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

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