 0.2| 1207 .4168 .41291.4090 .4052 .4013 .3974 3 936 3897 .3859 01.4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247 -0.0 | .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641

Number in the table represents P(Z<z) 。 z | | ༴ ཀླུ ༔ ༔ ༔ ༔ ཀྵ = ཀྵ ཀྵ བྱཱ ཀྵ ཀྵ བྱཱ ཀྵ བསྶཀྑ .9222 .9265 0.00 0.01་ 0,020.030.040.050.060,070.080.09 5000 .5040 .50305120 .51605199 5239 .527953195359 5398 .5438 54785517 .5557 .5596 .5636 .5675 .5714 5753 5793 .5832 .5B71 5910 .5948 5967 ,6026 .6064 .6103 .6141 03 1617967j7 ,6255 .6293 .6331636864066449.6406507 0.4， .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 6915 enངག .6985 .7019 ,7054 .7088 7123 7157 7190 .7224 .7257 .7291 .7324 .7357 .7389 7422 7454 .7486 7517 7549 ,7580 .7642 .7673 .7704 7734 .7764 .7794 .7823 .7852 738179107939 7967 .7995 -8023 .8051 .8078 .81068133 ,8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 8413 .8438 .8461 8485 .B508 .8531 .8554 .B577 8 599 .8621 .8643 .0665 .86868708 .8729 .8749 .8770_ .8790 .8810 .8830 .B849 .B869 .BBB8 .8907 .B925 .B944 .8962 .B980 .8997 .9015 13 | .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 .9192 .9207 .9236 .9251 .9279 .9292 .9306 .9319 1.5 | .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 9 441 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 17， .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 -9641 .9649 .9656 .9664 .9671 -9678 -9666 .9693 -9699 -9706 .9713 ,9719 .9726 9 732 .9738 .9744 .9750 .9756 .9761 9767 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9608 -9812 .9817 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9654 .9857 -9861 .9864 -9568 .9871 .9875 -3878 -9881 .9884 .9887 .9890 .9893 .9896 .9898 .9904 .9906 .9909 .991 .9913 .9916 | .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 T.9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 .9965 .9966 .9967 .9968 .9969 .9970 ,9971 .9972 .9973 .9974 .9974 .9975 .9976 .9977 .9977 -9978 .9979 -3980 .9981 .991 .9982 .9982 .9903 .9984 .9984 .9985 .9985 .99B6 .99e6 -9987 .9987 -9987 .9968 .9988 .9989 .9989 .9989 -9990 .9990 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993 .9993 .9993 -9994 .9994 .9994 -9994 .9994 .9995 .9995 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9996 .9998 .9998 .9998 .999 .9998 .9998 999B .9998 .9998 9998 .9998 .9998 .9999 .9999 .9999 .9999 9999 .9999 9 999 .9999 .9901 .9979 ཐ ཋ 。aཏྟཱཀྐཤྩཎཱཀྐ .9995

x=265 M = 269 x= 280 The total area under the normal curve above represents the total probability which is equal to 1. The area to the left of x=265= P = 0.4129 The area to the left of x=280 = P = 0.729). The shaded region between x=265 and x2 280 represents the probability that the pregnancy would last between 265 and 280 days. - ② gives the area of the shaded region. (2) - P = 0.7291 – 0.4129 = 6.3162. gn order to find the percentage, Multiply the probability Value with 100, 0.3162 X100 = 31-62%. Percentage of pregnancies should dast between 265 and 280 days = 31.62%.