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STAT 2300 NOTESCHAPTER 5: CONTINUOUS RANDOM

VARIABLES

Bill Welbourn Chapter 5: STAT 2300 – Summer 2009

5.1 Continuous Probability Distributions 5.3: The Normal Distribution

Outline

1 5.1 Continuous Probability DistributionsTerminology

2 5.3: The Normal Distribution

Terminology

Definition

Continuous RVRecall from Section 4.1, a continuous random variable mayassume any numerical value in one or more intervals.Use a continuous probability distribution to assignprobabilities to intervals of values.The probability of a continuous probability distribution taking asingle value (within its support) is zero.The curve f (x) is the continuous probability distribution of thecontinuous random variable x , provided that the area under thecurve, f (x), corresponding to a specified interval, represents theprobability of x taking a value within the interval.Other names for a continuous probability distribution:

probability curve; orprobability density function.

Terminology

Definition

Terminology

Definition

Terminology

Definition

Terminology

Definition

Terminology

Definition

Terminology

Definition

Terminology

Properties of a Continuous RV

Properties of f (x): f (x) is a continuous function such thatf (x) ≥ 0 for any value x within the support of the RV x .The total area under the curve of f (x) must equal one.

Essential notion: An area under a continuous probabilitydistribution is a probability.

Terminology

Area and Probability

The blue-colored area under the probability curve f (x), from the valuex = a, to x = b, is the probability that x could take any value withinthe interval, [a,b].

Symbolically, P(a ≤ x ≤ b).Equivalently, P(a < x < b). To see this, it holds

P(a ≤ x ≤ b) = P(x = a)︸︷︷︸=0

+P(a < x < b) + P(x = b)︸︷︷︸=0

= P(a < x < b).

Terminology

Area and Probability

The blue-colored area under the probability curve f (x), from the valuex = a, to x = b, is the probability that x could take any value withinthe interval, [a,b].

Symbolically, P(a ≤ x ≤ b).Equivalently, P(a < x < b). To see this, it holds

P(a ≤ x ≤ b) = P(x = a)︸︷︷︸=0

+P(a < x < b) + P(x = b)︸︷︷︸=0

= P(a < x < b).

Terminology

Distribution Shapes

Symmetrical and Rectangular

The uniform distribution: Section 5.2.

Symmetrical and Bell-ShapedThe normal distribution: Section 5.3.

SkewedEither left or right: Section 5.5 for the right-skewed exponentialdistribution.

Terminology

Distribution Shapes

Terminology

Distribution Shapes

Outline

1 5.1 Continuous Probability Distributions

2 5.3: The Normal DistributionPropertiesNormal ProbabilitiesThe Standard Normal Distribution, N(0,1)The Standard Normal TableFinding Normal ProbabilitiesInverse Normal CalculationsThe Tolerance IntervalLinear Interpolation (Optional)

Properties

The Normal Probability Distribution, N(µ, σ)

The normal probability distribution, with parameters µ and σ,is defined by the mathematical function

f (x) =1√

2πσ2e−

x−µσ )

, for all x ∈ (−∞,∞),

where µ ∈ (−∞,∞), and σ > 0. Here, µ and σ are therespective mean and standard deviation of the normaldistribution, π ≈ 3.14159, and e ≈ 2.71828.Suppose the random variable x is normally distributed with(specified) values of µ and σ. Then, we write, x ∼ N(µ, σ).

Properties

The Normal Probability Distribution, N(µ, σ)

The normal probability distribution, with parameters µ and σ,is defined by the mathematical function

f (x) =1√

2πσ2e−

x−µσ )

, for all x ∈ (−∞,∞),

where µ ∈ (−∞,∞), and σ > 0. Here, µ and σ are therespective mean and standard deviation of the normaldistribution, π ≈ 3.14159, and e ≈ 2.71828.Suppose the random variable x is normally distributed with(specified) values of µ and σ. Then, we write, x ∼ N(µ, σ).

Properties

Properties of the Normal Distribution

The normal curve issymmetrical and bell-shaped.The normal curve issymmetrical about its mean.The normal curve assumes itsmaximum value when x = µ.The area under either half(left/right of µ) of the normalcurve is 1/2.The area under the normalcurve is one.

Properties

Properties of the Normal Distribution (cont.)

