Example: Uniform Distribution

Chapter 6. Continuous Random Variables

A continuous random variable has a uniform distribution if its values are spread evenly over a certain range.

Reminder:

Continuous random variable takes infinitely many values Those values can be associated with measurements on a continuous scale (without gaps or interruptions)

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Density Curve

Example: Uniform Distribution

A density curve is the graph of a continuous probability distribution. It must satisfy the following properties:

A continuous random variable has a uniform distribution if its values are spread evenly over a certain range.

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Example: voltage output of an electric generator is between 123 V and 125 V. The actual voltage level may be anywhere in this range.

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1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve cannot fall below the x-axis.) 4

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Using Area to Find Probability

Area and Probability

Given the uniform distribution illustrated, find the probability that a randomly selected voltage level is greater than 124.5 volts.

Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.

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Shaded area represents voltage levels greater than 124.5 volts. The area corresponds to probability: P = 0.25. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

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Standard Normal Distribution

Standard Normal Distribution

The standard normal distribution is a bellshaped probability distribution with  = 0 and  = 1. The total area under its density curve is equal to 1.

Standard normal distribution has the following properties: 1. Its graph is bell-shaped 2. It is symmetric about its center 3. Its mean is equal to 0

( = 0)

4. Its standard deviation is 1 ( = 1) 7

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Examples

Standard Normal Distribution: Areas and Probabilities

Thermometers are supposed to give readings of 0ºC at the freezing point of water.

Probability that the standard normal random variable takes values less than z is given by the area under the curve from the left up to z (blue area in the figure)

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Since these instruments are not perfect, some of them give readings below 0ºC and others above 0ºC. Assume the following for the readings: • The mean is 0ºC • The standard deviation is 1ºC • They are normally distributed

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Example 1

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Look at Table A-2

If one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.27º. µ=0

σ=1

P (z < 1.27) = ???

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Using StatCrunch

Using StatCrunch

1.27

P(z –1.23) P = 0.89065146 P ≈ 0.8907

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Example 2

Example 3 A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees.

P (z > 1.23) = 0.8907

σ=1

µ=0

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P ( –2.00 < z < 1.50) = ???

Thus, there is a 89.80% change the picked thermometer will read above -1.23 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

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Notation

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P(a < z < b) = P(z < b) – P(z < a)

P(z < a) denotes the probability that the z score is less than a.

P(z > a)

P(a < z < b)

P(z < b)

P(z < a)

denotes the probability that the z score is greater than a.

Example:

P(a < z < b) denotes the probability that the z score is between a and b.

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P(-1.5 < z < 0.7) = P(z < 0.7) – P(z < -1.5) P(-1.5 < z < 0.7) = 0.7580364 – 0.0668072 P(-1.5 < z < 0.7) = 0.6912 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

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Finding z Scores When Given Probabilities

Finding z Scores When Given Probabilities

5% or 0.05

(z score will be positive)

1.645

Finding the z-score separating 95% bottom values from 5% top values. 25

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Finding z Scores When Given Probabilities

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Finding z Scores When Given Probabilities - cont

Enter the probability on the right box then hit Compute

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The z-score will be in the left box.

(One z score will be negative and the other positive)

Example: For top 5% (i.e. 0.05) z-score= 1.6449

Finding the Bottom 2.5% and Upper 2.5% 27

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Normal distributions that are not standard

Notation We use za to represent the z-score separating the top a from the bottom 1-a.

All normal distributions have bell-shaped density curves.

Examples: z0.025 = 1.96, z0.05 = 1.645

A normal distribution is standard if its mean  is 0 and its standard deviation  is 1. A normal distribution is not standard if its mean  is not 0, or its standard deviation  is not 1, or both.

Area = a

We can use a simple conversion that allows us to standardize any normal distribution so that Table A-2 can be used.

Area = 1-a za Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

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Converting to a Standard Normal Distribution

Conversion Formula Let x be a score for a normal distribution with mean  and standard deviation  We convert it to a z score by this formula:

z=

z=

x– 

x–µ



(round z scores to 2 decimal places) 31

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Example – Weights of Passengers

Finding x Scores When Given Probabilities

Weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. If one passenger is randomly selected, what is the probability he/she weighs less than 174 pounds?

P(x < 174) = ???

