Презентация на тему Calculating the probability of a continuous random variable – Normal Distribution. Week 9 (1)

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exam statistics

any value in an interval on the real line

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DR SUSANNE HANSEN SARAL

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Ch. 5-

x

b

μ

a

x

b

μ

a

x

b

μ

a

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Ch. 5-

Any normal distribution, F(x) (with any mean and standard deviation combination) can be transformed into the standardized normal distribution F(z), with mean 0 and standard deviation 1







We say that Z follows the standard normal distribution.

Z

f(Z)

0

1

using the Standard Normal Table
For

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FIGURE 2.9 – Normal Distribution

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using the Standard Normal

Step 2

Look up the probability from the table of normal curve areas

Column on the left is Z value

Row at the top has second decimal places for Z values

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Inc.
TABLE 2.10 – Standardized Normal Distribution (partial)
For Z =

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= P(z > -2) = .9772

In probability terms, a

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The Standard Normal Table
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P( z

To find the probability of: P (z > 1) and P (z < -1) we will use the complement rule:

P( z > 1.0) = 1 - .8413 = 0.1587

Finding the probability of z-scores

P ( z < + 0.55) = 0.7088 or 70.88 %
P (z > + .55) = 1.0 – 0.7088 = 0.2912 or 29.12%
P ( z > - 0.55) = 0.7088 or 70.88 %
P ( z < - 0.55) = 1.0 - .7088 = 0.2912 or 29.12 %
P ( z < + 1.65) = 0.9505 or 95.05 %
P (z > + 1.65) = 1.0 – 0.9505 = 0.0495 or 4.96 %
P( z > - 2.36) = .9909 or 99.09 %
P ( z < + 2.36) = .9909 or 99.09 %

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of bottles that:
 

than 505 ml
 

> -1.05 ) =
P (z < - 3.34) =

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< + 1.05) = 0.8531 or 85.31 %
P (z

Exercise: Find the probability of z-scores and draw a graph of the probability

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Example
 
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FIGURE 2.10
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Inc.
FIGURE 2.10
From the table for Z = 1.25 area P(z

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Inc.
FIGURE 2.10
From the table for Z = 1.25 area =

The probability is about 0.89 or 89 % that Haynes will not violate the contract

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2.10
P(z > 1.25)
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the company will not finish in 125 days and

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FIGURE 2.10

From the table for Z = 1.25 area P(z > 1.25) =
1 – P(z < 1.25) = 1 - 0.8944 =
0.1056 or 10.56 %

P(z > 1.25)

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or less, Haynes will get a bonus of $5,000
What

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or less, bonus = $5,000
Probability of bonus? P

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FIGURE 2.11

Because the distribution is symmetrical, equivalent to Z = 1.25 P(z < 1.25) so area = 0.8944

0.8944

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= $5,000
Probability of bonus?
Haynes Construction Company
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FIGURE 2.11

P(z < -1.25) = 1.0 – P(z < 1.25)
= 1.0 – 0.8944 = 0.1056
The probability of completing the contract in 75 days or less is about 11%

Because the distribution is symmetrical, equivalent to Z = 1.25 so area = 0.89435

0.8944

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and 125 days?
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FIGURE 2.12
P(110

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and 125 days?
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FIGURE 2.12
P(110

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and 125 days?
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FIGURE 2.12
P(110

P(110 ≤ X < 125) = 0.8944 – 0.6915
= 0.2029
The probability of completing between 110 and 125 days is about 20%

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