Chapter 4 Solutions

4.9:

a: 1 - .6306 = .3694 or 36.94%

b: 2525 or 25.25%

c: 9088 or 90.88%

d: 1 - .0334 = .9666 or 96.66%

e: .9088 - .6306 = .2782 or 27.82%

f: .2525 -.0334 = .2191 or 21.91%

g: .6306 - .2525 = .3781 or 37.81%

4.10:

a: 1 - .5517 = .4483

b: .8708 - .5517 = .3191

4.11

(a) In Table 3, the area closest to .8 is .7995, corresponding to z = .84. Thus the 80^th percentile y* must satisfy the equation .84 = (y* - 176)/30, which yields y* = (30)(.84) + 176 = 201.2

(b) The 20^th percentile y* satisfies the equation .84 = (y* - 176)/30, which yields y* = (30)(-.84) + 176 = 150.8

4.22a: You want to include the rectangle 40 in the region because the statement says “40 or more bristles”. This means that the continuity correction should be “Y > 39.5” in order to include Y = 40. Since the number of bristles is discrete, and you are using a continuous (normal) model, you have to be aware that Pr(Y >= 40) = Pr(Y = 40) + Pr(Y = 41) + Pr(Y = 42) + …… Each of these probabilities are represented by the area of some rectangle (compare this with the rectangles shown in Figure 4.31 on page 145. The example and scale are different but the idea is the same.) You want to capture the area of the entire rectangle Pr(Y = 40), so you have to start measuring the area from the boundary at Y = 39.5 on. So, when you take the area under the normal curve, you should start at 39.5, not at 40. Then Pr(Y > 39.5) = Pr(Z > (39.5-38.5)/2.9) =(approx) Pr(Z > .34) = 1 - .6331 = .3669.

4.22b: Here you want to use the continuity correction Pr(Y = 40) = Pr(39.5 < Y < 40.5) (approx) = Pr(.34 < Z < .69) = .7549 - .6331 = .1218 or 12.18%.

4.22c: Since the number of bristles is discrete, and you are using a continuous (normal) model, you have to be aware that Pr(35 < = Y <= 40) = Pr(Y = 35) + Pr(Y = 36) + Pr(Y = 37) + Pr(Y = 38) + Pr(Y = 39) + Pr(Y = 40). Each of these probabilities is represented by the area of some rectangle (compare this with the rectangles shown in Figure 4.31 on page 145. The example and scale are different but the idea is the same.) You want to capture the area of the entire rectangle Pr(Y = 35), so you have to start measuring the area from the boundary at Y = 34.5 on. So, when you take the area under the normal curve, you should start at 34.5, not at 35. Similarly you want to end with the entire rectangle Pr(Y=40), so you end at the boundary Y = 40.5. Then Pr(35 <= Y <= 40)) = Pr(34.5 < Y < 40.5) = (approx) Pr(-1.38 < Z < .69) = .7549 - .0838 = .6711 or 67.11%.

4.23

(a) This is the same as part (c) of Exercise 4.22.

Pr{35 <= Y <= 40} = Pr{34.5 <= Y <= 40.5} = Pr{(34.5-38.5)/2.9 < Z < (40.5 – 38.5)/2.9}. This is the probability that Z is between –1.38 and +.69,

which is .7549 - .0838 = .6711.
(b) Pr{35 < Y < 40} = Pr{(35-38.5)/2.9 < Z < (40 – 38.5)/2.9}. This is the probability that Z is between –1.21 and +.52, which is .6985 - .1131 = .5854. This does not agree very well with the answer from part (a).

4.24 [a] P(Y < 6) = P(Y < 6.5) = P(Z < -1.3/2.3) = P(Z < -.57) = .2843 = 28.43%

[b] P(Y = 6) = P(5.5 < Y < 6.5) = P(-1 < Z < -0.5652) = P(Z < -0.5652) - P(Z < -1) =0.1273 = 12.73%

[c] P(8 < Y < 11) = P(7.5 < Y < 11.5) = P(-.3/2.3 < Z < 3.7/2.3) = .4980 = 49.80%

4.25a: z = (6.55 – 6.85)/.42 = -.71; Table 3 gives .2389;

z = (6.45 – 6.85)/.42 = -.95; Table 3 gives .1711;

.2389 - .1711 = .0678.

4.25b: z = (8.05 – 6.85)/.42 = 2.86; Table 3 gives .9979

.9979 - .1711 = .8265.