There are an infinite number of possible normal distributions.The (numerical) specification of the mean (µ) and standarddeviation (σ) parameters, uniquely identifies a normaldistribution and its shape.The highest point of the normal curve is located at its mean.mean = median = mode. That is, all of the measures of centraltendency (review Section 2.2) equal each other.

Properties

The curve is symmetrical about its mean.The left and right halves of the curve are mirror images of eachother.The tails of any normal curve extend to infinity in both directions.

The tails get closer to the horizontal axis as the curve extends offto ±∞, but the curve never touches the axis (Recall, in Section5.1, we require f (x) ≥ 0 for all x).

The area under the normal curve to the right of the mean equalsthe area under the normal curve to the left of the mean.

The area under each half of the curve is 1/2.

Properties

The Position and Shape of the Normal Curve

Two Normal Curves with Distinct Means and Standard Deviations

The mean, µ, positions the peak of the normal curve over thereal axis.The variance, σ2, measures the width (or spread) of the normalcurve.

Properties

The Position and Shape of the Normal Curve

Two Normal Curves with Distinct Means and Standard Deviations

The mean, µ, positions the peak of the normal curve over thereal axis.The variance, σ2, measures the width (or spread) of the normalcurve.

Normal Probabilities

Suppose x is a normally distributed random variable with mean,µ, and standard deviation, σ. That is, x ∼ N(µ, σ).The probability that x could take any value in the range betweentwo given values, say a and b, where a < b, is P(a ≤ x ≤ b).

P(a ≤ x ≤ b) is the area coloredin blue under the normal curve, forx lying within the interval, [a,b].

Three Important Probabilities for the NormalDistribution (Empirical Result)

The 68%− 95%− 99.7% RuleApproximately 68.26% of all possible observed values of x liewithin (plus or minus) one standard deviation of µ. That is,

P(|x − µ| ≤ σ) = P(µ− σ ≤ x ≤ µ+ σ) = 0.6826.

Approximately 95.44% of all possible observed values of x liewithin (plus or minus) two standard deviations of µ. That is,

P(|x − µ| ≤ 2σ) = P(µ− 2σ ≤ x ≤ µ+ 2σ) = 0.9544.

Approximately 99.73% of all possible observed values of x liewithin (plus or minus) three standard deviations of µ. That is,

P(|x − µ| ≤ 3σ) = P(µ− 3σ ≤ x ≤ µ+ 3σ) = 0.9973.

P(|x − µ| ≤ σ) = P(µ− σ ≤ x ≤ µ+ σ) = 0.6826.

P(|x − µ| ≤ 2σ) = P(µ− 2σ ≤ x ≤ µ+ 2σ) = 0.9544.

P(|x − µ| ≤ 3σ) = P(µ− 3σ ≤ x ≤ µ+ 3σ) = 0.9973.

P(|x − µ| ≤ σ) = P(µ− σ ≤ x ≤ µ+ σ) = 0.6826.

P(|x − µ| ≤ 2σ) = P(µ− 2σ ≤ x ≤ µ+ 2σ) = 0.9544.

P(|x − µ| ≤ 3σ) = P(µ− 3σ ≤ x ≤ µ+ 3σ) = 0.9973.

Three Important Probabilities for the NormalDistribution (Visual Result)

The 68%− 95%− 99.7% Rule

The Standard Normal Distribution, N(0, 1)

Standardizing a Normal Random Variable

The N(0,1) Distribution

Suppose x is normally distributed with mean, µ, and standarddeviation, σ (i.e., x ∼ N(µ, σ)). The random variable z, given by

z =x − µσ

is normally distributed with mean zero and standard deviationone (i.e., z ∼ N(0,1)). The normal distribution with mean zeroand standard deviation one is called the standard normaldistribution.Because the mean (µ) and standard deviation (σ) each carry thesame units as x , it follows that z is “unit–less”.

Standardizing a Normal Random Variable

The N(0,1) Distribution

Suppose x is normally distributed with mean, µ, and standarddeviation, σ (i.e., x ∼ N(µ, σ)). The random variable z, given by

z =x − µσ

is normally distributed with mean zero and standard deviationone (i.e., z ∼ N(0,1)). The normal distribution with mean zeroand standard deviation one is called the standard normaldistribution.Because the mean (µ) and standard deviation (σ) each carry thesame units as x , it follows that z is “unit–less”.

