MA 113
Exam II Key
Fall 2014
D. Weiner
1-4. Let X be the weight in pounds of a random rabbit of that breed, so X ~ N(7.8, 2.20)
(note that weights are normally distributed).
1. want P( X > 7) = P( (X 7.8)/2.2 > (7 7.8)/2.2) = P(Z > -0.36), shade, look up
Range rule of thumb
Whats the average height of women in
this class?
Whats the standard deviation?
What are the max and min heights of
women in this class?
s range/4 i.e. most values are within 2
standard deviations of the mean
Exam 1:
Some Review Que
Math 113 Elementary Statistics
Review Questions for Test 2
Answering the questions below should help with the sorts of problems weve been
considering. Given that the problems of Chapter 8 use the methods of Chapter 7, the key
is really to know if youre de
Some Desperate Statistics Exercises
All populations concern individuals living on Wisteria Lane. Fill in the slots marked ?
Person
Population
percentile
Gabby
female heights
38th
Carlos
male weights
55th
Lynette
IQ scores
?
Bree
female weights
Susan
ditzi
Statistics Problems on p,e,n,T and also on
1.
What percentage of BU students weigh over 160 pounds? Stan polled 225 BU students, of whom 150
weighed over 160 pounds.
2.
In #1, be more accurate; then, in #1, be more confident.
3.
In #1, what if Stan polle
Table: Chi-Square Probabilities
http:/people.richland.edu/james/lecture/m170/tbl-chi.html
Table: Chi-Square Probabilities
The areas given across the top are the areas to the right of the critical value. To look up an area on the left,
subtract it from one
HW2_skeleton
Harrel Blatt
July 8, 2016
3.1
13. Mean cost: $2025.25; median cost: $2209; no mode cost.
14. Mean:41.41 Median: 40.76 No mode
15. Mean:3670 Median:3830 Mode:4940
16. Mean:266 Median:266 Mode:260
17.
a)
mean is bigger than median
b)
mean is eq
HW5S_skeleton
Harrel Blatt
July 28, 2016
10.1
9.
Right-tailed, u
10.
left-tailed, p
11.
Two-tailed, sigma
12.
right-tailed, p
13.
Left-tailed, u.
14.
Two-sided, sigma
15.
Ho: p=0.105
H1: p>0.105
Type I error: The sample evidence led the researcher to beli
MA113 Spring 2016
Homework 2 Solutions
2.
a)
P(D) = (20 + 30 + 10)/500 = 60/500 = 0.12
b)
P(A) = 200/500 = 0.4
c)
P(D and (A or C) = 30/500 = 0.06
d)
P(D|(A or C) = 30/400 = 0.075
4.
d)
a)
P(No Ins) = (208+157)/1875 = 365/1875 = 0.19
b)
P(F and No Ins) =
Homework 7 Solutions
26.
H0: = 6
H1: 6
= 0.05
X - 0
t=
s
n
, df=19
Reject H0 if t < -2.093 or if t > 2.093
t=
X - 0
5 .4 6
s
0.55 / 20
n
= -4.88
Reject H0 since 4.88 < -2.093. We have significant evidence, = 0.05, to show that 6,
p<0.01.
32.
n=15,
X
=6.1
Homework 6 Solutions
4.
90% CI
s = 3.4
Z
n=
E
1
2
E=1
= 31.3
2
(1.645) (3.4)
=
1
2
n = 32
6.
n = 40
= 14.6
X
X Z1
s = 2.8
s
n
2
14.6 (1.96)
2 .8
40
14.6 + 0.87
(13.73, 15.47)
8.
a)
Z
n=
E
1
2
b)
= 8.01, n = 9.
2
(1.645) (8.6)
=
5
n = 50
= 58.
HW4S_skeleton
Harrel Blatt
July 22, 2016
9.1
You can use the following syntax to check your
answers. Note the answers you get in R will not be
EXACTLY the same you get by hand but they should
be pretty close.
If your \(x = 542\) (ie the number of successe
HW1S_skeleton
Harrel Blatt
July 4, 2016
1.1
1. Statistics is the science of collecting, organizing, summarizing, and analyzing information to draw
conclusions or answer questions. In addition, statistics is about providing a measure of confidence in any
c
HW3Sum_skeleton
Harrel Blatt
July 13, 2016
7.1
31.
shadenorm(mu = 62, sig = 18, below = 44, col = "blue", dens = 200)
Interpretation 1. 15.87% of the cell phone plans in the United States are less than $44.00 per month.
Interpretation 2. The probability i