Transcript chapter 9

Chapter 9
Statistics
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9.1
Frequency Distributions;
Measures of Central
Tendency
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Figure 1
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Figure 2
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Your Turn 1
Find the mean of the following data: 12, 17, 21, 25, 27, 38, 49.
Solution: The mean of the n numbers x1 , x2 ,......xn is
x
x
.
n
Let x1  12, x2  17, and so on. Here n  7, since there are 7 numbers.
12  17  21  25  27  38  49
x
7
189

7
 27
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Your Turn 2
Find the mean of the following grouped frequency.
Interval Midpoint, Frequency Product x
x
f
f
9 - 12
0-6
3
2
6
7-13
10
4
40
14-20
17
7
119
21-27
24
10
240
28-34
31
3
93
35-41
38
1
38
Total = 27
Total = 536
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Your Turn 2 continued
Solution : A column for the midpoint of each interval has been
added. The numbers in this column are found by adding the
endpoints of each interval and dividing by 2. For the
interval 0–6, the midpoint is (0 + 6)/2 = 3. The numbers in the
product column on the right are found by multiplying each
frequency by its corresponding midpoint. Finally, we divide the
total of the product column by the total of the frequency column
to get
x
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 xf 536

 19.85.
n
27
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Your Turn 3 and Example 6 (c)
Your Turn: Find the median of the data 12, 17, 21, 25, 27, 38, 49.
Solution: The median is the middle number; in this case, 25. (Note
that the numbers are already arranged in numerical order.) In this
list, three numbers are smaller than 25 and three are larger.
12, 17, 21, 25, 27, 38, 49.
6 (c) Find the median for each list of numbers 47, 59, 32, 81, 74, 153.
Solution: First arrange the numbers in numerical order, from
smallest to largest 32, 47, 59, 74, 81, 153.
There are six numbers here; the median is the mean of the two
middle numbers. Median  59  74  133  66 1
2
2
2
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Figure 3
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9.2
Measures of Variation
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Your Turn 1
Find the range, variance, and standard deviation for the list of
numbers: 7, 11, 16, 17, 19, 35.
Solution: The highest number here is 35; the lowest is 7. The
range is the difference between these numbers, or 35 − 7 = 28.
The mean of the numbers is
Number x
Square of the
7  11  16  17  19  35 105

 17.5.
6
6
2
2

x

nx
Variance  s 2 
n 1
2301  6(17.5) 2 463.5

 92.7.

6 1
5
Continued
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Number x2
7
49
11
121
16
256
17
269
19
361
35
1225
Total = 2301
Your Turn 1 continued
The standard deviation s is
 x 2  nx 2
s
n 1
s  92.7
 9.628.
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Figure 4
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Figure 5
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Your Turn 2
Find the standard deviation for the grouped frequency
distribution.
Interval
x
x2
f
fx2
0-6
3
9
2
18
7-13
10
100
4
400
14-20
17
289
7
2023
21-27
24
576
10
5760
28-34
31
961
3
2883
35-41
38
1444
1
1444
Total = 27 Total =
12,528
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Your Turn 2 continued
 fx 2  nx 2
Standard deviation s 
n 1
12,528  27(19.85) 2

27  1
Mean was calculated on slide 12.
 8.52.
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9.3
The Normal
Distribution
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Figure 6
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Figure 7
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Figure 8
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Figure 9
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Figure 10
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Figure 11
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Figure 12
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Figure 13
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Figure 14
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Figure 15
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Figure 16
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Your Turn 2
Find a value of z satisfying the following conditions.
(a) 2.5% of the area is to the left of z. (b) 20.9% of the area is to
the right of z.
Solution: (a) Use the table backwards. Look in the body of the
table for an area of 0.0025, and find the corresponding value of
z using the left column and the top column of the table. You
should find that z = −1.96.
(b) If 20.9% of the area is to the right, 79.1% is to the left. Find
the value of z corresponding to an area of 0.7910. The closest
value is z = 0.81.
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Your Turn 3
Find the z-score for x = 20 if a normal distribution has a mean
35 and standard deviation 20.
Solution:
z

x

Here  is the mean and  is the standard deviation.
20  35
20
 0.75
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Figure 17
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Figure 18
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Figure 19
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Your Turn 4
Dixie Office Supplies finds that its sales force drives an average
of 1200 miles per month per person, with a standard deviation
of 150 miles. Assume that the number of miles driven
by a salesperson is closely approximated by a normal distribution.
Find the probability that a sales person drives between 1275 and
1425 miles.
Solution: We will find the z-score for both x1 = 1275 and x2=1425.
For x1  1275,
1275  1200
150
 0.500
z1 
For x2  1425,
1425  1200
150
 1.500
z2 
Continued
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Your Turn 4 continued
From the table, z1 = 0.500 leads to an area of 0.6915, while
z2 = 1.500 corresponds to 0.9332.
A total of 0.9332 − 0.6915 = 0.2417 or 24.17 %, of the drivers
travel between 1275 and 1425 miles per month. The
probability that a driver travels between 1275 miles and 1425
miles per month is 0.2417.
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Figure 20
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9.4
Normal Approximation
to the Binomial
Distribution
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Your Turn 1
Suppose a die is rolled 12 times. Find the mean and standard
deviation of the number of sixes rolled.
Solution: Using n =12 and p = 1/ 6, the mean is
1
  np  12    2.
6
The standard deviation is
 1  1 
  np(1  p)  12  1    1.291.
 6  6 
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Figure 21
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Figure 22
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Figure 23
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Figure 24
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