Relationship Between the Random Variables x and z

z measures the number of standard deviations that x is from itsmean, µ.

The algebraic sign of z indicates the position (left or right) of x inrelation to µ.z is positive if x > µ (x lies to the right of µ on the x − axis).Mathematically,

x > µ ⇐⇒ z =x − µσ

z is negative if x < µ (x lies to the left of µ on the x − axis).Mathematically,

x < µ ⇐⇒ z =x − µσ

x > µ ⇐⇒ z =x − µσ

x < µ ⇐⇒ z =x − µσ

x > µ ⇐⇒ z =x − µσ

x < µ ⇐⇒ z =x − µσ

Relationship Between the Random Variables x and z(cont.)

The Standard Normal Table

Because of the relationship between the random variables x and z(i.e., z = (x − µ)/σ), there is only the need to reference one normaltable. Namely, the standard normal table.

The standard normal table (as depicted by the textbook authors)is a table that lists the area under the standard normal curve, tothe right of the mean (z = 0) up to the z value of interest.See Table 5.1 on page 213.See Table A.3 within Appendix A, page 616.Not all standard normal tables display the same areas. Alwayslook at the accompanying figure (or, perhaps the table caption)for guidance on how to use the table.

The Standard Normal Table (cont.)

The values of z (accurate to the nearest hundredth decimalposition) in the table range from 0.00 to 3.09 in increments of0.01.

Accuracy of z to the tenth decimal position are listed in the far leftcolumn.Accuracy of z to the hundredth decimal position are listed acrossthe top of the table.

The areas under the normal curve to the right of the mean, up toany value of z, are given in the body of the table.

Calculating Probabilities Using the Standard NormalTable

Determine P(0 ≤ z ≤ 0.5)

Find the area listed in the table corresponding to a z value of0.50.Starting from the top of the far left column, go down the columnuntil you reach the value “0.5”.Read across the row, z = 0.5, until you reach the column headedby “.00”.The cell which intersects the row, z = 0.5, and column, .00,corresponds to the desired probability lookup.As listed in the table, we find

P(0 ≤ z ≤ 0.5) = 0.1915.

Calculating Probabilities Using the Standard NormalTable (cont.)

A Subset of Table 5.1, pg. 213z 0.00 0.01 0.02 0.03 0.04 0.05

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

Sample Probability Calculations

P(0 ≤ z ≤ 0.5)

= 0.1915.

P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5)

= 0.1915.

P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5)

= 0.1915.

P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5)

= 0.1915.

P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33)

= 0.1293.

P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33) = 0.1293.P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33) = 0.1293.P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33) = 0.1293.P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33) = 0.1293.P(0 ≤ z ≤ 0.15)

= 0.0596.

0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.01990.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.05960.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.09870.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.13680.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.17360.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088

P(0 ≤ z ≤ 0.5) = 0.1915.P(0 ≤ z ≤ 0.33) = 0.1293.P(0 ≤ z ≤ 0.15) = 0.0596.

Figure 5.9, pg. 214

The Symmetry of the Standard Normal Distribution

Calculate: P(−2.53 ≤ z ≤ 2.53)

First, write a probability statement in terms of Table 5.1. So, findP(0 ≤ z ≤ 2.53).It is, P(0 ≤ z ≤ 2.53) = 0.4943.By the symmetry (about its mean) of the normal distribution, itfollows that P(−2.53 ≤ z ≤ 0) = P(0 ≤ z ≤ 2.53).Thus,

P(−2.53 ≤ z ≤ 2.53) = P(−2.53 ≤ z ≤ 0) + P(0 < z ≤ 2.53)

= 0.4943 + 0.4943= 2(0.4943)

= 0.9886= 2P(0 ≤ z ≤ 2.53).

Calculate: P(−2.53 ≤ z ≤ 2.53)

P(−2.53 ≤ z ≤ 2.53) = P(−2.53 ≤ z ≤ 0) + P(0 < z ≤ 2.53)

= 0.4943 + 0.4943= 2(0.4943)

= 0.9886= 2P(0 ≤ z ≤ 2.53).

