4.5[a] P(Y > 80) = P(Z > -8/7) = P(Z < 8/7) = .8735 = 87.35%
[b] P(Y > 90) = P(Z > 2/7) = P(Z < -2/7) = .3875 = 38.75%
[c] P(Y < 75) = P(Z < -13/7) = .0316 = 3.16%
[d] P(75 < Y < 90) = P(-13/7 < Z < 2/7) = P(Z < 2/7) - P(Z < -13/7) = .5808 = 58.08%
[e] P(90 < Y < 100) = P(2/7 < Z < 12/7) = P(Z < 12/7) - P(Z < 2/7) = .3443 = 34.43%
[f] P(75 < Y < 80) = P(-13/7 < Z < -8/7) = P(Z < -8/7) - P(Z < -13/7) = .0949 = 9.49%
4.6 [a] P(Y > 90) = P(Z > 2/7) = P(Z < -2/7) = .3875 = 38.75% (Same as 4.5[b])
[b] P(75 < Y < 90) = P(-13/7 < Z < 2/7) = P(Z < 2/7) - P(Z < -13/7) = .5808 = 58.08% (Same as 4.5[d])
4.8 [a] The 60th percentile is the yield that will give a left tail of 60%. This has a z-score of Z = 0.2533.
(Notice that P(Z < 0.25) = .5987 and P(Z < 0.26) = .6026. Interpolating one third of the way gives us and estimate of P(Z < .02533) = .6000).
This means the the yield is .2533 SDs above the mean, or Y = 2.533(7) + 88 = 89.77
[b] The 35th percentile has a z-score of Z = -0.3853 because P(Z < -0.3853) = .3500. Thus Y = -0.3853(7) + 88 = 85.30
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 80th percentile y* must satisfy the equation .84 = (y* - 176)/30, which yields y* = (30)(.84) + 176 = 201.2
(b) The 20th percentile y* satisfies the equation .84 = (y* - 176)/30, which yields y* = (30)(-.84) + 176 = 150.8
4.13a: 1 - .9394 = .0606
4.13b: .3669
4.13c: .7257 - .1003 = .6254
4.14: To get the middle 90%, you chop off 5% off the left (z_{0.05}) and 5% off the right (z_{0.95}). This will get you the range from 2.31 cm to 4.05 cm.
4.16:
Y = run time of 3700 runners of 1996
mu = 230 minutes, sigma = 36 minutes
(a) percent greater than 200 minutes. For Y = 200, Z = (200 - 230) / 36 = -0.83,
= Pr{Y > 200}
= Pr{Z > -0.83} since (200 - 230) / 36 = -0.83
= 1 - 0.2033 = 0.7967 = 79.67%
(b) 60th percentile of the times - (from the normal table, find area nearest 60% - gives z = 0.25)
z = 0.25, so Y = 230 + (0.25)*(36) = 239 min.
(c) 240 minutes = 4 hours. More runners than expected finished in just under 4 hours; fewer than expected finished in just over 4 hours. This may be due to some runners pushing themselves to break the 4 hour mark.
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.
4.27: In Table 3, the areas closest to .95 are .9495, corresponding to z = 1.64, and .9505, corresponding to z = 1.65.
Interpolating between z = 1.64 and z = .165, we have z = 1.645. The 95th percentile y* satisfies the equation
1.645 = (y* .38)/.03
which yields y* = (.03)(1.645) + .38 = .4294 mm.
4.28a: z = -2; area = .0228 or 2.28%
4.28b: Pr{no thin-shelled eggs} = (.9772)^12 = .7582.
1 - .7582 = .2418, so 24.18% of the boxes will contain at least one thin-shelled egg.
4.38: Using continuity correction, Pr{80 <= Y <= 140} = Pr{79.5 < Y < 140.5} = Pr{ -1.281 < Z < 2.531} = .9943 - .1001 = .8933
4.39: From Table 3, the probability that a single child has a score of 80 or less is, using continuity correction, Pr(Y < = 80) = Pr(Y < 80.5) = Pr(Z < -1.21875) = .1115,
Pr{one 80 or less, four higher than 80} = 5C1(.1115)^1(.8885)^4 = .3474
4.41a: Using
the continuity correction,
Pr(Y
> 10) = Pr(Y > 10.5) = Pr(z > (10.5
7.3)/11.1) =
normalcdf((10.5 7.3)/11.1,99999999) = 0.3866.
The
answer may differ slightly if you use Table 3 (Ans =
0.3859)
4.41b: Using
the continuity correction,
Pr(Y
> 20) = Pr(Y > 20.5) = Pr(z > (20.5
7.3)/11.1) =
normalcdf(20.5,99999999,7.3,11.1) = 0.1172
The
answer may differ slightly if you use Table 3 (Ans =
0.1170)
4.41c: Using
the continuity correction,
Pr(5 < Y < 15) = Pr(5.5 < Y < 14.5) =
normalcdf(5.5,14.5,7.3,11.1) = 0.3061
The
answer may differ slightly if you use Table 3 (Ans =
0.7422 0.4363 = 0.3059)
4.42: Using the continuity correction,
Pr(Y
< 0) = Pr( Y < -0.5) =
normalcdf(-9999999,-0.5,7.3,11.1) = 0.2411
The
answer may differ slightly if you use Table 3 (Ans
=0.2420)
4.43: Pr(0
< Y < 15) = 0.7549 0.2546 = 0.5003.
Thus we expect (400)(0.5003), or about 200
observations to fall between 0 and 15.
Note that beats per minute Y, which is measured in beats per minute, is
a discrete random variable. Since a discrete random variable is being modeled by a continuous
curve (normal curve), the continuity correction should be applied. Thus the corrected solution is Pr(0.5 < Y < 14.5) = 0.4717; hence we expect np = 400(.4717) = 189, or so observations.
4.44: The
IQR is 14.80 (-0.2) = 15.0. An outlier
on the high end of the distribution is any point greater than 14.80 + 15.0 =
39.8. An outlier on the low end of the
distribution is any point less than 0.2 15 = 15.2.
4.45
I histogram looks kind of normal so it goes with (b) (or possibly with (c)).
II histogram has a right skew, so by Figure 4.29(a) and (b) (p.142) it goes with (d).
III histogram has long tails (see Figure 4.27 on p.141), so it goes with (a).
4.?? (This was the practice quiz which was distributed in class on 1 February) The Grade Point Average (GPA) of students in a certain college follow the normal curve with an average of 2.50 and standard deviation of 0.50.
(a) What percentage of the students have GPA's between 2.10 and 2.90?
For Y = 2.10, Z = (2.10 - 2.50) / 0.50 = -0.80
For Y = 2.90, Z = (2.90 - 2.50) / 0.50 = 0.80, so ans = 0.7881 - 0.2119 = 0.5762 = 57.62 %
(b) What percentage of the students have GPAs lower than 1.80?
For Y = 1.80, Z = (1.80 - 2.50) / 0.50 = -1.40, so ans = 0.0808 (8.08%)
(c) To make the dean's list one needs a GPA of 3.4 . What percentage of the students
make the dean's list.?
For Y = 3.4, Z = (3.4 - 2.5) / 0.50 = 1.80, so ans = 1 - .9640 = .0359 (3.59%)
(d) Find the 90th percentile GPA.
From the table (finding 90% in the center of the table), Z = 1.28, so
ans = Y = 2.50 + (1.28)*(0.50) = 3.14 (gpa score)
(e) Find the GPA that only 2% of the students exceed.
Looking for the 98th percentile. From the table, Z = 2.05, so
ans = Y = 2.50 + (2.05)*(0.50) = 3.525 (gpa score)