Transcript 2.2: Normal Distributions

Homework for 2.1 Day 1:
41, 43, 45, 47, 49, 51
1)
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To use the 68-95-99.7 rule to estimate the percent of
observations from a Normal Distribution that fall in an
interval
Use the standard Normal distribution to calculate the
proportion of values in a specified interval
Use the standard Normal distribution to determine a z-score
from a percentile
Use Table A to find the percentile of a value from any Normal
distribution
Make an appropriate graph to determine if a distribution is
bell-shaped
Use the 68-95-99.7 rule to assess normality of a data set
Interpret a normal probability plot
A. A specific type of density curve is a Normal Curve
B. The distributions they describe are called Normal
distributions
C. Characteristics:
Have the same overall shape: symmetric, single
peaked, bell-shaped
b) Mean-𝜇 and standard deviation is 𝜎
c) Mean is in the center and so is median (think back to
previous section and symmetrical denisity curves)
d) 𝜎 controls the spread
a)
 On the board
Keep in mind: these are
special properties of
normal distributions NOT
ALL DENSITY CURVES!!!!
 Example: The mean of MLB batting averages is 0.261
with a standard deviation of 0.034. Suppose that the
distribution is normal (this is key to know)with 𝜇 =
0.261 𝑎𝑛𝑑 𝜎 = 0.034
a) Sketch a normal density curve for this distribution of
batting averages. (Take notice of points that are
1,2,or 3 standard deviations away from the mean)
b) What percent of the batting averages are above
0.329?
c) What percent of the batting averages are between
0.193 and 0.295?
Page 114
 Z-Scores!!
 Check out the box on page 115
 Now, let’s look up our Table A in the back of the book…
 These are the values of the z scores, the area under the
curve to the left of z
 Example: Suppose we wanted to find the proportion
of observations in a Normal distribution that were
more than 1.53 standard deviations above the mean.
 Ok, so we want to know what proportion of
observations in the standard Normal distribution are
greater than, z=1.53
 First, find the area to the left of z=1.53 in table A
 What is that value?
 Z=1.53
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0.9370
Now, this is the value to the left of 1.53, or you can
think of it as ≤1.53
If I want to know greater than 1.53 I should…
1-0.9370=0.0630
Why?
AREA UNDER THE CURVE ALWAYS = 1
Example: Find the proportion of observations from the
standard Normal distribution that are between -0.58 and
1.79
 Look up the z-scores for those values:
 Z=-0.58→ 0.9633
 Z=1.79 →0.2810
 Values between: I should subtract them!
 Answer: 0.6823
 Look at the beige box on page 120
 Example: In the 2008 Wimbledon tennis tournament,
Rafael Nadal averaged 115 MPH on his first serves.
Assume that the distribution of his first serve speeds is
Normal with a mean of 115 mph, and a standard
deviation of 6 mph. About what portion of his speeds
would you expect to exceed 120 mph?
Conclude: About 20% of Nadal’s first serves will travel
more than 120 mph.
 What percent of Rafael Nadal’s first serves are between
100 and 110 mph?
Conclude: About 20% of Nadal’s first serves will travel
between 100 and 110 mph
 The heights of three-year old females are
approximately Normally distributed with a mean of
94.5 cm and a standard deviation of 4 cm. What is the
third quartile of this distribution?
 Hint: Look at the z-chart in reverse!!
Conclude: The third quartile of three-year old females’
heights is 97.18 cm
Plot the data- you can use a dotplot, stemplot, or
histogram. See if the shape is a
bell
2. Check to see if the data follows the 68-95-99.7
Rule
A. Find the mean and standard deviation
B. Calculate 1, 2, and 3 standard deviations to the
right and left
C. Find the percent of the data that lies between
those standard deviations
1.
12.9, 13.7, 14.1, 14.2, 14.5, 14.5, 14.6,
14.7, 15.1, 15.2, 15.3, 15.3, 15.3, 15.3, 15.5,
15.6, 15.8, 16.0, 16.0, 16.2, 16.2, 16.3,
16.4, 16.5, 16.6, 16.6, 16.6, 16.8, 17.0,
17.0, 17.2, 17.4, 17.4, 17.9, 18.4
No space in the fridge?
4
Frequency
3
2
1
0
12.9 14.1 14.5 14.7 15.2 15.5 15.8 16.2 16.4 16.6
Usable Capacity
17
17.4 18.4
 Combined with the graph, and the
fact that these numbers are
extremely close to our rule, we have
good evidence to believe this is a
Normal Distribution