Can someone help me please with number 6 I dont understand how my teacher got th

Question

Can someone help me please with number 6 I dont understand how my teacher got the answer it makes no sense to me. I have attached the chart to reference the number values. and the answer key but I dont know how she got that answer.

Can you please explain it to me? Thanks!!

TABLE A.2 Normal distribution The tabled entries represent the proportion p of the total area under the curve that is in the tail of the normal curve, to the right of the indicated value of z. (Example: .0694 or 6.94% of the area is to the right of z-148. This is found by using the z-1.4 row, and the 0.08 column, of the table.) If the value of z is negative, the tabled entry corresponding to the absolute value of z represents the area less than z. (Example: .3015 or 30.15% of the area is to the left of z -0.52 and this is found by using z = +0.52 in the table.) 0.0 5000 4960 4920 4880 4840 4801 4761 47214681 4641 0.2 4207 4168 412940904052 4013 3974 3936 3897 .3859 0.3 3821 .3783 3745 .3707 .3669 3632 3594 35573520 3483 0.4 3446 3409.3372 3336 3300 3264 3228 3192 3156 3121 0.5 3085 .3050 3015 2981 2946 .2912 .28772843 .2810 2776 0.6 .2743 2709 2676 2643 .2611 2578 2546 2514 2483 2451 0.7 .2420 .2389.2358 .2327 2297 .2266 .2236 .2206 .2177 .2148 0.8 2119 .2090 .2061 2033 2005 1977 .1949 19221894 .1867 0.9 .1841 .1814 1788 1762 1736 1711 1685 1660 .1635 .1611 1.0 .1587 .1562.1539 1515 1492 1469 1446 .1423 1401 1379 1.1 1357 1335 .1314 1292 .1271 .1251 .1230 1210 1190 1170 1.2 .1151 .1131.1112 1093 .1075 .1056 .1038 .1020.1003 .0985 1.3 0968 .0951 0934.0918 0901 0885 .0869 0853 0838 0823 1.4 .0808 0793 0778 0764 .0749 .0735 0721 0708 0694 0681 1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 ,0559 1.610548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 1.7 0446 0436 0427 0418 .0409 0400392 0384 0375 0367 1.8 .0359 .0351 .0344 0336 .0329 0322 .0314 0307 0301 0294 19.0287 .0281 .0274.0268 .0262 .0256 .0250 .0244 0239 .0233 2.0 .0228 .0222 0217 .0212 .0207 .0202 0197 0192 .0188 0183 2.1 .0179 .0174 .0170 .01660162 .0158 .0154 0150 0146 .0143 2.21.0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 2.31.0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 2.4 0082 0080 0078 0075 0073 .0071 .0069 0068 .0066 0064 2.5 0062 .0060 .005900570055 .0054 .0052 0051 .0049 0048 2.6 .0047 .0045 .0044.0043.0041 .0040 .0039 0038 0037 .0036 2.7 0035 .0034 .0033 0032 0031 .0030 .0029 0028 0027 .0026 8 2.9 0019 .0018 .0018 0017 .0016 0016 .0015 .0015 .0014 .0014 .0 .003 .0013 0013 0012 0012 0011 0011 0011 .0010 .0010 Adapted with rounding from Table il of Fisher and Yates 1974

Explanation / Answer

6)

P( -2.1 < Z < 1.01) = P( Z < 1.01) - P( Z < -2.1)

= P(Z < 1.01) - (1 - P(Z < 2.1))

In Z table, find probabilities for Z < 1.01 and Z < 2.1 we get

= 0.8438 - ( 1 - 0.9821)

= 0.8259.

= 82.59%

(Given table represents probabilities right to the z value.

So by using above z table,

Given in table , P(Z > 1.01) = 0.1562

Therefore P(Z < 1.01) = 1 - 0.1562 = 0.8438

And P( Z < -2.1) = P( Z > 2.1) = 0.0179.

P(-2.1 < Z < 1.01) = P( Z < 1.01) - P( Z < -2.1)

= 0.8438 - 0.0179

= 0.8259

= 82.59% )

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