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S TANDARD N ORMAL C ALCULATIONS Section 2.2. N ORMAL D ISTRIBUTIONS Can be compared if we measure in...
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Transcript of S TANDARD N ORMAL C ALCULATIONS Section 2.2. N ORMAL D ISTRIBUTIONS Can be compared if we measure in...
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STANDARD NORMAL CALCULATIONSSection 2.2
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NORMAL DISTRIBUTIONS
Can be compared if we measure in units of size σ about the mean µ as center.
Changing these units is called standardizing.
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STANDARDIZING AND Z-SCORES
If x is an observation from a distribution that has mean µ and standard deviation σ, the standardized value of x is
A standardized value is often called a z-score.
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HEIGHTS OF YOUNG WOMEN The heights of young women are
approximately normal with µ = 64.5 inches and σ = 2.5 inches. The standardized height is
𝑧=h h𝑒𝑖𝑔 𝑡−64.5
2.5
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HEIGHTS OF YOUNG WOMEN A woman’s standardized height is the
number of standard deviations by which her height differs from the mean height of all women. For example, a woman who is 68 inches tall has a standardized height
𝑧=68−64.5
2.5=1.4
or 1.4 standard deviations above the mean.
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HEIGHTS OF YOUNG WOMEN A woman who is 5 feet (60 inches) tall has a
standardized height
𝑧=60−64.5
2.5=−1.8
or 1.8 standard deviations less than the mean.
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STANDARD NORMAL DISTRIBUTION
The normal distribution N(0,1) with mean 0 and standard deviation 1.
If a variable x has any normal distribution N(µ,σ) with mean µ and standard deviation σ, then the standardized variable
has the standard normal distribution.
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THE STANDARD NORMAL TABLE
Table A (front of your book) is a table of areas under the standard normal curve. The table entry for each value z is the area under the curve to the left of z.
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USING THE Z TABLE
Back to our example of women 68 inches or less. We had a z-score of 1.4. To find the proportion of observations
from the standard normal distribution that are less 1.4, locate 1.4 in Table A.
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USING THE Z TABLE
Back to our example of women 68 inches or less. We had a z-score of 1.4. To find the proportion of observations from the
standard normal distribution that are less 1.4, locate 1.4 in Table A.
What does this mean?
About 91.92% of young women are 68 inches or shorter.
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Find the proportion of observations from the standard normal distribution that are greater than -2.15.
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Find the proportion of observations from the standard normal distribution that are greater than -2.15.
Proportion = 0.0158 Remember, Table A gives us what is less than
a z-score.
1 – 0.0158 = .9842
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STEPS FOR FINDING NORMAL DISTRIBUTION
Step 1: State the problem in terms of the observed variable x.
Step 2: Standardize x to restate the problem in terms of a standard normal curve. Draw a picture of the distribution and shade the area of interest under the curve.
Step 3: Find the required area under the standard normal curve, using Table A and the fact that the total area under the curve is 1.
Step 4: Write your conclusion in the context of the problem.
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CHOLESTEROL PROBLEM The level of cholesterol in the blood is important because high cholesterol
levels may increase the risk of heart disease. The distribution of blood cholesterol levels in a large population of people of the same age and sex is roughly normal. For 14-year old boys, the mean is µ = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation is σ = 30 mg/dl. Levels above 240 mg/dl may require medical attention. What percent of 14-year-old boys have more than 240 mg/dl of cholesterol?
Step 1: State the Problem.
Level of cholesterol = x x has the N(170,30) distributionWant the proportion of boys with cholesterol level x > 240
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CHOLESTEROL PROBLEM The level of cholesterol in the blood is important because high cholesterol
levels may increase the risk of heart disease. The distribution of blood cholesterol levels in a large population of people of the same age and sex is roughly normal. For 14-year old boys, the mean is µ = 170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation is σ = 30 mg/dl. Levels above 240 mg/dl may require medical attention. What percent of 14-year-old boys have more than 240 mg/dl of cholesterol?
Step 2: Standardize x and draw a picture𝑥>240
𝑥−17030
>240 −170
30
𝑧>2.33
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CHOLESTEROL PROBLEM Step 3: Use the Table (z > 2.33)
0.9901 is the proportion of observations less than 2.33.
1 – 0.9901 = 0.0099
About 0.01 or 1%
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CHOLESTEROL PROBLEM Step 4: Write your conclusion in the context of the
problem.
Only about 1% of boys have high cholesterol.
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WORKING WITH AN INTERVAL
What percent of 14-year-old boys have blood cholesterol between 170 and 240 mg/dl?Step 1: State the problem
We want the proportion of boys with Step 2: Standardize and draw a picture
170 240x
170 ≤ 𝑥≤ 240170 −170
30≤𝑥−170
30≤
240 − 17030
0 ≤ 𝑧≤ 2.33
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WORKING WITH AN INTERVAL
What percent of 14-year-old boys have blood cholesterol between 170 and 240 mg/dl?Step 3: Use the table
z < 0 0.5000 z < 2.33 0.9901
0 < z < 2.33 0.9901 – 0.5000 = 0.4901
0 ≤ 𝑧≤ 2.33
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WORKING WITH AN INTERVAL
What percent of 14-year-old boys have blood cholesterol between 170 and 240 mg/dl?Step 4: State your conclusion in context
About 49% of boys have cholesterol levels between 170 and 240 mg/dl.
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FINDING A VALUE GIVEN A PROPORTION
Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?
Find the SAT score x with 0.1 to its right under the normal curve. (Same as finding an SAT score x with 0.9 to its left)
µ = 505, σ = 110
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FINDING A VALUE GIVEN A PROPORTION
Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?
USE THE TABLE!!! Go backwards.
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FINDING A VALUE GIVEN A PROPORTION
Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?
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FINDING A VALUE GIVEN A PROPORTION
Scores on the SAT Verbal test in recent years follow approximately the N(505,110) distribution. How high must a student score in order to place in the top 10% of all students taking the SAT?z-score = 1.28Unstandardize:
𝑥=505+ (1.28 ) (110 )=645.8
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Homework
• 2.19 p. 95• 2.21, 2.22, 2.23 p. 103