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Chapter 10
Introduction to Statistics
10.1 Frequency Distributions; Measures Of Central Tendency
Population, variable, sample
Example 1 (p. 572 – 573)
Grouped frequency distribution
Histogram, Frequency polygon
Example 2 (p. 573 – 574)
Stem and leaf
Example 3, 4 (p. 575)
10.1 Frequency Distributions; Measures Of Central Tendency
• Summation notation (sigma notation)
x1 + x2 + x3+ ….+ xn =
• The mean (arithmetic average)
The mean of the n numbers x1, x2, x3, … xn is
Examples 5, 6 (p. 576, 577)
Example 7 (p. 578)
n
iix
1
xnx
n
xxx n ...21
MEAN OF A GROUPED DISTRIBUTION
• The mean of a distribution where x represents the midpoints, f the frequencies, and n=
• Sample mean
• Population mean
isf ,
xnxf )(
MEDIAN AND MODE
• Median: The middle entry in a set of data
arranged in either increasing or decreasing order.
If there is an even number of entries, the median is defined to be the mean of the two center entries.
Example 8 (p. 580)• Mode: the most frequent entry. If all entries have
the same frequency, there is no mode.
Example 9, 10 (p. 581, 582)
10.2 MEASURES OF VARIATION
• Range of a list of numbers: max – minExample 1: (p. 587)
• Deviations from the mean of a sample of n
numbers x1, x2 , x3, … xn, with mean is:
x1 –
x2 –
…
xn –
x
x
x
x
10.2 MEASURES OF VARIATION
• Deviations from the mean of a sample of n
numbers x1, x2 , x3, … xn, with mean is:
x1 –
x2 –
…
xn –
Example 2: (p. 587)
x
x
x
x
10.2 MEASURES OF VARIATION
SAMPLE VARIANCE
• The variance of a sample of n numbers x1,
x2 , x3, … xn, with mean , is s2 =
Shortcut formula: s2 =
Population variance: 2 =
2
1)(
n
xxx
2)(
n
xx
1
22
n
xnx
10.2 MEASURES OF VARIATIONSAMPLE STANDARD DEVIATION
• The standard deviation of n numbers x1, x2 , x3, …, xn, with mean , is
Example 3: (p. 590)
x
11
)( 222
n
xnx
n
xxs
10.2 MEASURES OF VARIATION
STANDARD DEVIATION FOR A GROUP DISTRIBUTION
•The standard deviation for a distribution with mean , where x is an interval midpoint with frequency f, and n =
Example 4: (p. 591-592)
x isf ,
1
22
n
xnfxs
10.3 NORMAL DISTRIBUTION• Continuous distribution
Outcome can take any real number
Figure 6
Figure 7
10.3 NORMAL DISTRIBUTION• Skewed distribution
The peak is not at the center
10.3 NORMAL DISTRIBUTION• Normal distribution
bell-shaped curve (4 basic properties)
• Normal curvesThe graph of normal distribution
4 Basic Properties Of Normal Distribution
1. The peak occurs directly above the mean2. The curve is symetric about the vertical
line through the mean.3. The curve never touches the x-axis4. The area under the curve is 1
The mean: Standard deviation: Standard normal curve
AREA UNDER NORMAL CURVEThere area of the shaded region under the normal curve from a to b is the probability that an observed data value will be between a and b.
Distribution of annual rainfall
Figure 11
Figure 12
Figure 13
Figure 14
ExamplesExample 1: Find the following areas under the
standard normal curve.
a)Between z = 0 and z = 1
b)Between z = -2.43 and z = 0
Example 2: Find the following areas under the standard normal curve:
a)Between .88 standard deviations below the mean and 2.35 standard deviation above the mean.
b)Between .58 standard deviations above the mean and 1.94 standard deviations above the mean.
c)The area to the right of 2.09 standard deviations above the mean.
Your Turn 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.
Z-score
• z-score
• If a normal distribution has mean and standard deviation , then the z-score for the number x is:
z = x
Example 3
Find the z-score for x = 20 if a normal distribution has a mean 35 and standard deviation 20.
Solution:
Here is mean and is the standard deviation. x
z
20 3520
0.75.
AREA UNDER NORMAL CURVE
The area under normal curve between x=a and x=b is the same as the area under the standard normal curve between the z-score for a and the z-score for b.
Examples
Example 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.
a)Find the probability that a salesperson drives between 1200 and 1600 miles per month.
b)Find the probability that a salesperson drives between 1000 and 1500 miles per month.
Examples
Example 5 The mean total cholesterol level for all Americans is 187 (mg/dl) and the standard deviation is 43 (mg/dl). Assuming total cholesterol levels are normally distributed, what is the probability that an randomly selected American has a cholesterol level higher than 250? If 200 Americans are chosen at random, how many can we expect to have total cholesterol level higher than 250?
Boxplots• First quartile Q1
• Second quartile Q2 (median)
• Third quartile Q3
• Five-number summary: minimum, Q1, Q2, Q3, maximum.
• Boxplots: graph displaying the five-number summary.
• Examples 6, 7 (p. 606, 607).
10.4 Normal Approximation to the Binomial Distribution
• The expected number of successes in n binomial trials is np, where p is the probability of success in a single trial
= np
• Variance and standard deviation
2 = np(1 – p) and = )1( pnp
• Suppose an experiment is a series of n independent trials, where the probability of a success in a single trial is always p. Let x be the number of successes in the n trials. Then the probability that exactly x success will occur in n trials is given by:
xnx ppx
n
)1(
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
112 2.
6np
The standard deviation is
1 1(1 ) 12 1 1.291.
6 6np p
Example 1
P(x=9)
In 2004, the Gallup Organization conducted a poll that asked “If you prefer to have a job outside the home, or would you prefer to stay at home and take care of the house and family?” They found that 54% of respondents answered “Outside the home”. Suppose we select 100 respondents at random.
a)Use the normal distribution to approximate the probability that at least 65 respondents would answer “outside the home”.
b)Find the probability of finding between 55 and 62 respondents who choose “Outside the home” in a random sample of 100.
Example 2
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