Math 3070-5: On-Line Homework Answer Key
Spring Semester 2002




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Answers to
HW #1 HW #2 HW #3 HW #4
HW #5 HW #6 HW #7 HW #8
HW #9

Answers to Homework 1 (Back to Top)

(1) 63

(2) 24

(3) 156

(10) 29

(12) (a) 7!; (b) 6!

(16) (a) 7.6.5=210; (b) 3.6.5=90; (c) To be bigger than 330, you have to draw 340 or more (why?). The number of ways to get a 3-digit number that starts with a 3 and is greater than 340 is 15. The number of ways to draw a 3-digit number greater than or equal to 400 is 3.6.5=90. Answer = 90+15=115.

(17) The seating has to be 

gb|gb|gb|gb|g
Answer: 5!.4!

(26) 9! divided by (3!.4!.2!) = 1260

Answers to Homework 2 (Back to Top)

(1) 0.1743

(4) (a) 0.55; (b) 0.83; (c) 0.27.

(7) (a) 0.054; (b) 0.444.

(8) (a) 84; (b) 40; (c) 15.

(9) (a) 0.384; (b) 0.512.

(11) 800.

(12) (a) 0.28; (b) 0.72.

Answers to Homework 3 (Back to Top)

(1) (a) 0.3125; (b) 12y(1-y)2 if 0 <= y <= 1; 0 else; (c) 0.25.

(2) (a) f(x) = 

0.4   if x=1
0.2   if x=3
0.2   if x=5
0.2   if x=7

(2)(b) P( 4 < X <= 7) = F(7) - F(4) = 0.4.

(3) (a) g(x) = 1 over (1+x)2, and h(y) = e-y; (b) 1/3.

(4) (a) 1/16; (b) 0.41 (approx)

(5)

Warning: e2 should be e-2 instead

(a) Here is the (approx) table for x=0,1,2,3,4,5,6:

x0123456
f(x)0.140.270.270.18 0.090.040.01

(c) Here is the (approx) table for x=0,1,2,3,4,5,6:

x0123456
F(x)0.140.41680.86 0.950.980.995

(9) (a) (1/50) e-x/50 if x>0; (b) 0.25 (approx)

(10) (a) f(x)=1/10 for x=1,2,....,10;

Answers to Homework 4 (Back to Top)

(2) E(XY)=1/3; E(X)=E(Y)=1/12; Cov(X,Y)=47/144.

(4) (a) E(X)=5.

(4) (b) E(X2)=50; Var(X)=25; SD(X)=5.

(4) (c) 125.

(12) (a) Cov(X,Y)=1; Var(X)=5; Var(Y)=3; Corr(X,Y)= 1 over the square root of 15.

Answers to Homework 5

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(7, p. 158) (a) Answer = The prob. of the being to right of (32-40)/6.3=-1.27 on the standard normal curve, which is about 0.898.

(7, p. 158) (b) Answer = The prob. of the being to left of (28-40)/6.3=-1.91 on the standard normal curve, which is about 0.0281.

(7, p. 158) (c) Answer = The prob. of the being between (37-40)/6.3=-0.48 and (49-40)/6.3 on the standard normal curve, which is about 0.61.

(18, p. 159) (a) Answer = The prob. of the being outside the interval [-1.3,1.3] on the standard normal curve, which is about 0.1936.

(18, p. 159) (b) Answer = The prob. of the being inside the interval [-0.52,0.52] on the standard normal curve, which is about 0.397.

(1, p. 178) The question is: if X is Binomial with n=100 and p=.49, what is the probability that X is between 482 and 510 ...... Answer =0.

(2, p. 178) The answer is the integral of 6 e-6x dx from 15 to infinity. Answer=e-90.

(5, p. 178) The question is: if X is Binomial with n=1000 and p=0.65, find the prob. that X is between 494 and 515. Note that E(X)=np=650 and Var(X)=np(1-p)=227.5, so that the SD of X = sigma is about 15. Since n is large, X is close to being normal. So, we are asked to find the prob. that a standard normal is between (494-650)/15 = -10.4, and (515-650)/15 = -9. This prob. is very close to 0.

(8, p. 223) You are asked to find the prob. that a sample avg. of n=36 objects is greater than 1458/36=40.5. Supposing that normal approximation is good here (n=36 only!), note that this sample avg. is approx. normal with mean 40 and Var = sigma/root(n) approx = 0.33. So, the answer is 0.064 = 6.4%.

Answers to Homework 6

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(2, p. 223) First compute the mean and the SD of X: the mean is 4, and the SD = sigma = about 1.63. Sample mean (X-bar) is therefore approx. normal with mean 4 and 

SD = sigma/root(n) = approx 0.02.
So the area between 4.1 and 4.4 is approx the area under the standard normal between (4.1-4)/0.02 = 5, and (4.4-4)/0.02 = 20. Thus, the answer is approx 0%.

(8, p. 223) See the solution to Homework #5.

(13, p. 224) The difference between the two sample averages is approximately normally distributed with mean 72-28=44 cm. Its SD is 

square root of ( sigma12 over n1 + sigma22 over n2),
which equals 3.066 = approx 3. The prob. that this sample average difference is at least 44.2 is the area to the right of (44.2-40)/3=1.4 under the standard normal curve. Answer: 1- 0.9192= approx 0.08 =8%.

