Lecture Statistical techniques in business and economics - Chapter 7: The normal probability distribution

When you have completed this chapter, you will be able to: Explain how probabilities are assigned to a continuous random variable, explain the characteristics of a normal probability distribution, define and calculate z value corresponding to any observation on a normal distribution, determine the probability a random observation is in a given interval on a normal distribution using the standard normal distribution, use the normal probability distribution to approximate the binomial probability distribution.

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THE

Normal

PROBABILITY DISTRIBUTION

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4.

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is

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Discrete or

Continuous Characteristics Characteristics

Variables

Also Recall that… Also Recall that…

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Probabilities are assigned to intervals of values !

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Range of Values Range of Values

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Probability Function

See Histo grams

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Probability Function

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A probability function can have any shape,

as long

as it is nonnegative (the curve is above the xaxis, and the total area under the curve is 1)

…that all probability functions are not normal !

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The arithmetic mean, median, and mode

of the distribution are equal and located at the peak.

The arithmetic mean, median , and mode

of the distribution are equal and located at the peak.

…thus half the area under the curve is above

the mean and half is below it

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Normal Probability Distribution

The normal probability distribution

is symmetrical about its mean.

The normal probability distribution

is symmetrical about its mean.

The normal probability

distribution is asymptotic,

i.e. the curve gets

closer and closer to

the xaxis

but never actually touches it.

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is symmetrical

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A zvalue is the distance between a selected value (designated X)

and the population mean, divided by the population Standard Deviation,

and the population mean,

divided by the population Standard Deviation,

µ ) ( X

z

Formula Formula

A A zvalue zvalue is, therefore, a location on the standard normal curve is, therefore, a location on the standard normal curve.

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Area = +0.5 Area = + 0.5

half is below the mean

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the mean of $3,000.

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 … Z becomes the metre stick for measuring

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7 23

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and the population mean µ,

divided by the population Standard Deviation,

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Transferring each value into z scores,

0 1.25

Z Z11= (6050)/8 = 1.25 = (60 50 )/8 = 1.25

µ ) ( X

z

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0.05

0.3944

…thus

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Transferring each value into z scores,

0 20 40 50 60 80 100

µ ) ( X

z

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… because both sides are symmetrical, the left side (the area between the mean and a negative z score 1.25) must

have the same area………… 0.3944

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Practise using the Normal Distribution Table Practise using the Normal Distribution Table

Look up 1.00 in Table Look up 1.00 in Table

Locate Area on the normal curve

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Normal Distribution Table

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279

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1.64

The Area between 1.64<Z<0 is 0.4495The Area between 1.64<Z<0 is 0.4495

Therefore, the Probability of 1.64<Z<0 is 44.95%Therefore, the Probability of 1.64<Z<0 is 44.95%

Because of symmetry, this area

is the same as between z of 0 and positive 1.64

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2.22

The Area between –2.22<Z<0 is 0.4868The Area between –2.22<Z<0 is 0.4868

Therefore, the Probability of –2.22<Z<0 is 48.68%Therefore, the Probability of –2.22<Z<0 is 48.68%

2.22

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2.96

The Area between –2.96<Z<0 is 0.4985The Area between –2.96<Z<0 is 0.4985

Therefore, the Probability of –2.96<Z<0 is 49.85%Therefore, the Probability of –2.96<Z<0 is 49.85%

2.96

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1.65

Look up 1.65 in Table Look up 1.65 in Table

1 2

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1

Look up – 2.00 then 1.00 in Table

2.00 0 1.00

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0.44 0 1.96

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Area = 0.5 Area = 0.5

half is below the mean

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0

A rea = 0.5

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A rea = 0.5

a1

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A rea = 0.5

a1

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Area = 0.5

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Area = 0.5

a1

A

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1

…the area between:

2 Look up 1.96 then 2.58 in Table Look up 1.96 then 2.58 in Table

The Area (aThe Area (a11) between 0<Z<1.96 is 0.4750) between 0<Z<1.96 is 0.4750

Therefore, the required Total Area is 0.0201Therefore, the required Total Area is 0.0201The Area (aThe Area (a22) between 0<Z<2.58 is 0.4951) between 0<Z<2.58 is 0.4951 Subtract

3 Adjust as needed

1.96 2.58

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a1

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68.26%

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The daily water usage per person in Newmarket, Ontario is normally distributed with a mean of 80 litres and

a standard deviation of 10 litres.

The daily water usage per person in Newmarket, Ontario is normally distributed

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robability that…

a person from Newmarket, selected at random, will use between 80 and 88 litres per day, is…?

10 80

80

0.00(X

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robability that…

The area under a normal curve between a zvalue of 80 and

a zvalue of 88 is 0.2881, therefore, we conclude that 28.81% of the residents will use between 80 to 88

litres per day!

a zvalue of 88 is 0.2881, therefore, we conclude that 28.81% of the residents will use between 80 to 88

litres per day!

3 Look up 0.80 in Table Look up 0.80 in Table A =.2881

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robability that…

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279

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robability that…

a zvalue of 92 is 0.5403, therefore, we conclude that

54.03% of the residents will use between 76 to 92 litres per day!

a zvalue of 92 is 0.5403, therefore, we conclude that

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Sketch given information onto the normal curve

1

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Now, use the table “ backwards ”

to find the zscore corresponding to a1=.35

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279

Determine the Z score from the given

areas

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Normal Distribution Table

Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279

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Using the Normal Curve

to Approximate the Binomial Distribution

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Use when np and n(1 p ) are both greater than 5! Use when np and n (1 p ) are both greater than 5!

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The probability of success

stays the same for each trial

The trials are independent Interested in the number of successes

The experiment consists of n Bernoulli trials

The outcomes are classified into

one of two mutually exclusive categories, …such

as success or failure

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Mean and Variance

of a Binomial Probability Distribution

Sketch given information onto the normal curve

1

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49.5

Locate 50 on the normal curve

1

48

If we want at least 50, we start at 49.5 on the normal curve!If we want at least 50, we start at 49.5 on the normal curve!

5 0

represented by the narrow strip,

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…find the probability that at most 50 students

in a class of 80 did their homework last night

Locate 50 on the normal curve

49.5

48

5 0

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The value 5 is subtracted or added ,

depending on the problem,

to a selected value

when a binomial probability distribution

(a discrete probability distribution) is being approximated by a

continuous probability distribution

Continuity Correction Factor

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showed that 15% of Canadian households owned a

video camera. For a sample of 200 homes , how many

25 5 . 0498

= n p (1 – p ) = (30)(1 15 ) = 25.5

= n p = ( 15 )( 200 ) = 30

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a1

What is the probability that less than 40

homes in the sample have video cameras?

What is the probability that less than 40 homes in

the sample have video cameras?

“less than” does not

include all of 40 – so use 39.5

“less than” does not include all of 40 – so use 39.5

2 Calculate the Z scores Calculate the Z scores

3 Look up 1.88 in Table Look up 1.88 in Table

z1

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…and much more!

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This completes Chapter 7

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