Download Lenarz Math 102 Exam #4 Form B December 13, 2012 Name

Lenarz
Math 102
Exam #4 Form B
December 13, 2012
Name:
Directions: Answer the following questions in the space provided. You may use a calculator.
If you do not show your work, you will receive no credit. The point value of each question is
indicated. WRITE YOUR FINAL ANSWER ON THE ANSWER LINE WHEN
APPROPRIATE. IF YOU DO NOT FOLLOW DIRECTIONS YOU WILL BE
PENALIZED!
Formulas:
P
x=
n
x
s
s=
Σ(x − x)2
n−1
z=
x−x
s
1. Use the histogram to answer the following questions:
(a) (1 point) How many people sent five messages per day?
6
(a)
(b) (1 point) How many people sent eight or more messages per day?
(b)
9 + 8 + 5 + 1 = 23
(c)
8
(c) (1 point) What is the mode of the distribution?
(d) (1 point) How many people were in the survey?
(d) 2 + 5 + 6 + 3 + 4 + 9 + 8 + 5 + 1 = 43
Math 102
December 13, 2012
Page 2
2. Consider the data set:
13, 24, 27, 45, 32, 29, 28, 39, 25, 21, 37, 36, 42, 34, 49
(a) (3 points) Find the mean. Round your answer to two decimal places.
32.07
(a)
Solution:
13 + 24 + 27 + 45 + 32 + 29 + 28 + 39 + 25 + 21 + 37 + 36 + 42 + 34 + 49
15
481
=
15
= 32.07
x =
(b) (3 points) Find the mode.
(b)
no mode
(c)
32
(c) (3 points) Find the median.
Solution: First sort the data:
13, 21, 24, 25, 27, 28, 29, 32, 34, 36, 37, 39, 42, 45, 49
Since there are 15 data values, the middle data value is the eighth data value
which is 32.
(d) (3 points) Find the five number summary.
(d)
13, 25, 32, 39, 49
Solution: Using the sorted data above,
13, 21, 24, 25, 27, 28, 29, 32, 34, 36, 37, 39, 42, 45, 49
{z
}
|
{z
}
|
lower half
upper half
the lower half has median 25, so Q1 = 25. The upper half has median 39, so
Q3 = 39. Thus the five number summary is
13, 25, 32, 39, 49
Math 102
December 13, 2012
Page 3
(e) (3 points) Draw a box-and-whiskers plot for the data.
Solution: .
10
15
20
25
30
35
40
45
50
3. (8 points) Calculate the mean and standard deviation for the following grouped data.
First complete the table. Round answers to two decimal places.
x − x (x − x)2
−4
16
(x − x)2 · f
16
x
3
f
1
xf
3
4
1
4
−3
9
9
6
2
12
−1
1
2
7
5
35
0
0
0
8
3
24
1
1
3
3 27
15 105
2
4
12
42
9
Totals
Solution:
105
x=
=7
15
r
s=
42
≈ 1.73
15 − 1
Math 102
December 13, 2012
Page 4
4. (6 points) A college course requires a mean score of 70 on seven exams to receive a C
for the course. Marcus currently has scores of 45, 55, 79, 65, 77, and 83. What score
does Marcus need to get on his last exam to get a C in the course?
Solution: Let x be the score on the last exam. Then for all seven exams to have a
mean of 70, we must have
45 + 55 + 79 + 65 + 77 + 83 + x
7
404 + x
7
404 + x
404 + x
x
x
= 70
= 70
=
=
=
=
70(7)
490
490 − 404
86
5. A vending machine is designed to dispense a mean of 7.5 ounces of coffee into an 8 ounce
cup. If the standard deviation of the amount of coffee dispensed is 0.2 ounces and the
amount is normally distributed, use Table 1 on the last page to find the percent of times
the machine will dispense
(a) (4 points) less than 7.3 ounces of coffee.
(a)
15.9%
Solution:
7.3 − 7.5
−0.2
=
= −1
0.2
0.2
Using the table on the last page we see this z-score corresponds to 0.341. This
means that 34.1% of the data is between z = −1 (our data point of x = 7.3)
and z = 0 (the mean x = 7.5). Thus the percentage of data that is less than
7.3 ounces (z = −1) is 50% − 34.1% = 15.9%.
z7.3 =
(b) (4 points) between 7.3 and 7.7 ounces of coffee.
(b)
68.2%
Solution: From the previous part, we know that 7.3 ounces corresponds to
0.341 in the table.
7.7 − 7.5
0.2
z7.7 =
=
=1
0.2
0.2
Using the table on the last page we see this z-score also corresponds to 0.341,
but this data point is to the right of the mean, rather than the left. So the
Math 102
December 13, 2012
Page 5
percent of times the machine will dispense between 7.3 and 7.7 ounces of coffee
is
34.1% + 34.1% = 68.2%
(c) (4 points) more than 7.7 ounces of coffee.
(c)
15.9%
Solution: From above we know that
z7.7 = 1
which corresponds to 0.341. This means that 34.1% of the data is between z = 0
(the mean x = 7.5) and z = 1 (our data point of x = 7.7). Thus the percentage
of data that is more than 7.7 ounces (z = 1) is 50% − 34.1% = 15.9%.
6. (5 points) A weighted mean is a mean in which some data are worth more than other
data. If x1 , x2 , . . . , xn are data points with weights w1 , w2 , . . . , wn respectively, the
weighted mean of the data is given by
x1 w1 + x2 w2 + · · · + xn wn
w1 + w2 + · · · + wn
One place where weighted means are used is in the calculation of grades. In a college
class a student has the following exams scores with the indicated weights
Exam 1
Exam 2
Exam 3
Final Exam
Score Weight
75
10
82
25
85
40
90
15
What (weighted) mean score does the student receive? Round your answer to one
decimal place.
6.
83.9
Solution:
(75)(10) + (82)(25) + (85)(40) + (90)(15)
7550
=
≈ 83.9
10 + 25 + 40 + 15
90