# Test results and answers to most test questions

```Test results and answers to most test questions
Test scores:
Mean = 39 (79%)
Median = 41 (83%)
Standard deviation = 6.87 (21%)
25
20
15
10
5
0
96
-1
00
91
-9
5
86
-9
0
81
-8
5
76
-8
0
71
-8
5
66
-7
0
61
-6
5
56
-6
0
51
-5
5
46
-5
0
41
-4
5
36
-4
0
Number of students
Histogram of test scores
Test score (percentage of maximum)
Approximate breakdown by letter grades: (keep in mind that these are only if I had to
give you letter grades based on just this tes. The letter grade assignments are very
approximate and can change significantly for the final grade):
above 85%: A or Aabove 65%: B, B- or B+
above 55%: C or Cabove 50%: D
below 50%: E
MIDTERM 1.
1. According to U.S. Department of Energy, the average american home uses energy on
four main areas, shown in the following figure:
Lights and
Appliances,
33%
Heating and
Cooling , 44%
Water
Cold Storage,
Heating , 14%
9%
a) (2 points) If we want to present the same information with a different type of graph,
which would be the appropriate one: pie chart, bar graph or a histogram?
bar graph, because the variable involved (types of energy usage) is qualitative
2. A study aims to find out whether there is a relationship between the age at which
babies begin to crawl and outside temperature at that time. In particular, they want to
predict crawling age from the outside temperature. Here are the data for 5 babies:
crawling age (weeks)
temperature (degrees, F)
30
66
31
73
30
72
32
63
34
48
a) (4 points) Find the mean crawling time and the mean temperature in this data set.
mean age 31.4 weeks
mean temperature 64.4 degrees
b) (2 points) Find the standard deviation for the temperature
s=10.06
c) (4 points) What are the explanatory and the response variables in this study?
explanatory: outside temperature
response: crawling age
crawling age, months
d) (3 points) Make a scatterplot
35
34
33
32
31
30
29
0
20
40
60
80
temperature, degrees
e) (2 points) Based on your scatterplot, do you think there is a positive association,
negative association or no association between the crawling age and the temperature?
Why?
negative association, because the overall trend is down to the right.
f) (4 points) The correlation between the crawling age and the temperature is
approximately
r = -0.89. State whether this value of r supports of the following statements and briefly
say why for each one of them.
“There is a strong positive association between the crawling age and the outside
temperature at that time”
No, the association is strong negative, because r is negative and close to -1
“Low outside temperatures delay child’s motor development”
No, because the value of r does not say anything about one of the variables causes the
other.
3. (3 points) Below are the numbers of home runs for 14 National League baseball teams
in early 1994:
19 29 31 42 46 47 48 48 53 55 55 63 66 67
a) (3 points) What is the 5-number summary of these data?
19 42 48 55 67
b) (2 points) Make a boxplot
70
60
50
40
30
20
c) (2 points) Make a stemplot
1|9
2|9
3|1
4|26788
5|355
6|367
d) (3 points) Describe the shape of the stemplot
left skewed with no outliers (also regular, asymmetric, one cluster, one maximum,
center between 40 and 50)
4. a) (3 points) In the standard Normal distribution what is the proportion of observations
that have -0.2 &lt; z ≤ 0.5 ?
From table A: to the left of 0.5 is .
to the left of -0.2 is
b) (3 points) In the standard Normal distribution what is the z value such that 92% of
observations are above this value?
92% or 0.92 are above z, which means that 0.08 are below z. Looking up 0.08 in the table
A backwards (closest value is 0.0793) find z=-1.41
5. A researcher measures the lifespan of elephants. He finds that the lifespan is Normally
distributed with the mean of 63 years and the standard deviation of 6 years.
a) (2 points) What is the variable measured in this study?
lifespan of elephants
b) (4 points) What is the proportion of elephants that live longer than 65 years?
Need to find proportion with X&gt; 65, where X (lifespan) is N(63,6)
Standardize: it is the same as z &gt; (65-63)/6 = 0.33.
Look up 0.33 in table A, get 0.