Download Elementary Statistics and Inference Elementary Statistics and

Elementary Statistics and
Inference
22S:025 or 7P:025
Lecture 23
1
Elementary Statistics and
Inference
22S:025 or 7P:025
Chapter 18
2
Chapter 18
A.
Introduction
ƒ
When we toss a coin several times, the tosses are
independent – the probability (chance) of tossing a
head is the same for each toss:
P( H ) =
1
2
ƒ Possible patterns of outcomes for 5 tosses of a
coin.
3
1
Chapter 18 (cont.)
HHHHH – 5
HHTTT
HHHHT
HTTHT
HHHTH
HTTTH
HTHTT
HHTHH – 4
THTTH – 2
HTHHH
TTHTH
THHHH
TTTHH
THHTT
TTHHH
TTHHT
THTHH
HTTHT
THHTH
THHHT
TTTTH
HTHHT – 3
TTTHT
HHTHT
TTHTT – 1
HHHTT
THTTT
HHTTH
HTTTT
HTHTH
HTTHH
TTTTT - 0
4
Chapter 18 (cont.)
„
Toss a coin 5 times – outcomes
Number of
Heads
Patterns
0
1
1
5
2
10
3
10
4
5
5
1
2 5 = 32 possible pattern
5
Chapter 18 (cont.)
„
The number of heads can be determined mathematically
– in “n” tosses of the coin the number of heads and tails
is:
n!
n ⋅ (n − 1) ⋅ (n − 2) − − − 1
=
X !(n − X !) X ( X − 1) − − − 1 ⋅ (n − X )(n − X − 1) − − − 1
If n=5,
n=5 X=3:
5!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
=
= 10 patterns for 3 heads
3!(2)! (3 ⋅ 2 ⋅1)(2 ⋅1)
If n=5, X=2:
5!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
=
= 5 patterns for 2 heads
4!1! (4 ⋅ 3 ⋅ 2 ⋅1)(1)
6
2
Chapter 18 (cont.)
„
We can approximate the probability of obtaining the
number of heads in “n” repetitions of tossing the coin (a
binomial event), using the Normal Curve approximation.
7
Chapter 18 (cont.)
B.
Probability Histograms
8
Chapter 18 (cont.)
9
3
Chapter 18 (cont.)
10
Chapter 18 (cont.)
„
As the number of tosses increases the probability
histogram approaches a normal distribution.
„
The sum on the pair of dice approaches a normal
distribution histogram as the number of repetitions
increases.
„
The product of the numbers on a pair of dice does not
approach a normal distribution as the number of tosses
increases.
11
Chapter 18 (cont.)
12
4
Chapter 18 (cont.)
13
Chapter 18 (cont.)
Exercise Set A – (pp. 312-315) #1, 2, 3, 4, 5, 6
14
Chapter 18 (cont.)
15
5
Chapter 18 (cont.)
C.
Probability Histograms and The Normal Curve
„
In tossing a coin, the count of heads (tails) probability
histogram approaches a Normal probability histogram.
16
Chapter 18 (cont.)
17
Chapter 18 (cont.)
18
6
Chapter 18 (cont.)
19
Chapter 18 (cont.)
D.
Normal Approximation
„
The Normal Curve represents the probability
distribution of a continuous variable and can be used to
approximate the probability of obtaining sums/counts in
a chance process.
Example: A coin is tossed 100 times. Estimate the
chance of obtaining
a) exactly 50 heads
b) between 45 and 55 heads inclusive (include
endpoints)
c) between 45 and 55 heads exclusive (exclude
endpoints)
20
Chapter 18 (cont.)
a) avg of the box = ½
SD of the box = ½
⎛1⎞
E (count of heads) = n ⋅ avg of the box = 100⎜ ⎟ = 50
⎝ 2⎠
⎛1⎞
SE (count of heads) = n ⋅ SD of the box = 100 ⎜ ⎟ = 5
⎝2⎠
7.97%
49.5
SE=5
50.5
X
M=50
Z=
Z=-.10
Z=.10
X − Mean
SD
About 7.97% of scores are between Z±.10. The probability
of exactly 50 heads ≈ 7.97%.
21
7
Chapter 18 (cont.)
b)
SE=5
72.87%
55.5
44.5
Inclusive – include
end points
X
M=50
Z=
ƒ
ƒ
− 5.5
= −1.10
5
Z=
Z=
5.5
= 1.10
5
X −M
SE
Percent of scores between ±Z=1.10 is 72.87%.
Probability of scores between 44.5 and 55.5 is .7287,
or 72.87% of scores.
22
Chapter 18 (cont.)
c)
SE=5
63.19%
54.5
45.5
Exclusive – exclude
end points
X
M=50
Z=
ƒ
ƒ
45.5 − 50
= −.90
5
Z=
54.5 − 50
= +.90
5
Z
Percent of scores between ±Z=.90 is 63.19%.
Probability of scores between 45 and 55 is .6319 or
63.19% of scores.
23
Chapter 18 (cont.)
Note: Ordinarily we would find such probabilities by finding
area between 45 and 55. This amounts to replacing
the area under the histogram between 45 and 55 by
the area under the Normal Curve between the
corresponding values in standard score units (Zscores).
) Keeping
p g track of the end p
points has an
official name – the continuity connection. See a
more detailed comment on page 318 of text.
Exercise Set B – (pp. 318-319) #1, 2, 3, 4, 5, 6
24
8
Chapter 18 (cont.)
#3. A coin is tossed 100 times. Estimate the chance of
obtaining 60 heads.
SE=5
M=50 59.5
Z=
59.5 − 50 9.5
=
= 1.8
5
5
X
60.5
Z=
60.5 − 50 10.5
=
= 2.1
5
5
Z
25
Chapter 18 (cont.)
„
Area between mean and 2.1 standard score units = 48.22.
„
Area between mean and 1.8 standard score units = 46.40.
„
The area (percent) between 1.8 and 2.1 standard score
units is 1.8%.
„
Probability of obtaining exactly 60 heads is about 1.82% or
.0182.
26
27
9
28
10