Seventy percent of the students applying to a university are accepted. Using the

Question

Seventy percent of the students applying to a university are accepted. Using the binomial probability tables, answer the following questions for a sample of 10 applicants.

A) only 1 is accepted

B) all 10 are accepted

C) at lease 9 are accepted

D) no more that 4 are accepted

E) Between 5 and 8 are accepted ( including 5 and 8)

Find the mean

continued TABLE 2 01 .20 8 0 ,923 ,663 430 272 168 00 .058 075 .279 385 .336 .267 383 198 003 .051 49 238 .294 .296 005 .033 .084 147 208 254 4 .000 .000 .005 .018 .046 .087 36 000 000 .000 .003 009 ,023 047 000 .000 .000 .000 .00 .000 .000 .000 .000 .000 000 .001 000 .000 .000 .000 .000 .914 .630 .387 232 134 040 083 299 .387 .368 302 225 56 .003 .063 72 .260 .302 300 .267 000 008 045 176 .234 .267 007 .028 066 72 000 .000 .001 005 017 ,039 074 000 000 000 00 003 009 .021 000 000 .000 .000 000 004 000 .000 .000 .000 .000 .000 9 .000 .000 .000 .000 .000 000 .000 100 0 904 599 349 97 107 .056 028 09 315 387 .347 .268 188 121 94 .276 .302 .282 233 .000 .010 .057 30 ,20 .250 .267 000 .011 040 088 146 .200 008 026 ,058 03 000 .000 .000 00 006 ,016 ,037 7 .000 .000 .000 .000 .00 003 .009 000 000 000 000 000 ,000 001 000 .000 .000 .000 .000 .000 10 .000 .000 .000 .000 .000 000 .000 032 259 .279 188 .08 022 .003 .000 021 216 272 .219 118 042 010 .000 014 176 252 238 154 .069 02 004 000 40 090 ,209 279 232 124 .041 00 010 060 161 25 .251 167 .074 02 006 040 121 25 .201 .042 002 .45 .008 055 157 257 .263 72 070 .002 005 034 11 260 213 041 008 .001 003 021 .076 66 .238 234 60 075 023 004 031 219 .273 109 03 004 002 018 64 .246 .246 164 018 002 010 .044 205 .246 205 044 .60 65 70 75 .80 .95 .00 000 .000 .000 .000 000 00 000 000 000 .000 008 003 000 .070 041 022 004 001 000 .000 000 172 24 08 047 023 009 .263 .232 188 136 .087 .046 .018 .005 000 .257 .279 279 .254 .208 47 084 .033 005 157 209 259 296 .311 .294 .238 149 051 .090 137 98 .267 .336 .385 .383 .279 .032 .058 68 272 430 .017 .663 00 000 000 000 000 000 000 .000 000 000 .000 041 .021 010 ,004 .001 000 .000 116 074 042 02 009 003 00 000 000 67 118 .074 039 017 005 .00 260 .251 172 17 066 .028 .25 272 .267 .234 176 107 .045 61 216 ,267 300 302 .260 172 034 0600 100 56 225 .302 368 387 .010 .02 .040 .075 .134 232 .387 001 .008 063 299 .630 000 000 000 000 000 000 000 .000 000 004 002 000 000 000 000 000 000 000 023 011 004 000 .000 000 000 075 .042 02 .009 .003 00 160 069 037 006 00 000 000 234 .20 154 03 058 026 008 .001 .238 .251 .238 .200 46 1.088 040 .011 .001 66 .215 .252 .267 .250 20 130 .057 .076 21 176 .233 .282 302 276 194 075 021 040 .072 121 88 .268 .347 .387 .315 .006 014 .028 .056 107 97 .349 .599

Explanation / Answer

Given, p = 0.7, n = 10

A) r = 1, look at the row where n = 10, r = 1 and column where p =0.7

P(X = 1) = 0.000

B) P(X=10) = 0.028

c) P(X >= 9) = P(X = 9) + P(X = 10) = 0.121 + 0.028 = 0.149

d) P(X < = 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0 + 0 + 0.001 + 0.009 + 0.037 = 0.047

e) P(5 <= X <= 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.103 + 0.200 + 0.267 + 0.233 = 0.803

f) Mean = E[X] = n*p = 10*0.7 = 7

g) std. deviation = sqrt(n*p*(1-p)) = sqrt(10*0.7*(1-0.7)) = 1.449

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