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Properties of a Normal Distribution 5.1 Introduction to Normal Distributions x • The mean, median, and mode are equal • Bell shaped and is symmetric about the mean • The total area that lies under the curve is one or 100%
Properties of a Normal Distribution
Means and Standard Deviations Curves with different means, same standard deviation
Inflection point
Means?
Inflection point
x • As the curve extends farther and farther away from the mean, it gets closer and closer to the x-axis but never touches it.
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Curves with different means, different standard deviations
• The points at which the curvature changes are called inflection points. The graph curves downward between the inflection points and curves upward past the inflection points.
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Determining Intervals
Empirical Rule
68%
About 68% of the area lies within 1 standard deviation of the mean 3.3 3.6 3.9 4.2
4.5 4.8 5.1
x
Example: An instruction manual claims that assembly time for a product is normally distributed with a mean of 4.2 hours and a standard deviation of 0.3 hour. Determine the interval in which 95% of the assembly times fall.
About 95% of the area lies within 2 standard deviations
95% of the data will fall within 2 standard deviations of the mean.
About 99.7% of the area lies within 3 standard deviations of the mean
4.2 – 2 (0.3) = 3.6 and 4.2 + 2 (0.3) = 4.8. 95% of the assembly times will be between 3.6 and 4.8 hrs.
The Standard Normal Distribution
Chptr 2: The Standard Score
Standard normal distribution: mean = 0, standard deviation = 1
The standard- or z-score, represents the number of standard deviations a random variable x falls from the mean:
Using z-scores any normal distribution can be transformed into the standard normal distribution.
Test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of: (a) 161 (b) 148 (c) 152
–4 –3 –2 –1 0 1
2 3
4
z
If a normal distribution is standardized using tables, then each value must be standardized to find probabilities.
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Cumulative Areas The total area under the curve is one.
Cumulative Areas Using a standard normal table, find the cumulative area for a z-score of –1.25
Sum left to right
z
–3 –2 –1 0 1 2 3
z
–3 –2 –1 0 1 2 3
Cumulative area is close to 0 for z-scores close to –3.49
Pg. A16: down the z column on the left to z = –1.2 and across to the cell under .05 = 0.1056, the cumulative area. The probability that z is at most –1.25 is 0.1056.
Cumulative area for z = 0 is 0.50 Cumulative area is close to 1 for z-scores close to 3.49
Finding Probabilities
Finding Probabilities
To find the probability that z is less than a given value, read the cumulative area in the table corresponding to that z-score.
To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1. Find P(z > –1.24).
Find P(z < –1.45) P (z < –1.45) = 0.0735
0.1075 0.8925
–3 –2 –1
0 1
2 3
z
Read down the z-column to –1.4 and across to .05 = 0.0735
z –3 –2 –1 0 1 2 3 The cumulative area (area to the left) is 0.1075. So the area to the right is 1 – 0.1075 = 0.8925. P(z > –1.24) = 0.8925
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Finding Probabilities
Summary
The probability that z is between two values: find the cumulative areas for each and subtract the smaller area from the larger.
To find the probability that z is less than a given value, read the corresponding cumulative area.
Find P(–1.25 < z < 1.17)
z
-3 -2 -1 0 1 2 3 To find the probability is greater than a given value, subtract the cumulative area in the table from 1.
–3 –2 –1 0 1 2 1. P(z < 1.17) = 0.8790
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z
2. P(z < –1.25) = 0.1056
3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734
Probabilities can’t be negative, so subtract smaller from larger
z
-3 -2 -1 0 1 2 3
To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. -3 -2 -1
*cdf*
0 1 2 3
z
Probabilities and Normal Distributions
Section 5.2 Normal Distributions Finding Probabilities
If a random variable, x, is normally distributed, then the probability that x will fall within an interval is equal to the area under the curve in the interval. Example: IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a person selected at random will have an IQ score less than 115.
100 115
To find the area, first find the standard score equivalent to x = 115
z
115 100 1 15
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Probabilities and Normal Distributions
Monthly utility bills in a city are normally distributed with a mean of $100 and a standard deviation of $12. A utility bill is randomly selected. Find the probability it is between $80 and $115.
