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Homework Solutions 06 - (W4150 Intro to Probability and...

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Solutions to Homework #6 (W4150 ‘Intro to Probability and Statistics’, S04) Sec. 6.4. (1) Use tables or built-in Excel function for normal distribution to answer each question here: (a) Area = Φ (1.43) = .9236 (b) Area = 1 - Φ (-.89) = .8133 (c) Area = Φ (-.65) - Φ (-2.16) = .2424 (d) Area = Φ (-1.39) = .0823 (e) Area = 1 - Φ (1.96) = .0250 (f) Area = Φ (1.74) - Φ (-.48) = .6435 Sec.6.4. (3) (a) From table A.3, k = -1.72 (b) Since P(Z > k) = .2946 then P(Z < k) = 1 – P(Z > k) = .7054; and from the same table we find k = .54 (c) The area to the left of z = .83 is found from table A.3 (or Excel) to be .1762. Therefore, the total area to the left of k is (.1762) + (.7235) = .8997 ° k = 1.28 Sec. 6.4. (4) (a) z = (17 – 30)/6 = -2.17 ° Area = 1 - Φ (-2.17) = 1 - .0150 = .9850 (b) z = (22 – 30)/6 = -1.33 ° Area = Φ (-1.33) = .0918 (c) z 1 = (32 – 30)/6 = .33, z 2 = (41 – 30)/6 = 1.83 ° Area = Φ (1.83) - Φ (.33) = .9664 - .6293 = .3371 (d) z = .84 ° x = 6(.84) + 30 = 35.4 (e) z 1 = -1.15 and z 2 = 1.15 ° x 1 = 6(-1.15) + 30 = 23.1 and x 2 = 6(1.15) + 30 = 36.9 Sec. 6.4. (6) Z 1 = [( μ - 3 σ ) - μ ]/ σ = -3; Z 2 = [( μ + 3 σ ) - μ ]/ σ = 3. Hence P( μ - 3 σ < X < μ + 3 σ ) = P(-3 < Z < 3) = Φ (3) - Φ (-3) = .9987 - .0013 = .9974 Sec. 6.4. (14) Everywhere here you’re supposed to respect the fact that the heights were measured within precision of + .25cm (i.e. 0.5cm in total) and therefore every time you recalculate the r.v. Z into standard normal you have to subtract .25cm from the actual value if you’re looking for
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