(6, p. 237) The total number, X, of hurricances in 2 years is the sum of two Poisson random variables, each with mu=6. So X is Poisson with parameter mu = 12 (why? Think of the Poisson-to-Binomial approximation.)

  1. P( X=15) equals 
     
( e-121215 ) over 15!.
  2. P ( X <= 9) equals 
     
sum from i=0 to i=9 of:
( e-1212i ) over i!.

Answers to Homework 7

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(6, page 236) Recall that the following has a Chi-squared distribution with (n-1)=19 degrees of freedom:

W = (n-1) S2 / sigma 2 .
Now if sigma2 really were 8, then the sample value of W would be W = (20-1) . 20 / 8 = 40.

On the other hand, the probability that a Chi2 random variable with 19 d.f. is 40 or more is essentially 0% (page 685, Table A.5). This amounts to saying that if sigma2 were 8, then it would be impossible to sample W and observe a value of 40 or more. In other words, if sigma2 were 8, we would never observe

S2 = W sigma2 / (n-1) = 20.
Since we did observe S2 to be 20, we can infer that sigma2 cannot be 8.

(8, page 236) We can use Table A.4 (pp. 683-684) to obtain the following.

  1. When v=14, t0.025= 2.145.
  2. When v=10, -t0.10= -1.372.
  3. When v=7, t0.995= -3.499.

(12, page 236) The value of the T-statistic is 

(27.5 - 30) over 5/ root(16) = -2.
On the other hand, the t0.025 with 15 d.f. equals 2.131. Thus, the interval [-t0.025 , t0.025] = [-2.131,2.131] does contain the T-statistic, and the company is satisfied with this month's performance.

(14, page 236) Unless the measured saturated fat contents are normally distributed, this question cannot be answered. If they are normally distributed, however, note that 

X-bar= .48 (approx), and S = 0.18 (approx).
Thus, if the cereal-makers are reporting the average of .5 gm accurately, the T-statistic has the observed value: 
T = (0.475-0.5) over 0.183/ root(8) = -0.386 (approx).
With 7 d.f., the prob of observing -0.386 or even less is somewhere between 0.3 and 0.4 (Table A.4, p. 683). This is a reasonable likelihood, and we have no axe to grind with their reported finding of mu approx 0.5 gm.

(15, page 236) (a) 2.71 (Table A.6*, p. 687); (b) 3.51 (Table A.6, p. 688).

(16, page 236) The SD of the first data set is 3.23 (approx), while that of the second data set is about 1.418. I.e., S1=3.23, S2=1.418. If sigma1=sigma2, i.e., if the theory is correct, then S12 over S22 has an F distribution with d.f.'s v1=9, and v2=7. The F-statistic, on the other hand, equals 

the square of (3.23/1.418)=5.19.
Thus, if the theory of equal-variances were right, the prob. of observing an F-value of 5.19 or more is somewhere between 0.01 and 0.05. Since this is unlikely, there is significant statistical evidence against the theory, i.e., the variances are most likely not equal.

(17, page 236) S12= 15750, and S22= 10520. Thus, the F-statistics is 

 F = 15750/10520 = 1.44 (approx).
With v1=4, v2=5, the odds of seeing this, if the variances were equal, would be a good deal bigger than 0.05 and we have no reason to doubt this theory of equal variances.

Answers to Homework 8 (Back to Top)

(13, page 252) X-bar is approx. 1.01; S is approx 0.03. So, the 99% CI is 

1.01 +/- t times 0.03/3,
where t equals 3.25 (8 d.f.). The CI is approx. (0.98,1.04)

(2, page 287) X-barD=-3.55 (difference = after - before); SDD = 2.78. Therefore, the 95% CI is 

-3.55 +/- t 2.78 / root(7),
where t = 2.447 (6 d.f.). Thus, the CI is (-6.12,-0.98). Therefore, there is not enough statistical evidence against the manufacturer's claim.

(3, page 287) The sample percentage is 28/75 = .37 (approx.). Applying normal approximation, we obtain (z=1.96) the following 95% CI for the true proportion sold by the McDonald's: 

.37 +/- 1.96 * root( (0.373 times 0.627)/75 )
	= (0.25,0.49).
Therefore, there is not enough statistical evidence against the McDonald's claim.

(5, page 287) If diff=pre - post, we have X-barD=40.6 and SD = 15.8 (approx). Thus, a 95% CI for the difference is 

40.6 +/- t times 15.8 / root(12),
where t=2.9 (11 d.f.). Therefore, a 95% CI for preburn - postburn is (27.4,53.8).

Answers to Homework 9 (Back to Top)

(4, page 304) For part (a), compute alpha as Pp=0.6 { X is not in [1,12]}, which is 

 
P{X=0} + ( 1- P{X <= 11} ) = 0 + [ 1- 0.9095] = 0.09,
from the binomial table. For part (b), when the alternative is p=0.5, 
beta = P{X <= 12} - P{X=0} = 0.9963.
When the alternative is p=0.7, beta = 0.8732. Part (c): No. Beta is much too high.

(1, page 325) z=(788-800) divided by 40/root(30) = 1.64. This yields a (two-sided) P-value of about 10% ... do not reject.

Last Update: May 1, 2002