Question

5. At a local high school 5000 juniors and seniors recently took an aptitude test. The results of the exam were normally distributed with mean = 450 and = 50. Calculate the following: a. The PERCENT of students to the nearest tenth of a percent that scored over 425 b. The number of students that scored more than 475 C. The probability of a student selected at random having scored between 400 and 575 A statistics instructor recorded the grades of his students on the final exam. The grades are: 6.

Answer 1

Given h=450 6. So XN Noronal distributions so 2-x- follows standard normal distribution 2 = x-450 a) pcx>425) = P(x-> 425-45°) =P (2> - / - Anea Ene shaded roegios % 2=0 P-%<22) + Ploca<02 PROC2</2) + 0.5 Cheve in o.lgist 0.5 from Note: ② Since symmetric Pry<2<) -proca<42 Procar) = 42 table =a.6915 = 69.15% ~169.2% b) P(x>475) = P(x-45o > 475-650) = P12 > Y =proc2400)-Ploc 2<42) 2=0 2= Ya 0.1915 - 0.5 - 0.3085 = 30.85% a. numbero SOUOX 30-85 studento = f (1543

<< sys-° ) opt ves < x < səs) =pt tv - so so So = PC -1 42 ) 2 - 20 - Р(-12 ) + Ploca-s) PC o<2412)+ Рcocces) 0.3413 + С. 4938 = ( 0- 835)

0.4177 04279 1948 Table - 1 0.03 0.00 0.07 0.04 0.01 0.08 0.03 0.02 0.00 0.06 0.0040 0000000 0 0000 0.016 00120 00350 00199 0 0239 0.0019 0.0270 00439 00478 0 0596 0 0517 OOS 0039 01 00075 0 0536 0 0714 0053 02 00793 00012 00071 0 0910 0.0011 00087 0.1026 0.1054 0.1103 0.1141 03 01255 0 1129 01293 0.1217 0.1480 0.1331 0.1360 0.1517 0.1406 0.1443 GINS 01591 0 1591 01620 01664 0 1700 1736 0 1772 0.1800 0.1844 0169 1919 0.1950 1983 0.2019 0.2054 0 2008 0.2123 0.2157 0.2100 0.2224 08 0.2257 0.2291 0 2324 02357 02380 07472 02454 02486 0 2517 0.2540 02580 07 02611 0 2642 0.2673027040.2734 02784 02794 02852 0.2823 0 2881 02910 0.2939 0.2967 0295 0.3023 0.3051 03078 0.3133 15 03186 03212 0 3230 0.3264 0.329 0.3315 0 3340 03365 0.338 10 0321303438 0 3461 0 3485 0350803531 03554 0.3577 0.3599 03621 03523 0 3665 0.3686 0 3708 03729 0 3740 0 3770 0 3700 03810 03630 1.2 03849 03069 0.3868 0 3907 0 3925 0.3944 03982 03980 0.3997 0.4015 13 0 4032 04049 0.4066 0.4082 0.4009 0.4115 0.4131 0.4147 0.4162 1.4 04192 0.4207 0.4222 0.4236 04251 0.4265 04292 0.4306 0.4319 1.5 0.4332 04945 0435 04370 0.432 04394 0.4406 0.4418 04429 0.4441 04452 04161 0.4474 0.4484 0.4495 0.1505 0.4515 0.4525 04535 0.4545 0.4554 0.4564 0.4573 04582 0.4591 0 4500 0 4500 4608 0.4616 0.4625 0.4633 0441 0 46.49 0.4856 04664 0.4671 04676 0.4686 04693 04670 0.4700 1.9 0.4713 0.4719 0.4726 0 4732 0.4738 0.4750 0.4756 0.4761 0.4767 20 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0 4812 0.4817 21 0.4821 0.4.20 0.4830 0.4834 0.4838 0.4842 0.4846 04850 04854 0.4857 22 0.4861 0.484 0.4868 0.4871 0.4875 0.4881 0.4878 04584 0.4890 0.4867 0.4893 23 0.4896 0.4895 0.4901 0.4904 0.4006 0.4909 0.4911 0.4913 0.4916 22 0.41 0.4920 04922 0.4925 0.4927 0.4929 0.4931 230 4938 0.4932 0.4934 0.4936 04940 0.4941 0.4943 0.4945 0.4946 0.1953 0.4949 0.4951 0.4952 04955 0.4956 0.4957 0.4959 04960 0.4961 04962 04963 0.4964 27 0.4965 0 4966 0.4967 0.496 0.4969 0.407004971 0.4974 28 0.4972 0.4974 0.4973 0.4975 0.4978 0.4977 0.4977 0.4978 0.4979 0.4979 04980 0.4981 29 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 3.0 4986 0.4987 04987 0.4988 0.4987 0.4988 0.4989 0.4989 0.4989 04990 0.4990 0.4873 Menn 22.24 03100 Areas under the Standard Normal Probability Distribution between the Mean and Positive Values of Example: To find the area under the curve between the mean and a point 2.24 to the right of the mean, look up the value opposite 22 and under 0,04 in the table: 04575 of the area under the curve lies between the mean anda z value of 2 24