371
Homework Statistics #4 Solution
Spring 2004
2.5 Construct a dotplot of the MAO data:
6
8
10 12 14 MAO (mols/108 platelets)
16
18
Construct a stem-leaf display of the MAO data: The decimal point is 1 digit(s) to the right of the | 0 | 4 0 | 57778889 1 | 0011234 1 | 59 2.25 Determine the mean litter size: y=
375 36
= 10.4
2.32 For the study of milk production in sheep data, determine: The median, the quartiles and IQR: By hand (textbook method): median(Q2 )=82.6, Q1 = 63.7, Q3 = 102.9, IQR = 39.2 In R: > sort(x) [1] 25% 44.4 50% 56.5 75% 63.7 65.6 75.1 82.6 89.8 91.5 102.9 108.1 110.1 > quantile(x,probs=c(0.25,0.5,0.75)) 64.65 82.60 97.20 > IQR(x) [1] 32.55
1
Statistics 371
Homework #4 Solution
Spring 2004
Construct a modified boxplot of these data:
100 110
2.42 Invent a sample of size 5 for which the deviations (yi - y ) are -3, -1, 0, 2, 2. Compute the standard devidation of your sample. One possibility is the sample: 17, 19, 20, 22, 22; here, s=2.1. Any sample chosen will have the same value fos s: 1 5-1
5
50
60
milk yield (liters) 70 80 90
s= .
(yi - y )2 =
i=1
1 ((-3)2 + (-1)2 + 02 + 22 + 22 ) = 2.1 4
2.47 For the lizard data, determine the quartiles, IQR, and range for the distance (in m) run in two minutes for the 15 animals. By hand (textbook method): Q1 = 26.4, Q3 = 37.5, IQR = 11.1, range=27.1. In R: > sort(y) [1] 18.4 22.2 24.5 26.4 27.5 28.7 30.6 32.9 32.9 34.0 34.8 37.5 42.1 45.5 45.5 > quantile(y,probs=c(0.25,0.75)) 25% 75% 26.95 > 36.15 IQR(y) [1] 9.2 3.6 If we take a random sample of two students from a college where 55% of the students are women, then: Pr(both students are women)=Pr(first is a women)Pr(second is a women)=0.55(0.55)=0.3025 And: Pr(at least one of the students is a woman) = Pr(exactly one is a woman OR both are women) = 1-Pr(both are men)=1-Pr(first is a man)Pr(second is a man)=1-(0.45)(0.45)=0.7975
2
Statistics 371
Homework #4 Solution
Spring 2004
3.8 A student takes a multiple choice test in which they have only learned 40% of the material and, if she doesn't know the answer , there is a 20% chance she will get it right by guessing. If a question is chosen at random: Pr(she gets question right)=Pr(she knows the answer AND gets it right) +Pr(she does not know the answer AND gets it right) =0.4(1)+(0.6)(0.2)=0.52 3.15 For the distribution of diameters of Douglas fir trees data: (a) Percentage of...