Transcript Chapter 10 - SaigonTech

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)
n
x1 + x2 + x3+ ….+ xn =  xi
i 1
• The mean (arithmetic average)
The mean of the n numbers x1, x2, x3, … xn is
x1  x2  ...  xn  x

x
n
n
Examples 5, 6 (p. 576, 577)
Example 7 (p. 578)
MEAN OF A GROUPED DISTRIBUTION
• The mean of a distribution where x represents
the midpoints, f the frequencies, and n=  f , is
 ( xf )
x
n
• Sample mean
• Population mean
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 – min
Example 1: (p. 587)
• Deviations from the mean of a sample of n
numbers x1, x2 , x3, … xn, with mean x is:
x1 – x
x2 – x
…
xn –
x
10.2 MEASURES OF VARIATION
• Deviations from the mean of a sample of n
numbers x1, x2 , x3, … xn, with mean x is:
x1 – x
x2 – x
…
xn –
x
Example 2: (p. 587)
10.2 MEASURES OF VARIATION
SAMPLE VARIANCE
• The variance of a sample of n numbers x1,
x2 , x3, … xn, with mean
Shortcut formula:
s2
=

Population variance: s2 =
x , is
s2
x 2  nx 2
n 1
 ( x  x)
n
2
=
 ( x  x)
n 1
2
10.2 MEASURES OF VARIATION
SAMPLE STANDARD DEVIATION
• The standard deviation of n numbers
x1, x2 , x3, …, xn, with mean x , is
s

( x x)2
n 1
Example 3: (p. 590)

x 2  nx 2
n 1
10.2 MEASURES OF VARIATION
STANDARD DEVIATION FOR A GROUP
DISTRIBUTION
•The standard deviation for a distribution
with mean x, where x is an interval
midpoint with frequency f, and n =  f , is
s
 fx
2
 nx
n 1
Example 4: (p. 591-592)
2
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 curves
The graph of normal distribution
4 Basic Properties Of
Normal Distribution
1. The peak occurs directly above the mean
2. The curve is symetric about the vertical
line through the mean.
3. The curve never touches the x-axis
4. The area under the curve is 1
The mean: m
Standard deviation: s
Standard normal curve: m0, s  1
AREA UNDER NORMAL CURVE
There 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
Examples
Example 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 m and
standard deviation s, then the z-score for
the number x is:
xm
z=
s
Example 3
Find the z-score for x = 20 if a normal distribution has a
mean 35 and standard deviation 20.
Solution:
z
xm
s
20  35

20
 0.75.
Here m is mean and s is the standard deviation.
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 zscore 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
m = np
• Variance and standard deviation
s2 = np(1 – p) and s =
np(1 p)
• 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:
n x
n x
  p (1  p)
 x
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
m  np  12    2.
6
The standard deviation is
 1  1 
s  np(1  p)  12  1    1.291.
 6  6 
Example 1
P(x=9)
Example 2
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.