Transcript Sample Proportions - Mr. Baca's Math Web Page

Normal Distribution
Z-scores put to use!
Section 2.2
Reference Text:
The Practice of Statistics, Fourth Edition.
Starnes, Yates, Moore
Today’s Objectives
• The 68-95-99.7 Rule
• State mean an standard deviation for The
Standard Normal Distribution
• Given a raw score from a normal
distribution, find the standardized “z-score”
• Use the Table of Standard Normal
Probabilities to find the area under a given
section of the Standard Normal curve.
The 68-95-99.7 Rule
• How many standard deviations do you
think it would take for us to have the entire
sample or population accounted for and
just have a .03% uncertainty?
• In other words, how many standard
deviations away from the mean
encompasses almost all objects in the
study?
The 68-95-99.7 Rule
• 3!
• The 68-95-99.7 Rule describes the percent of
observations fall within 1,2 or 3 standard
deviations. Look at the visual:
The 68-95-99.7 Rule
• So,
– Approximately 68% of the observations fall
within
of the mean µ
– Approximately 95% of the observations fall
within 2 of the mean µ
– Approximately 99.7% of the observations fall
within 3 of the mean µ
The 68-95-99.7 Rule
• If I have data within 2 standard deviations,
then I'm accounting for 95% of
observations
• Question: what percent is in the left tail?
You Try!
• The distribution of number of movies AP Statistic
students watch in two weeks is close to normal.
Suppose the distribution is exactly Normal with
mean µ= 6.84 and standard deviation = 1.55 (this
is non fiction data)
• A) Sketch a normal density curve for this distribution of movies
watched. Label the points that are one, two, and three SD away
from the mean.
• B) What percent of the movies is less that 3.74? Show your work!
• C) What percent of scores are between 5.29 and 9.94? Show work!
• Remember: Always put your answers back into context!
Break!
- 5 Minutes
Standardizing Observations
• All normal distributions have fundamentally the
same shape.
• If we measure the x axis in units of size σ about a
center of 0, then they are all exactly the same
curve.
• This is called the Standard Normal Curve
– We abbreviate the normal dist. As N( µ,
)
• To standardize observations, we change from x
values (the raw observations)  z values (the
standardized observations) by the formula:
z
x

The Standard Normal Distribution
• Notice that the z-score formula always
subtracts μ from each observation.
– So the mean is always shifted to zero
• Also notice that the shifted values are
divided by σ, the standard deviation.
– So the units along the z-axis represent
numbers of standard deviations
• Thus the Standard Normal Distribution is
always N(0,1).
Example!
• The heights of young women are:
N(64.5, 2.5)
• Use the formula to find the z-score of a
woman 68 inches tall.
68  64.5
z
 1.4
2.5
• A woman’s standardized height is the
number of standard deviations by which
her height differs from the mean height of
all young women.
Normal Distribution Calculations
• What proportion of all young women are less
than 68 inches tall?
– Notice that this does not fall conveniently on one of the σ
borders
– We already found that 68 inches corresponds
to a z-score of 1.4
• So what proportion of all standardized
observations fall to the left of z = 1.4?
• Since the area under the Standard Normal
Curve is always 1, we can ask instead, what is
the area under the curve and to the left of z=1.4
– For that, we need a table!!
The Standard Normal Table
• Find Table A of the handout
– It is also in your textbook in the very back
• Z-scores (to the nearest tenth) are in the left
column
– The other 10 columns round z to the nearest hundredth
• Find z = 1.4 in the table and read the area
– You should find area to the left = .9192
• So the proportion of observations less than z =
1.4 is about 92%
– Now put the answer in context: “About 92% of all
young women are 68 inches tall or less.”
What about area above a value?
• Still using the N(64.5, 2.5) distribution, what
proportion of young women have a height of
61.5 inches or taller?
• Z = (61.5 – 64.5)/2.5 = -1.2
• From Table A, area to the left of -1.2 =.1151
– So area to the right = 1 - .1151 = .8849
• So about 88.5% of young women are 61.5”
tall or taller.
What about area between two
values?
• What proportion of young women are between
61.5” and 68” tall?
• We already know 68” gives z = 1.4 and area to
the left of .9192
• We also know 61.5” gives z = -1.2 and area to
the left of .1151
• So just subtract:
.9192 - .1151 = .8041
• So about 80% of young women are between
61.5” and 68” tall
– Remember to write your answer IN CONTEXT!!!
Given a proportion, find the observation x
• SAT Verbal scores are N(505, 110). How high must you
score to be in the top 10%?
• If you are in the top 10%, there must be 90% below you
(to the left).
• Find .90 (or close to it) in the body of Table A. What is
the z-score?
– You should have found z = 1.28
• Now solve the z definition equation for x
z
x

x  505
110
x  1.28 110  505
x  645.8
1.28 
• So you need a score of at least 646 to be in the top 10%.
How to Solve Problems
Involving Normal Distribution
• State: Express the problem in terms of the observed
variable x
• Plan: draw a picture of the distribution and shade the
area of interest under the curve.
• Do: Preform the calculations
– Standardize x to restate the problem in terms of standard normal
variable z
– Use Table A and the fact that the total area under the curve is 1
to find the required area under the standard normal curve
• Conclude: Write your conclusion in context of the
problem.
• Lets look at TB pg 120 “Tiger on the Range”
Today’s Objectives
• The 68-95-99.7 Rule
• State mean an standard deviation for The
Standard Normal Distribution
• Given a raw score from a normal
distribution, find the standardized “z-score”
• Use the Table of Standard Normal
Probabilities to find the area under a given
section of the Standard Normal curve.
Homework
TB Pg 131: 41-74 (multiples of 3)