A random sample of n 100 observations is selected from a population with -30 and

Question

A random sample of n 100 observations is selected from a population with -30 and o 23 Approximate the probabilities shown below b. P(22.1x268) d Pka 27 0) a. P(x2 20) Normal Curve Areas c. Plxs 28 2) A'ick the icon to view the table of nonnal curve areas a.P(x228|-(Round to three decimal places as needed) 0000 3 0636075714 1203 979 293 16641700 1255 1879 2224 1554 1915 1950 985 20192064 2257 2291 25N0211 1591 2123 2157 | 210) 2304 2357 2 238924 2517 264 2910 2909 HITH 31.33 M621 37293749 4222 4236 425 425 429 4319 42924300 4418 409 44 33243454357430 x 4515425 A554 4573 46L6 4750 | 4756 | 4% 4778 4783 4 479 47984 4817 488144 4936 4949 49 938940494 4953 4955 6 4967 4967 49 4975 4976 4977 4977 497 4984 4984 985 1.1. 11 Enter your answer n the answer box and then dick Check An ·0 00 01 0203 04 060 , ol

Explanation / Answer

mean of sample is 30 and standard error is 23/sqrt(100)=2.3

thus

a) P(x>=28)=P(z>(28-30)/2.3)=P(z>-0.87) or P(z<0.87), from normal table we get 0.8078

b) P(22.1<=x<26.8)=P((22.1-30)/2.3<z<(26.8-30)/2.3)=P(-3.43<z<-1.39)=P(1.39<z<3.43)=P(z<3.43)-P(z<1.39) = 1-0.9177=0.0823

c) P(x<28.2)=P(z<(28.2-30)/2.3)=P(z<-0.78) or 1-P(z<0.78)=1-0.7823

d) P(x>27)=P(z>(27-30)/2.3)=P(z>-1.3) or P(z<1.3) = 0.9032

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