Övningar enligt 2a upplagan.

Le1
2.1: 1/15
2.3: 0.45
2.4: (a) 0.4 (b) 0.2 (c) 0.2 (d) 0.1 (e) 0.8333 (f) 0.7143
2.5: 0.72
2.6: 0.6
2.7: 0.66
2.8: 0.9508
2.11: (a) 0.8488 (b) 0.4774 (c) 0.0012
2.14: (a) 0.006 (b) 0.000068 (c) 0.000051 (d) 0.000018
2.16: (a) 0.049 (b) 25/49
2.17: 1/3
2.32:

Le2
3.1: (a) p(0)=0.42, p(1)=0.46, p(2)=0.12
3.2: E(Y)=200 dollars och Var(Y)=110000 dollars^2
3.4: E(X)=3.5 och Var(X)=35/12
3.8: E(X)=1.5 och Var(X)=1.25
3.9: 1/16
3.10: 0.28
3.11: (a) p(x,y)=1/36 om x=y, 1/18 om x<y, 0 om x>y (b) beroende (c) p(1)=11/36, p(2)=9/36, p(3)=7/36, p(4)=5/36, p(5)=3/36, p(6)=1/36 (d) 2/9
3.12: (a) beroende (b) beroende
3.13: (a) p(0)=0.2, p(2)=0.7, p(4)=0.1 (b) p(-2)=0.3, p(0)=0.4, p(2)=0.3 (c) p(0)=0.8, p(1)=0.1, p(4)=0.1
3.16: (a) beroende (b) 0.6
3.19: (a) E(100X)=0 och Var(100X)=40000 (b) E(100Y)=0 och Var(100Y)=40000 (c) E(50X+50Y)=0 och Var(50X+50Y)=20000

Le3
3.5: (a) 19/33 (b) 5/3
3.20: (a) 0.0596 (b) 0.986
3.21: 0.245
3.23: (a) 0.005 (b) 0.00314
3.24: (a) 0.0328 (b) 0.4096
3.25: (a) 0.1111 (b) 0.1937
3.27: (a) 0.945 (b) 0.061
3.28:
3.29: 0.0923
3.30: 0.0166
3.31: (a) 0.2061 (b) 1.25
3.35: 9/13
3.37: 0.827

Le4
4.2: (a) 0.003 (b) 0.217
4.6: 55/6
4.7: 0.875
4.8: 0.264
4.9: (a) 0.269 (b) 0.666
4.10: 0.4764
4.11: (a) 0.6 (b) 15
4.14: (a) 4 och 0.2 (b) 0.353
4.16: (a) 0.8944 (b) 0.8944 (c) 0.1056 (d) 0.7888 (e) 1 (f) 0 (g) 0.84
4.17: (a) 0.1611 (b) 0.1611 (c) 0.8389 (d) 0.3222 (e) 1 (f) 0 (g) 1.28
4.18: (a) 0.9965 (b) 0.3802 (c) 0.6233 (d) 0.2321 (e)1 f) 0.8413, (g) -2.12
4.19: (a) 0.9974 (b) 1
4.29: 0.567
4.30: 0.8693
4.32: (a) 4 (b) 6

Le5
4.23: 0.9918
4.24: 0.2033
4.25: (a) 0.4544 (b) 0.9997
4.26: 0.3226
4.27: 0.6318
4.28: 0.1151
4.31: (a) 0 (b) 38
5.1: 0.01
5.2: (a) 0.3693 (b) 1 (c) 5 (d) 2 (e) 0.7351 (f) −0.5902 (g) 4.0026
5.6:

Le6
6.1: (a) p00 = p01 = p11 = p12 = p21 = p22 = 0.5 anc p02 = p10 = p20 = 0 (c) 0.125
6.3: (a) P = ((0.6 0.4), (0.2 0.8)) row-wise (b) 0.28
6.9: (a) 0.0579 (b) 0.0001
6.12: (a) 3 seconds (b) 0.000027 (c) 0.9699
6.14: 0.2810, 20 seconds, 17.32 seconds
6.17: (a) 0.945 (b) 0.061
6.18: 0.0162
6.19: (a) 3.75 seconds (b) 0.4129, i.e. observing 50 or more messages is not unlikely
6.20: 0.884
6.21: 0.215
6.22: (a) 0.735 (b) 7500 dollars and 3354 dollars
6.23: (a) 0.003 (b) 0.6 minutes and 0.12 minutes^2
6.24: 0.544

Le7
8.1: (b) min(X) = 37, Q1_hat = 43, M_hat = 50, Q3_hat = 56, max(X) = 60, and min(X) = 21, Q1_hat = 35, M_hat = 39, Q3_hat = 46, max(X) = 53.

