Download Statistic for the day: The width of train tracks is 4 feet 8.5 inches. Why?

Jan 21 Statistic for the day:
The width of train tracks is
4 feet 8.5 inches.
Why?
Assignment: Read Chapter 9
Exercises from Chapter 8: 16, 18
These slides were created by Tom Hettmansperger
and in some cases modified by David Hunter
Research Question 1: How high
should I build my doorways so
that 99% of the people will not
have to duck?
Secondary Question 2: If I built my
doors 75 inches (6 feet 3 inches) high,
what percent of the people would
have to duck?
Histogram of Height, with Normal Curve
Frequency
30
20
10
0
60
70
80
Height
Question 2
Question 1
Find the value at Question 1 so that 99% of the distribution
is below it.
The value at Question 2 is 75; find the amount of distribution
above it.
Z-Scores: Measurement in
Standard Deviations
Given the mean (68), the standard deviation
(4), and a value (height say 75) compute
Z = (75-mean) / SD = (75-68) / 4 = 1.75
This says that 75 is 1.75 standard deviations
above the mean.
Answer to Question 2: What percent of people would
have to duck if I built my doors 75 inches high?
Recall: 75 has a Z-score of 1.75
From the standard normal table in the book: .96 or
96% of the distribution is below 1.75. Hence, .04
or 4% is above 1.75.
So 4% of the distribution is above 75 inches.
Histogram of Height, with Normal Curve
Frequency
30
20
4% in here
96% in here
(FIND IN TABLE)
10
0
60
70
Height
75
(GIVEN)
80
Question 2
The value at Question 2 is 75; find the amount of distribution
above it. Convert 75 to Z = 1.75 and use Table 8.1 in book.
Question 1: What is the value so that 99% of the
distribution is below it? Called the 99th percentile.
1. Look up the Z-score that corresponds to the 99th
percentile. From the table: Z = 2.33.
2. Now convert it over to inches:
2.33 = (h – 68)/4
h = 68 + 2.33(4) = 77.3
Since 77 inches is 6 feet 5 inches, 99% of the distribution
is shorter than 77 inches and they will not have to duck.
Histogram of Height, with Normal Curve
Frequency
30
20
99% in here
(GIVEN)
10
0
60
70
80
Height
(FIND):
Question 1
77.3 inches is the 99th percentile
Find the value at Question 1 so that 99% of the distribution
is below it. Look up Z-score for 99th percentile and convert
it back to inches.
Compare Heights of Females and Males
Stat 100 students Sp01
Height
80
70
60
Female
Male
Sex
Heights in Inches ( red circle is my doorway 77 inches)
85
78
75
(6-3)
65
Lakers
n=14
PSU BB
n=12
Steelers
n=30
Stat 100
Males
n=78
Shaquille O’Neal is 7 feet 1 inch
or 85 inches tall. How many
people in the country are taller?
1. We will assume that heights are normally distributed
with mean 68 inches and standard deviation 4 inches.
2. O’Neal’s Z-score is Z = (85-68)/4 = 4.25. In other words
O’Neal is 4.25 standard deviations above the mean!
We would generally consider him from a different population.
3. There is .000011 above 4.25 standard deviations.
3. There are roughly 250 million people in US.
48.8% are over the age of 20.
That is 122 million.
4. Hence, there should be roughly
.000011 times 122 million
or 1342 people taller than Shaquille O’Neal
Who Is the Tallest Person in Class?
What is your Z-Score?
What is your percentile?
How many PSU students are taller?
Suppose someone claims to have
tossed a coin 100 times and got
70 heads. Would you believe
them?
We need to know what the distribution of
the number of heads in 100 tosses looks
like for a fair coin.
 We need the mean and standard deviation
for this distribution.

Toss a coin 100 times
Repeat 500 times and form a histogram
90
80
Frequency
70
60
50
40
30
20
10
0
35
45
55
Number of heads
65
1. What is the mean?
2. What is the standard deviation?
3. Let’s suppose the smooth version is bell shaped.
So the distribution of the number
of heads in 100 tosses of a fair
coin is:








Roughly normal, mean about 50, SD about 5
What is the Z-score of 70?
Ans: 4
What is the percentile?
Ans: .999968 or 99.9968%
Now do you believe them?
NO
Weighted coin is a BETTER explanation