Normal Distribution Activity in Statistical Methods | STA 2023, Study notes of Data Analysis & Statistical Methods

Material Type: Notes; Professor: Murphy; Class: Statistical Methods; Subject: STA: Statistics; University: Valencia Community College; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/04/2009

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Normal Distribution Activity (in-class)
Practice Demo:
Suppose that 33 percent of women believe
in the existence of aliens. If 100 women
are selected at random, what is the
probability that more than 45 percent of
them will say that they believe in aliens?
SET UP:
Role #1: “100 women selected”
“45 percent of them”
Role #2:
p
ˆ
(
)
pp =
ˆ
µ
( )
n
pp
pSD )1(
ˆ
=
Role #3:
(
)
=p
ˆ
µ
0.33
( )
=100
)33.01(33.0
ˆ
pSD
0.047
Role #4:
p
ˆ
.19 .24 .28 .33 .38 .42 .47
pf3
pf4
pf5
pf8
pf9

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Practice Demo:

Suppose that 33 percent of women believe

in the existence of aliens. If 100 women

are selected at random, what is the

probability that more than 45 percent of

them will say that they believe in aliens?

SET UP:

Role #1: “100 women selected”

“45 percent of them”

Role #2: p ˆ μ ( p ˆ ) = p ( )

n

p p SD p

Role #3: μ ( p ˆ) =0.33 ( ) ≈

SD p ˆ 0.

Role #4:

p ˆ

1.Suppose family incomes in a town are

normally distributed with a mean of

$1,200 and a standard deviation of $

per month. What is the probability that a

family has an income between $1,

and $2,250?

SET UP:

Role #1: No sample of size greater

than one was taken.

Family incomes are the

population.

Role #2: X μ σ

Role #3: μ = 1200 σ = 600

Role #4:

X

3.Find the area under the curve between

the z-scores of -2 and 1.

SET UP:

Role #1: No sample of size greater

than one was taken.

“z-scores”

Role #2: z μ = 0 σ= 1

Role #3: μ = 0 σ = 1

Role #4:

Z

4.Adult nose length is normally distributed

with mean 45mm and standard deviation

6mm. Find the probability that the

sample mean nose length is between

44mm and 46mm for random samples of

36 adults.

SET UP:

Role #1: “samples of 36 adults”

“ sample mean nose length”

Role #2: x μ ( x ) =μ ( )

n

SDx

σ

Role #3: μ ( x ) = 45 ( ) = =

SD x 1

Role #4:

x

6.A restaurateur anticipates serving about

180 people on a Friday evening, and

believes that about 20% of the patrons

will order the chef’s steak special. How

many of those meals should he plan on

serving in order to be pretty sure of

having enough steaks on hand to meet

customer demand? Justify your answer,

including an explanation of what “pretty

sure” means to you.

SET UP:

Role #1: “serving about 180 people”

“20% of the patrons”

Role #2: p ˆ μ ( p ˆ ) = p ( )

n

p p SD p

Role #3: μ ( p ˆ) =0.2 ( ) ≈

SD p ˆ 0.

Role #4:

p ˆ

7.In this example we will be interested in the

heights of northern European males. We take

such a person and reduce him to a single number

via the usual operations for measuring someone's

height. Then we model the height of northern

European males as a normal population with mu

= 150 cm and sigma = 30 cm. If we sample one

northern European male, what's the probability

that his height will fall outside of 140 and 170?

In other words, what are the chances that he'll be

either below 140, or he'll be above 170 in height?

That's what we mean by the word "outside."

SET UP:

Role #1: “sample one northern European…”

Heights of northern European

males are the population.

Role #2: X μ σ

Role #3: μ = 150 σ = 30

Role #4:

X