39. Proide an appropriate response use the Standard Nrmal Table to find the prot

Question

39. Proide an appropriate response use the Standard Nrmal Table to find the proto Assume that the heights of women are normaly distributed with of 2.5 inches. The cheerleaders for a local professional basketbal lean If a woman is randomly selected, what is the probability that her heigt is must be between 65 5 and 68.0 inches between 65 5 and 68.0ches? a mean of 63.6 inches and a standard deviation A 0.7881 O B. 09608 C. 0 1844 O D. 03112 hrtps:'ditemprod pearsoncmg.comapilv 1pritmot 40. Find the 2-scores for which 98% of the distrbutors area les between-zand OA (0.99,0.99) B. (-2.33,2.33) c. (-1.645, 1.645) OD, (-1,06,1 .96) 3: Standard Normal Table (Page 1) 20.09 0.00 0.07 0.06 0.s 0.04 0.03 0.02 0.01 0.0 -3.40.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 .30.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0 0005 0.0005 -3.2 0.0005 0.00050.0005 0 0008 0.0006 0.0006 0.0008 0.0008 0.0007 0.0007 3.1 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 00009 0.0010 3.00.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0 0012 0.0013 0.00130.0013 2.9 0.0014 0.0014 0.0015 0.00150.0016 0.0016 0.0017 0.001800018 0.0019 2.800019 0.00200.0021 0.0021 0.0022 0.0023 0.0023 0.0024 00025 0.0026 2.7 0.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 00034 0.0035 -2.60.0036 0.0037 0.0038 0.0039 00040 0.0041 0.0043 0.0044 0.00450.0047 -2.50.00480.0049 0.0051 0.0052 0.0054 0.0055 0.0057 0.0059 0.0060 0.0082 -2 2.4 0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0 0078 0 0080 0 0082 2.310.0084 0.0087 0.0089 0.0091 0.0094 0.0098 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 0.0138 0.0139 - 21 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 001660.0170 0.0174 0.0179 -2.0 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0 0228 -1.90.0233 0.0239 0.0244 0.0250 0.0258 0.0262 0.0268 0.0274 0.0281 0.0287 -1.800294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0338 0.0344 0.0351 0 0359 -1.7 0.0367 0.03750 0384 0.032 0.0401 0409 0.0418 00427004360.0446 1.6 0.0455 0.0465 0.0475 0.0485 0.0495 0.0505 0.0516 0.0526 0.0537 0.0548 -1.5 0.0559 0.0571 0.0582 0.0594 0.0605 0.0618 0.0630 0.0643 0.0655 0.0668 - 1.4 0.0681 0.0894 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 0.10200.1038 0. 1058 0.1075 0.1093 0.1112 0.1131 0.1151 -1.10.1170 0.1190 0.12100.1230 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 0.16350.1660 0.1685 0.1711 0.1736 0.1762 0.1788 0.18140.1841 -0.8 0.1867 0.1894 0.1922 0.1949 0.1977 0.2005 0.2033 0.2061 0 2090 0.2119 -0.7 0.2148 0.21770.2208 0.2236 0.2266 0.2296 0.2327 0.23580 2389 0.2420 0.60.2451 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 0.3158 0.3192 0.3228 0.3264 0.3300 0.3336 0.3372 0.3409 0.3446 0.3 0.3483 0.3520 0.35570.3594 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 4920 0.4960 0.5000 -0.0 0.4641 0.4681 0.4721 0.4761 0.4801 0.4840 0.4880 0. z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00

Explanation / Answer

1) probability of height between 65.5 and 68 =P(65.5<X<68)=P((65.5-63.6)/2.5<Z<(68-63.6)/2.5)=0.9608-0.7764

=0.1844

option C

40)

option B is correct

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