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Complete and post the following problems. 1. We have a population that is normally distributed. A sample of 10 gives an average of 4.00 and a standard deviation of 1.873. What is the probability of pulling a single sample between 3 and 5? Use the posted Table for the Student t distribution. The pdf file is located directly beneath the assignment document. Use that table to solve these two problems. Use linear interpolation to solve (basic algebra) these two problems. 2. From the above problem; what is the probability of pulling a single sample less than 1.6? Table A-2 The , distribution The entry in TOW * (degrees of freedomi under column beading P (probability) is the value 1. for which POSISION = e p 0.05 0.1 0.2 0.25 0.3 0.4 0.45 0.475 0.49 0.495 A I 158 325 .727 1,000 1.376 3.08 6.31 12.71 31.82 63.66 2 142 289 617 816 1.06) 1.89 2.92 4,30 6.96 9.92 3 137 277 584 765 978 1.64 2.35 3.18 4.54 5.84 4 134 271 569 .741 941 1.53 2.13 27% 3.75 4.60 3 132 267 559 727 920 1.48 2:02 2.57 3.36 4.03 6 131 265 553 318 906 1.44 1.94 2.45 3.14 3.71 7 .130 263 549 711 896 1.42 1.90 2.36 3.00 3.50 8 130 262 .546 706 889 1.40 1.86 2,31 2,90 3.36 9 129 261 543 703 883 1.38 1.83 2.26 1.82 3.25 10 129 260 142 700 879 1.37 1.81 2.23 2.76 3.17 11 .129 260 540 697 876 1.36 1.80 2.20 272 3,11 12 128 259 .539 695 873 1.36 1.78 2.18 2,68 3.06 13 12M 259 538 694 870 1.35 1.77 216 2.65 3.01 14 128 258 537 692 868 1.34 1.76 2.14 2.62 2.98 15 128 258 536 691 866 1.34 1.75 2.13 2.60 2.95 16 128 258 535 690 865 1.34 1.75 1.12 2.58 2.92 17 128 257 534 689 863 1.33 1.74 2.11 2.57 2.90 18 127 297 534 688 862 1.33 1.73 2,10 2.55 2.88 19 127 257 .533 .688 861 1.33 1.73 1.09 2.54 2.86 20 127 257 .533 687 860 1.32 1.72 2.09 2.53 2.84 21 127 257 532 686 859 1.32 1.72 2.08 2.53 2.83 22 127 256 532 686 .558 1.32 1.72 2.07 251 2.62 23 127 256 532 685 858 1.32 1.71 2.07 2.50 2.81 24 127 256 531 685 837 1.32 1.71 2.06 2.49 2,80 25 127 256 531 694 856 1.33 1.71 2.06 2.48 2.79 26 127 256 531 684 856 1.32 1.71 2.06 2.48 2.78 27 127 356 531 684 855 1,31 1.70 2.05 2.47 2.77 28 127 256 530 683 855 1.31 1.70 2.05 2.47 2.76 29 127 256 .330 683 .854 1.31 1.70 204 2.46 2.76 30 127 256 530 683 .854 1.31 1.70 2,04 2.46 2.75 40 126 255 529 681 851 1.30 1.68 2.02 2.42 2.70 60 136 254 527 679 848 1.30 1.67 2.00 2.39 2.66 120 .126 254 536 677 845 1.29 1.66 1.98 2.36 2.62 - .126 253 524 674 842 1.28 1.645 1.96 1.33 2.58 Sownvel R. A Fisher and F. Yates, Tollies for anuf Medical published by Longman Group Ltd., London (previously published by Hiver and Bnyd. and by permission of the authors and publishen.

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1.
         
Given that mean = 4.00 and standard deviation (SD) = 1.873

then, for P(3 < x < 5)

t1 =( 3-mean)/SD/(sqrt(n)) = (3-4)/(1.873/sqrt(10)) = -1.69
t2 =( 5-mean)/SD/(sqrt(n)) = (5-4)/(1.873/sqrt(10)) = 1.69

Thus Probability P(3 < x < 5) = P(-1.69 < t < 1.69) .
Since t distribution is symmetric, therefore

P(-1.69 < t < 1.69) = 2P(0 < t < 1.69)

Also sample size is 10. Thus we will check probability and t -value at k = 10-1 = 9 in the table...

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