Calculate: P(−2.53 ≤ z ≤ 2.53)

P(−2.53 ≤ z ≤ 2.53) = P(−2.53 ≤ z ≤ 0) + P(0 < z ≤ 2.53)

= 0.4943 + 0.4943= 2(0.4943)

= 0.9886= 2P(0 ≤ z ≤ 2.53).

Calculate: P(−2.53 ≤ z ≤ 2.53)

P(−2.53 ≤ z ≤ 2.53) = P(−2.53 ≤ z ≤ 0) + P(0 < z ≤ 2.53)

= 0.4943 + 0.4943= 2(0.4943)

= 0.9886= 2P(0 ≤ z ≤ 2.53).

Calculate: P(z ≥ −1)

It holds,

P(z ≥ −1) = P(−1 ≤ z ≤ 0) + P(z ≥ 0)

= P(0 ≤ z ≤ 1) + 0.5= 0.3413 + 0.5= 0.8413.

Calculate: P(z ≥ 1)

Method OneIt holds,

P(z ≥ 1) = 1− P(z < 1)

= 1− [P(z < 0) + P(0 ≤ z < 1)]

= 1− (0.5 + 0.3413)

= 0.5− 0.3413= 0.1587.

Method TwoIt holds,

P(z ≥ 1) = P(z ≥ 0)− P(0 ≤ z < 1)

= 0.5− 0.3413= 0.1587.

Finding Normal Probabilities

General ProcedureFormulate the problem in terms of x values.Calculate the corresponding z values, and restate the problem interms of these z values.Find the required areas under the standard normal curve byusing the table.Note: It is always useful to draw a picture showing the requiredareas before using the normal table.

Example 5.2, pg. 217

ProblemWe want the probability that the mileage of a randomly selectedmidsize car will be between 32 and 35 mpg.

Solution SetupLet x be the random variable of mileage of midsize cars, in mpg.Given: x ∼ N(33 mpg,0.7 mpg).We need to find, P(32 mpg ≤ x ≤ 35mpg).

Example 5.2 (cont.)

For x = 32The corresponding z value is found by

z =x − µσ

=32− 33

0.7= −1.43.

z =x − µσ

=35− 33

0.7= 2.86.

Solution

P(32 mpg ≤ x ≤ 35 mpg) = P(−1.43 ≤ z ≤ 2.86)

= P(−1.43 ≤ z ≤ 0) + P(0 < z ≤ 2.86)

= P(0 ≤ z ≤ 1.43) + 0.4979= 0.4236 + 0.4979= 0.9215.

Example 5.2 (cont.)

z =x − µσ

=32− 33

0.7= −1.43.

z =x − µσ

=35− 33

0.7= 2.86.

Solution

P(32 mpg ≤ x ≤ 35 mpg) = P(−1.43 ≤ z ≤ 2.86)

= P(−1.43 ≤ z ≤ 0) + P(0 < z ≤ 2.86)

= P(0 ≤ z ≤ 1.43) + 0.4979= 0.4236 + 0.4979= 0.9215.

Example 5.2 (cont.)

z =x − µσ

=32− 33

0.7= −1.43.

z =x − µσ

=35− 33

0.7= 2.86.

Solution

P(32 mpg ≤ x ≤ 35 mpg) = P(−1.43 ≤ z ≤ 2.86)

= P(−1.43 ≤ z ≤ 0) + P(0 < z ≤ 2.86)

= P(0 ≤ z ≤ 1.43) + 0.4979= 0.4236 + 0.4979= 0.9215.

Inverse Normal Calculations

Finding z Values on a Standard Normal Curve

Notation, pg. 221For the standard normal distribution, let zα denote the point onthe horizontal axis which yields a right-hand tail area equal to α.The figure shown below, demonstrates how to find zα forα = 0.025.

The Point z0.025 = 1.96

Finding z Values on a Standard Normal Curve

Notation, pg. 221For the standard normal distribution, let zα denote the point onthe horizontal axis which yields a right-hand tail area equal to α.The figure shown below, demonstrates how to find zα forα = 0.025.

The Point z0.025 = 1.96

ProblemWe want to know how many︸︷︷︸

Quantity = ?

blank VHS tapes to stock (at the

beginning of the week) so that there is only a 5 percent chance︸︷︷︸

Given probability

stocking out during the week.