Normal Distribution
Normal Distribution: = 100; = 12 100 115
P(80 < x < 115)
SAME
SAME
Find P(x < 115). Standard Normal Distribution
Application Example
P(–1.67 < z < 1.25) Find P(z < 1).
0 1 From Standard Normal Table: P(z < 1) = 0.8413, so P(x 30 the sampling distribution of will be normal mean standard deviation
Find the z-score for a sample mean of 70:
z
2.14
There is a 0.0162 or 1.62% probability that a sample of 60 sockeye will have a mean length greater than 70 cm. What is probability that 1 fish will be > 70 cm? P(z>0.28) = 1-0.6103 = 0.3897 39%
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Application Central Limit Theorem A long time ago, the mean price of gasoline in California was $1.164 per gallon. What is the probability that the mean price for a sample of 38 gas stations in California is between $1.169 and $1.179? Assume the standard deviation = $0.049.
Since n > 30 the sampling distribution of
will be normal
Application Central Limit Theorem P( 0.63 < z < 1.90) = 0.9713 – 0.7357 = 0.2356
z
mean standard deviation
Calculate the z-score for sample values of $1.169 and $1.179.
.63 1.90 The probability is 0.2356 that the mean for the sample is between $1.169 and $1.179. Hint: drawing the distribution, values, and area of interest will help keep calculations clear.
Central Limit Theorem Section 5.5 Creature Cast Central Limit Theorem video http://vimeo.com/75089338
Normal Approximation to Binomial Distributions
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Binomial Distribution Characteristics • There are a fixed number of independent trials, n. • Each trial has 2 outcomes, Success or Failure. • The probability of S on a single trial is p and the probability of F is q. In total:
p+q=1
• We can find the probability of exactly x successes out of n trials. Where x = 0 or 1 or 2 … n. • x is a discrete random variable representing a count of the number of S’s in n trials.
Application 34% of Americans have type A+ blood. If 500 Americans are sampled at random, what is the probability at least 300 have type A+ blood? Using Chapter 4 you could calculate the probability that exactly 300, exactly 301… exactly 500 Americans have A+ blood type and then add the probabilities (but this should drive you crazy). Alternatively…use normal curve probabilities to approximate binomial probabilities. If np 5 and nq 5, then the binomial random variable x is approximately normally distributed with mean
= np and standard deviation
Why np 5 and nq 5?
0
1
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5
n=5 p = 0.25, q = .75 np =1.25 nq = 3.75
Binomial Probabilities The binomial distribution is discrete with a probability histogram graph. The probability that a specific value of x will occur is equal to the area of the rectangle with midpoint at x. Example: If n = 50 and p = 0.25 find
n = 20 p = 0.25 np = 5 nq = 15
0.111
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n = 50 p = 0.25 np = 12.5 nq = 37.5 0
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40
Add the areas of the rectangles with midpoints at x = 14, x = 15, and x = 16: 0.111 + 0.089 + 0.065 = 0.265
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0.089
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0.065
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Correction for Continuity
Correction for Continuity
Use the normal approximation to the binomial distribution to find .
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Values for the binomial random variable x are 14, 15 and 16.
Normal Approximation to the Binomial Use the normal approximation to the binomial to find . Find the mean and standard deviation using binomial distribution formulas:
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The interval of values under the normal curve is
To ensure the boundaries of each rectangle are included in the interval, subtract 0.5 from a left-hand boundary and add 0.5 to a right-hand boundary.
Application A survey of Internet users found that 75% favored government regulations of “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation. Since np = 150 5 and nq = 50 5, can use the normal approximation to the binomial distribution.
Adjust the endpoints to correct for continuity P(13.5 ≤ x 16.5). Convert each endpoint to a standard score:
The binomial phrase of “fewer than 140” means up to 139: 0, 1, 2, 3…139. Use the correction for continuity to translate to the continuous variable in the interval . Find P(x< 139.5).
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Application A survey of Internet users found that 75% favored government regulations of “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation. Use the correction for continuity P(x