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µ = 172 and σ = 29

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x − µ 174 – 172 z= σ = = 0.069 29 P(z < 0.069) = 0.5279

1. Use StatCrunch to find the z score corresponding to the given probability (the area to the left). 2. Use the values for µ, , and the z score found in step 1, to find x: x = µ + (z • )

P(x < 174) = 0.5279

(If z is located to the left of the mean, be sure that it is a negative number.) 33

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Example – Lightest and Heaviest

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Central Limit Theorem

The weights of taxi passengers have a normal distribution with mean 172 lb and standard deviation 29 lb. Determine what weight separates the lightest 99.5% from the heaviest 0.5%?

The Central Limit Theorem tells us that the distribution of the sample mean x for a sample of size n approaches a normal distribution, as the sample size n increases.

µ = 172 and σ = 29

P(x < ???) = 0.995 P(z < 0.995) = 2.575 x = µ + σ*z = 172 + 2.575*29 = 246.7 Separating weight: 247 ib

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Central Limit Theorem – cont.

Central Limit Theorem

Conclusions:

Given:

1. The distribution of the sample mean x will, as the sample size increases, approach a normal distribution.

1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation .

2. The mean of that normal distribution is the same as the population mean µ.

2. A random sample of size n is selected from the population.

3. The standard deviation of that normal distribution is  n. (So it is smaller than the standard deviation of the population.) 37

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Practical Rules:

Formulas

1. For samples of size n larger than 30, the distribution of the sample mean can be approximated by a normal distribution.

the mean

µx = µ

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the standard deviation

x = n

2. If the original population is normally distributed, then for any sample size n, the sample means will be normally distributed. 3. We can apply Central Limit Theorem if either n>30 or the original population is normal.

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Example: Water Taxi Passengers

Example – cont

Assume the population of taxi passengers is normally distributed with a mean of 172 lb and a standard deviation of 29 lb

a) Find the probability that if an individual man is randomly selected, his weight is greater than 175 lb.

a) Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb.

µ = 172 σ = 29 P(x > 175) = ???

b) Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb.

n = 20 µx = µ σx = σ/ 𝒏 P(x > 175) = ???

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µ = 172

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σ = 29

x − µ 𝟏𝟕𝟓 −𝟏𝟕𝟐 Z = σ = 𝟐𝟗 = 0.10

P(z > 0.10) = 0.4602

P(x > 175) = 0.4602 41

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Example – cont

Example - cont

b) Find the probability that 20 randomly selected men will have a mean weight that is greater than 175 lb (so that their total weight exceeds the safe capacity of 3500 pounds). µ = 172 σ = 29 n = 20 Z=

a) Find the probability that if an individual passenger is randomly selected, his weight is greater than 175 lb.

P(x > 175) = 0.4602

x − µx x −µ 𝟏𝟕𝟓 −𝟏𝟕𝟐 = σ = = 0.4626 𝟐𝟗 σx 𝒏

b) Find the probability that 20 randomly selected passengers will have a mean weight that is greater than 175 lb.

𝟐𝟎

P(z > 0.4626) = 0.3228

P(x > 175) = 0.3228 It is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate from the mean.

P(x > 175) = 0.3228 Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

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Review

Approximation of a Binomial Distribution with a Normal Distribution

Binomial Probability Distribution If np  5 and nq  5

1. The procedure must have a fixed number of trials, n. 2. The trials must be independent.

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3. Each trial must have all outcomes classified into two categories (commonly, success and failure). 4. The probability of success p remains the same in all trials (the probability of failure is q=1-p)

Then µ = np and  =

npq

and the random variable has

Solve by binomial probability formula or calculator

a

distribution. (normal)

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Procedure for Using a Normal Distribution to Approximate a Binomial Distribution

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Continuity Correction When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a whole number x in the binomial distribution by representing the number x by the interval from x – 0.5 to x + 0.5 (that is, adding and subtracting 0.5).

1. Verify that both np  5 and nq  5. If not, you cannot use normal approximation to binomial. 2. Find the values of the parameters µ and  by calculating µ = np and  = npq. 3. Identify the discrete whole number x that is relevant to the binomial probability problem. Use the continuity correction (see next).

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at least 8 (includes 8 and above)

Example: Finding the Probability of “At Least 122 Men” Among 213 Passengers

more than 8 (doesn’t include 8) at most 8 (includes 8 and below) fewer than 8 (doesn’t include 8)

The value 122 is represented by the interval (121.5,122.5)

exactly 8

The values “at least 122 men” are represented by the interval starting at 121.5. Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

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