Solution SetupLet x be the random variable representing weekly demand.Let st represent the number of tapes, so that there is only a 5%probability of weekly demand exceeding this number of tapes.We need to determine the value of st , such thatP(x ≥ st) = 0.05.Given: x ∼ N(100 tapes,10 tapes).

Quantity = ?

beginning of the week) so that there is only a 5 percent chance︸︷︷︸Given probability

Quantity = ?

Example 5.5, pg. 221 (cont.)

Figure 5.20, pg. 222

SolutionIn panel (a) of Fig. 5.20, st is located on the horizontal axisunder the right-tail of the normal curve, N(100,10).The z value corresponding to st is given by

z =st − µσ

=st − 100

10. (1)

In panel (b) of Fig. 5.20, the right-tail area is 0.05, so the areaunder the standard normal curve to the right of the mean, z = 0,is 0.5− 0.05 = 0.45.

z =st − µσ

=st − 100

10. (1)

z =st − µσ

=st − 100

10. (1)

Solution (cont.)Use the standard normal table to find the value of z (outer portion ofthe table), corresponding with a table entry of 0.45 (inner portion ofthe table).

But do not find 0.45 (this value is not given in the table); insteadfind values which bracket (flank) that of 0.45.For a table entry (area) of 0.4495, z = 1.64.For a table entry (area) of 0.4505, z = 1.65.For an area of 0.45, use the z value midway between the valuesof 1.64 and 1.65 (see the linear interpolation slides at the end ofthis module).So, z0.05 = 1.645 (approximately).

Solution (cont.)Since z0.05 = 1.645, by (1), it follows that

1.645 =st − 100

Solving for st , we have

st = 100 + (1.645)(10) = 116.45.

Rounding up, 117 tapes should be stocked so that the probabilityof running out will not be more than 5 percent.

1.645 =st − 100

st = 100 + (1.645)(10) = 116.45.

1.645 =st − 100

st = 100 + (1.645)(10) = 116.45.

The Tolerance Interval

Finding a Tolerance Interval

Given a normal distribution, N(µ, σ), and a specified value of α,α ∈ (0,1), consider computing a tolerance interval (denoted by[µ± kσ] ) which contains 100(1− α)% of the measurements for thisnormal distribution.

Steps for Finding k

Let [x1, x2] denote the tolerance interval.It follows that

z2 =x2 − µσ

=µ+ kσ − µ

σ= k .

But, (α/2)% of the measurements must fall to the right of x2 (andso also z2).Hence, z2 = k = zα/2.

Steps for Finding k

z2 =x2 − µσ

=µ+ kσ − µ

σ= k .

Steps for Finding k

z2 =x2 − µσ

=µ+ kσ − µ

σ= k .

Steps for Finding k

z2 =x2 − µσ

=µ+ kσ − µ

σ= k .

Steps for Finding k

z2 =x2 − µσ

=µ+ kσ − µ

σ= k .

Tolerance Interval Example

90% Tolerance IntervalFrom the previous slide, k = zα/2.Here, 1− α = 0.90, so that α = 0.10.We previously found (using linear interpolation) thatz0.05 ≈ 1.645.Therefore, [µ± 1.645σ] is a 90% tolerance interval for a normaldistribution with mean, µ, and standard deviation, σ.

Linear Interpolation (Optional)

Point Slope Formula, y = m(x − x0) + y0

Consider the two points, (x0, y0) and (x1, y1), which lie on theline.The slope, m, satisfies the equation

m =y1 − y0

x1 − x0

It can be shown that any third point on the line, say (x2, y2),satisfies

x2 = x0 + (y2 − y0)(m)−1, for m 6= 0. (2)

m =y1 − y0

x1 − x0

x2 = x0 + (y2 − y0)(m)−1, for m 6= 0. (2)

m =y1 − y0

x1 − x0

x2 = x0 + (y2 − y0)(m)−1, for m 6= 0. (2)

ApplicationFrom Table 5.1 (pg. 213), we find

P(0 ≤ z ≤ 1.64) = 0.4495 and P(0 ≤ z ≤ 1.65) = 0.4505.(3)

Let (x0, y0) = (1.64,0.4495) and (x1, y1) = (1.65,0.4505),corresponding to the probabilities given by (3). Consider theordered pair, (x2, y2 = 0.45), where x2 satisfies

P(0 ≤ z ≤ x2) = 0.45.

By (2), it is