Question

Use Table A to find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question. (Round your answers to four decimal places.) (a) z < −0.46 (b) z > −0.46 (c) z < 1.65 (d) −0.46 < z < 1.65

Here is Table A

TABLE A Standard Normal cumulative proportions z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002 3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003 3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010 2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 2.7 .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 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 1.7 .0446 .0436 .0427 .0418 .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 .0708 .0694 .0681 1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823 1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985 1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170 1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379 0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611 0.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867 0.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148 0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451 0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776 0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121 0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483 0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859 0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247 0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641

Answer 1

(a) z < −0.46

using excel function NORMSDIST(z)

setting z = -0.46

we get

proportion = NORMSDIST(-0.46)

= 0.3228

(b) z > −0.46

using excel function NORMSDIST(z)

setting z = -0.46

we get

proportion = 1- NORMSDIST(-0.46) .......(subtracting from 1 for right tailed)

=1- 0.3228

= 0.6772

(c) z < 1.65

using excel function NORMSDIST(z)

setting z = 1.65

we get

proportion = NORMSDIST(1.65)

= 0.9505

(d) −0.46 < z < 1.65

proportion = P(z<1.65) - P(z<-0.46)

using the proportion values calculated in above parts, we get

Required proportion = 0.9505 - 0.3228

= 0.6278