Homework 3 Solution
Chapter 6, #17
a) According to the model, 68% will be within 1 standard deviation of the
mean, or in other words, in the range 3.66-4.93 to 3.66+4.93, which
is (-1.27, 8.59). This means that 100-68 = 32% is outside this
interval. Because the normal curve is symmetric, half must be above
and half below, and so we'd expect 16% to fall below -1.27 hours.
b) For these data, that's rather silly, since one can't watch a negative
number of hours of television, although it might feel that way.
c) The data are skewed right, and so the normal model does not provide a
good match.
Chapter 11, #12
For each question, there are 4 choices and only one of them is correct. So
to simulate answering a question on this exam, we'll let the numbers 0 and
1 represent a right answer, 2,3,4,5,6,7, represent wrong answers, and we'll
ignore 8 and 9's. (There are other ways of doing this, of course, as
long as 25% of the numbers you use represent the right answer.)
There are 6 questions on the exam, so we'll repeat each selection of a random
number 6 times. This constitutes one trial.
Our outcome variable is the number of "correct"s out of 6 attempts. We'll
count this at the end of the trial.
And we'll repeat the whole thing 10 times.
For the sake of variety, we'll begin at row 30.
9453473 are the first non-ignored six numbers, and they translate to: ignore
wrong wrong wtong wrong wrong wrong. So the number of corrects is 0
for Trial 1.
55982135 are the next six non-ignored numbers: wrong wrong ignore ignore
wrong right wrong wrong. So Trial 2 has 1 correct.
7026087 ----> 2 rught
93685162 ---> 1
11937182 3
635413 1
86956463 0
719719840 3
9744530 1
160177 3
So our "data" are 0,1,2,1,3,1,0,3,1,3
The average 1.5. The most we got right was 3. And so it is very
unlikely that she could get all 6 correct. To be certain of how often
this could occur, just by chance, we coudl do many more simulations.
Repeating this for 8 more trials we get the following data set (each number
is the outcome of one trial):
0,1, 2,,1,2,4