Solution for Homework 3
3.2 (a) The density curve forms a 1*1 square, which has area 1.
(b) 20%. The region is a rectangle with height 1 and base width 1 - 0.8 = 0.2.
(c) 60%. The region is a rectangle with height 1 and base width 0.6.
(d) 50%. The region
General addition rule
For any two events A and B, p(A or
B) = p(A) + p(B) - p(A and B)
Conditional Probability
p(B/A) = p(A and B)
p(A)
General Multiplication Rule
For any two events A and B
p(A and B) = p(A) x p(B/A),
or
p(A and B) = p(B) x p(A/B)
Indepe
Solutions of homework 10
t = 2.015.
(b) t = 2.518.
16.3 (a)
16.9 (a) df = 24.
(b) t = 1.12 is between 1.059 (p =0.15) and 1.318 (p = 0.10).
(c) Double the value of P because Ha is two-sided: The P-value is between 0.30and 0.20.
(d) t = 1.12 is not signifi
Solutions of homework 6
4.2 Water temperature is explanatory, and weight change (growth) is the responsive
variable. Both are quantitative.
4.6 (a) Graph is omitted. Speed is explanatory.
(b) The relationship is curved low in the middle, higher at the ext
Solutions of homework 9
14.2 (a) If
= 115, the distribution is approximately Normal with mean
= 115 and standard deviation /
25 = 6.
(b) The actual result lies out toward the high tail of the curve, while 118.6 is fairly close to the middle. If
= 115, obs
Solutions of homework 8
10.19. In the long run, the gambler earns an average of 94.7 cents per bet. In other words, the
gambler loses (and the house gains) 5.3 cents for each $1 bet.
10.22. (a) Normal with mean 123 mg and standard deviation / 3 0.0462 mg.
Solutions of homework 7
5.30 Note that y = 46.6 + 0.41. We predict that Octavio will score 4.1 points above the
mean on the final exam: y = 46.6 + 0.41( x + 10) = 46.6 + 0.41 x + 1.4. (Alternatively,
because the slope is 0.41, we can observe that an incre
Solution of homework 4
3.24
(a) About 5%: x < 240 corresponds to z < (240-266)/16 = 1.625. Table A gives
5.16% for 1.63 and 5.26% for 1.62.
(b) About 55%: 240 < x < 270 corresponds to (240-266)/16 < z < (270-266)/16,
which is 1.625 < z < 0.25. The area to
Solution of homework 5
10.24
(a) Let X be Shelias measured glucose level. P (X > 140) = P(Z > (140-125)/10) = P
(Z > 1.5) = 0.0668
(b) If x is the mean of four measurements (assumed to be independent), then x has a
N(125, 15/ 4 ) = N(125, 7.5) distributio
Solution for Homework 11
Extra Problem Sheet:
9 (a) X = amount of cereal in the small bowl
Y= amount of cereal in the large bowl
T = X + Y (total amount)
E (T) = 4.0 and SD (T) = 0.5.
P (T > 4.5) = P (Z > 1) = 0.1587
(b) D = Y X (difference)
E (D) = 1 and
Solution for Homework 12
20.2 (a) The percentages, respectively, are 11/20 = 55%, 68/91 74.7%, and 3/8 =
37.5%.
(b) Some (but not too much) time spent in extracurricular activities seems to be
beneficial.
20.4 (a)
<2
2 to 12
>12
Total
C or better
13.78
62
Math 171 Prelim #1 Solutions
1. (a) Let X be the monthly income of a randomly chosen male Cornell graduate. Let Y
be the monthly income of a randomly chosen female Cornell graduate. The total
monthly income is then X + Y and the dierence is X Y . (Note th
LECTURE 1
Solution to homework one
Sec 4.2 problem 4,8
4: Use T for tails and H for heads.
(a): thesamplespacc is {(H,H,H) H H,T) (H,T, H), (H, T,T), (T, H, H) (T,H,T), (T, T, H), (T,T, T)}
D.
(T, T, H), (T, T, T)}. 7
7 7
(b): The events include {(
Math 171 Final Exam (Spring 2004) Solutions
1. (a) Let X denote the players net winnings. Then, X = 7 3 or X = 0 3 with corresponding probabilities
P (X = 4) = 0.4 and P (X = 3) = 0.6.
Thus, by denition,
E(X) = 4 P (X = 4) + 3 P (X = 3) = 4 0.4 3 0.6 = 0.
What does correlation mean?
Has no units
It is the strength of a linear
association between 2
quantitative variables
o
Remember:
r = zx z y
n 1
The correlation line goes through (0,0)
Has a slope of r
The differences at each point between
the observed val