Chapter 3 Homework Problems

Graded: 3.2 (see normalPlot), 3.4, 3.18 (use qqnormsim from lab 3), 3.22, 3.38, 3.42 #3.2

#3.2(a)
1-pnorm(q = -1.13, mean = 0, sd = 1)
## [1] 0.8707619
#(b)
pnorm(q = .18, mean = 0, sd = 1)
## [1] 0.5714237
#(c)
1-pnorm(q = 8, mean = 0, sd = 1)
## [1] 6.661338e-16
#(d)
pnorm(q = .5, mean = 0, sd = 1)-pnorm(q = -.5, mean = 0, sd = 1)
## [1] 0.3829249

3.4

#3.4
#(a)X~N(4313,583), X~N(5261,807)
#(b)
1-pnorm(q = 5513, mean = 5261, sd = 807)
## [1] 0.3774186
1-pnorm(q = 4948, mean = 4313, sd = 583)
## [1] 0.1380342

Mary Finished Faster than 37.7% of female runners while Leo finished faster than only 13.8% of male runners

3.18

The data does indeed follow a normal distribution based on the results below

pnorm(q = (61.52+4.58), mean = 61.52, sd = 4.58)-pnorm(q = (61.52-4.58), mean = 61.52, sd = 4.58)
## [1] 0.6826895
pnorm(q = (61.52+(2*4.58)), mean = 61.52, sd = 4.58)-pnorm(q = (61.52-(2*4.58)), mean = 61.52, sd = 4.58)
## [1] 0.9544997
pnorm(q = (61.52+(3*4.58)), mean = 61.52, sd = 4.58)-pnorm(q = (61.52-(3*4.58)), mean = 61.52, sd = 4.58)
## [1] 0.9973002

3.22

  1. the probability that the first defect is on the 10th trial is the probability that the first 9 trials are succesfful and the 10th trial is defective, 1.67%
  2. the probability of no defects in a 100 is the probability of a successful trial occuring a 100 times in a row
  3. you would expect 50 trials before the first defect with a standard deviation of 1.96%
  4. you would expect 20 trials before first defect with a standard deviaiton of 4.75%
  5. decreasing the probability of an event raises the mean number outcomes and lowers the standard deviation
#(a)
defect_on_10<-(.98)^9*.02
defect_on_10
## [1] 0.01667496
#(b)
no_defect_in_100<-(.98)^100
no_defect_in_100
## [1] 0.1326196
#(c)
expected = 1/.02
std_dev = .98*(.02)
expected
## [1] 50
std_dev
## [1] 0.0196
#(d)
expected_2 = 1/.05
std_dev_2 = .05*(.95)
expected_2
## [1] 20
std_dev_2
## [1] 0.0475

3.38

(a)the probability of having 2 boys out of 3 kids is 38.2% (b)In order to have two boys this can occur in three different ways either the 1st and 2nd can be boys, the 2nd and 3rd, OR the 1st and 3rd, each of these has a probability of (.49)(.51)(.51)so when you add up all three of those scenarios you get 3.49.51*.51 which also gives us 38.2% (c)In order to calculate the probability of having three boys out of 8 children would be much more tedious to calculate manually because there would be a lot more scenarios to go through

#a
dbinom(2,3,.51)
## [1] 0.382347
#b
3*.49*.51*.51
## [1] 0.382347

3.42

(a)the probability of getting the third successful attempt on the 10th attempt is 3.9% (b)the probability of tenth serve being successful is 15% this is independent of any past results (c)because the probability in a was the entire scenarion where as the probability in b was only the second component of the scenario

dbinom(2,9,.15)*.15
## [1] 0.03895012