8.2: (a) X_bar = 17.9540, s^2 = 9.9682, s= 3.1573 (b) 0.4465
8.3:
8.4: 0.9930
8.5:

pop<-matrix(data=c(seq(1790,2010,10),3.9,5.3,7.2,9.6,12.9,17.1,23.2,31.4,38.6,50.2,63.0,
                   76.2,92.2,106.0,123.2,132.2,151.3,179.3,203.3,226.5,248.7,281.4,308.7),
            ncol = 2,nrow = 23)

plot(pop,xlim=c(1790,2010))

8.6:

popinc<-matrix(data=NA,ncol=2,nrow=22)
for(i in 1:22){
  popinc[i,1]<-pop[i+1]
  popinc[i,2]<-pop[i+1,2]-pop[i,2]
}

plot(popinc,xlim=c(1800,2010))
mean(popinc[,2])
median(popinc[,2])
var(popinc[,2])

8.7:
poprel<-matrix(data=NA,ncol=2,nrow=22)
for(i in 1:22){
  poprel[i,1]<-pop[i+1]
  poprel[i,2]<-(pop[i+1,2]-pop[i,2])/pop[i,2]
}

plot(poprel,xlim=c(1800,2010))
mean(poprel[,2])
median(poprel[,2])
var(poprel[,2])

8.9: (a) X_bar = 49.6, M_hat = 47.5, Q1_hat = 43, Q3_hat = 52, and s = 23.5 (b) 105 is an outlier (c) X_bar = 43.4, M_hat = 45, Q1_hat = 43, Q3_hat = 51, and s = 13.9 (d) note the change in sample mean and standard deviation whereas the quartiles do not change that much (robust).

Le8
9.1: (b) 5/8 (c) 0.171
9.2: (b) 5/20
9.3: (a) a_hat = min(x_i) and b_hat = max(x_i) (b) lambda_hat = 1/X_bar (c) mu_hat = X_bar (d) sigma_hat = sqrt(sum_i (x_i - mu)^2/n) (e) mu_hat = X_bar and sigma_hat = sqrt(sum_i (x_i - X_hat)^2/n)
9.4: 2.1766

Le9

9.7: (a) 37.7 ± 1.5 or [36.2, 39.2] (b) H0 : μ = 35 vs HA : μ > 35. Z = 2.9348 ≥ z_0.01 = 2.326. Therefore, reject H0 in favor of HA.
9.8: (a) 42± 1.225 or [40.775, 43.225] (b) 0.1790
9.9: (a) 50± 33.7 or [16.3, 83.7] (b) H0 : μ = 80 vs HA : μ ̸= 80. t = −2.598 belongs to the acceptance region (−2.920; 2.920), therefore, H0 is not rejected. (c) [11.6, 89.4] (thousand dollars)
9.10: (a) 0.12 ± 0.047 or [0.073, 0.167] (b) H0 : p ≤ 0.1 (or H0 : p = 0.1) vs HA : p > 0.1. Z = 0.9428. Therefore H0 is not rejected in either case.
9.12: (a) 0.62 ± 0.054 or [0.566, 0.674] (b) 0.2358
9.13: 0.0062
9.16: (a) −0.02 ± 0.043 or [−0.063, 0.023] (b) H0 : p1 = p2 is not rejected because the 98% confidence interval contains 0
9.17: 0.0325 or 3.25% for pHat1 = 45%, 0.0312 or 3.12% for pHat2 = 35%, 0.0450 or 4.50% for pHat1 − pHat2 = 10%

Le10
10.31: (a) 4.5 and 2.25
10.32: (b) 4/13
10.33: (b) 0.4 and 0.02, (c) 0.22, (d) reject H0
10.34: (a) 17.6611, (b) [16.7657,18.5565], (c) reject H0
10.35: (a) 17.9540, (b) [17.0234, 18.8846], (c) reject H0
10.36: (a) N(5.5, 0.255), (b) N(5.575, 0.245), 5.575 and 0.060, (c) [5.095, 6.055]
10.37: 0.5145
10.39:
10.40: (a) 0.9 and 9/1100, (b) [0.7743, 1], (c) reject H0

Le11
11.1: y_hat = 0.009 + 0.000246x and 0.107 seconds
11.3: (a) y_hat = 24.77 − 0.0000395x
11.5: (a) b0 = 13.3586, b1 = 2.3569, b2 = 4.0926 (2nd edition), b0 = 1.5741, b1 = 2.3569, b2 = 4.0926 (3rd edition)
11.8: (a) y_hat = 19.50 + 17.36x, (b) 88.94
11.10: (a) y_hat = −32.1+ 1.36x, (c) 260.4 and 267.2 million people
11.11: (a) b0 = 5.8700, b1 = 0.0057, and b2 = 0.0068, (b) 320.2 and 335.0 million people
11.12: (a) the quadratic model is more appropriate, (b) b0 = 7.716, b1 = 1.312, and b2 = −0.024, (d) 25.53 and 28.41 lbs

Le12
11.2: (c) [0.4547, 0.6703]. This interval does not contain 0, therefore, the slope is significant (significantly
different from 0) at a 1% level of significance.
11.3: (b) R2 = r2 = 0.0289. This model can explain only 2.89% of the total
variation of the number of miles per gallon. This is a rather poor fit. The number
of miles per gallon should probably depend on many factors other than mileage, (c) [16.74, 30.04] and [22.78, 24.00]
11.4: (b) there is significant evidence against the null hypothesis H0 : β1 = 1.8 (1000 dollars), (c) [50.40; 67.02], (d) It is wrong to say that the actual investment will be between 50,400 and 67,020
with probability 0.95. Rather, 95% refers to the long run proportion of prediction
intervals constructed by the same procedure as we used in (c). That is, about 95%
of prediction intervals computed from a large number of independent samples will
contain the actual 2017 investment
11.5: (b) 52.8047 thousand dollars, (c) Our prediction
will decrease by b2 = 4.0926 thousand dollars