Question

The standard deviation of the scores on a skill evaluation test is 320 points with a...

The standard deviation of the scores on a skill evaluation test is 320 points with a mean of 1434 points.

If 338 tests are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 43 points? Round your answer to four decimal places.

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Answer #1

Given that the scores has mean, say overline{x} = 1434 and S sigma = 320.

For sample size, n = 338 (large), by Central Limit Theorem,

overline{x}sim N(mu ,sigma/sqrt{n} )

Hence, The probability that the mean of the sample would differ from the population mean by less than 43 points

= Pl--μ < 43

= (r-μ) 43 P(

=  43 320/V338......(Since, N(0, 1 7l and Z is the standard normal variate)

=  P(Z <2.47)

From Standard normal tables,

STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score 0.0 50000 .5039950798 .51197 .51595 51994 .5239252790.53188 .53586 0.153983 5438054776 .55172 .55567.559625635656749.57142 57535 0.2 57926 .58317.58706 59095 .59483.5987 60257 60642 0.3 6179 62172.62552 62930 63307 63683 64058 6443164803 65173 0.4 65542 65910 66276 66640 67003 6736467724 68082 68439 68793 0.5 69146 6949769847 .70194 .70540 .70884 .71226 .71566.71904 72240 7 3 7 1 0.7 75804 76115 76424 76730 .77035 .77337 77637.77935 .78230 78524 0.8 .78814 .79103.79389.79673 .7995580234 80511 8078581057 81327 82639 82894 83147 83398 83646 83891 1.0 84134 84375 84614 84849 .85083 85314 85543 8576985993 86214 1 86433 86650 86864 87076 87286 8749387698 8790088100 88298 1.2 88493 88686 88877 89065 .8925 89435 896178979689973 90147 1.3 90320 90490 90658 .90824 90988 91149 91309 91466.91621 .91774 1.4 .91924 92073 92220 92364 92507 92647 92785 92922 93056 93189 1.5 93319 93448 93574 93699 93822.93943 94062 94179.94295 94408 1.6 94520 94630 94738 .94845 94950 .9505395154 95254.95352 95449 1.7 95543.95637 95728 .958189590795994 96080 9616496246 96327 1.8 96407 96485 96562 .96638 9671296784 96856 96926.96995 97062 2 1 2.0 97725 .97778 .9783 97882 .97932 97982 98030 9807798124 .98169 2.198214 .9825798300 98341 .98382 98422 9846 9850098537 .98574 2.2.98610 .98645.98679 98713.9874598778 98809 98840.98870 98899 2.3 98928 98956 98983 99010 99036 99061 99086 .99111 99134 99158 2.5.99379 .99396 .99413 99430 .99446 9946 99477 99492.99506 .99520

P(Z <2.47) = 0.99324 simeq 0.9932

Hence, the probability that the mean of the sample would differs from the population mean by less than 43 points is 0.9932

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