Statistik 1 6 distribusi probabilitas normal

l

GOALS

1. List the characteristics of the normal probability distribution.

2. Define and calculate z values.3. Determine the probability an observation will lie

between two points using the standard normal distribution.

4. Determine the probability an observation will be above or below a given value using the standard normal distribution.

5. Compare two or more observations that are on different probability distributions.

6. Use the normal distribution to approximate the binomial probability distribution.

Chapter SevenThe Normal Probability DistributionThe Normal Probability Distribution

Distribusi Probabilitas KontinyuDistribusi Probabilitas Kontinyu

Distribusi probabilitas kontinyu: didasarkan pada variabel acak kontinyu biasanya diperoleh dengan mengukur.

Dua bentuk distribusi probabilitas kontinyu:Distribusi probabilitas seragam (uniform)Distribusi probabilitas normal

Normal Distribution

Importance of Normal Distribution

1. Describes many random processes or continuous phenomena

2. Basis for Statistical Inference

Characteristics of a Normal Probability Distribution

The normal curve is bell-shaped and has a single peak at the exact center of the distribution.

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.

The normal probability distribution is unimodal=only one mode.

It is completely described by Mean & Standard Deviation

The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.

Normal Probability distribution

e=2.71828

Do not worry about the formula, you do not need to calculate using the above formula.

2

2

( )

21( )

2

x

P x e

Normal Probability distribution

The density functions of normal distributions with zero mean value and different standard deviations

Symmetric

Mean=median = mode

Unimodal

Bell-shaped

Asymptotic

Difference Between Normal Distributions

x

x

x

(a)

(b)

(c)

Normal Distribution Probability

Probability is the area under the curve!

c dX

f(X) A table may be constructed to help us find the probability

Infinite Number of Normal Distribution Tables

Normal distributions differ by mean & standard deviation.

Each distribution would require its own table.

X

f(X)

Distribusi Probabilitas Normal Standar

Distribusi Normal Standar adalah distribution normal dengan mean= 0 dan standar deviasi=1.

Juga disebut z distribution.

A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is:

X

z

The Standard Normal Probability Distribution

Any normal random variable can be transformed to a standard normal random variable

Suppose X ~ N(µ, 2)Z=(X-µ)/ ~ N(0,1)

P(X<k) = P [(X-µ)/ < (k-µ)/ ]

Standardize the Normal Distribution

Z

= 0

z = 1

Z

Because we can transform any normal random variable into standard normal random variable, we need only one table!

Normal Distribution

Standardized Normal Distribution

X

XZ

Transform to Standard Normal DistributionA numerical example

Any normal random variable can be transformed to a standard normal random variable

x x- (x-)/σ x/σ

0 -2 -1.4142 0

1 -1 -0.7071 0.7071

2 0 0 1.4142

3 1 0.7071 2.1213

4 2 1.4142 2.8284

Mean

2 0 0 1.4142

std 1.4142 1.4142 1 1

Standardizing Example

ZZ

= 0

Z = 1

.12

Normal Distribution


X = 5

= 10

6.2

12.010

52.6

XZ

Data of weight of children under 5 years old in Kg. Mean ()= 5 and What is the probability of children weighted at 5 Kgs to 6.2 Kgs?

Obtaining the Probability

ZZ

= 0

Z = 1

0.12

Z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

0.0478

.02

0.1 .0478

Standardized Normal Probability Table (Portion)

ProbabilitiesShaded Area Exaggerated

Example P(3.8 X 5)

Z Z = 0

Z = 1

-0.12

Normal Distribution

0.0478


Shaded Area Exaggerated

X = 5

= 10

3.8

12.010

58.3

XZ

Example (2.9 X 7.1)

0

Z

= 1

-.21 Z.21

Normal Distribution

.1664

.0832.0832



5

= 10

2.9 7.1 X

ZX

ZX

2 9 510

21

7 1 5

1021

..

..

Example (2.9 X 7.1)

0

Z

= 1

-.21 Z.21

Normal Distribution

.1664

.0832.0832



5

= 10

2.9 7.1 X

ZX

ZX

2 9 510

21

7 1 5

1021

..

..

Example P(X 8)

ZZ

= 0

Z

= 1

.30

Normal Distribution


.1179

.5000 .3821


ZX

8 5

1030.

X = 5

= 10

8

Example P(7.1 X 8)

z = 0

Z = 1

.30 Z.21

Normal Distribution

.0832

.1179 .0347



ZX

ZX

71 510

21

8 5

1030

..

.

= 5

= 10

87.1 X

Normal Distribution Thinking Challenge

You work in Quality Control for GE. Light bulb life has a normal distribution with µ= 2000 hours & = 200 hours. What’s the probability that a bulb will lastbetween 2000 & 2400 hours? less than 1470 hours?

Solution P(2000 X 2400)

ZZ

= 0

Z

= 1

2.0

Normal Distribution

.4772


ZX

2400 2000

2002 0.

X = 2000

= 200

2400

Solution P(X 1470)

Z Z= 0

Z = 1

-2.65

Normal Distribution

.4960 .0040

.5000


ZX

1470 2000

2002 65.

X = 2000

= 200

1470

Finding Z Values for Known Probabilities

Z .00 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

.1179 .1255

Z Z = 0

Z = 1

.31

.1217 .01

0.3 .1217

Standardized Normal Probability Table (Portion)

What Is Z Given P(Z) = 0.1217?


Finding X Values for Known Probabilities

Z Z = 0

Z = 1

.31X = 5

= 10

?

Normal Distribution Standardized Normal Distribution

.1217 .1217


1.810)31.0(5 ZX

EXAMPLE 1

The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200. What is the z-value for a salary of $2,200?

$2,200 $2,0001.00

$200

Xz

EXAMPLE 1 continued

A z-value of 1 indicates that the value of $2,200 is one standard deviation above the mean of $2,000.

A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.

50.1200$

200,2$700,1$

X

z

What is the z-value of $1,700 ?

Areas Under the Normal Curve

About 68 percent of the area under the normal curve is within one standard deviation of the mean. ± P( - < X < + ) = 0.6826

About 95 percent is within two standard deviations of the mean. ± 2 P( - 2 < X < + 2 ) = 0.9544

Practically all is within three standard deviations of the mean. ± 3 P( - 3 < X < + 3 ) = 0.9974

Areas Under the Normal Curve

Between:± 1 - 68.26%± 2 - 95.44%± 3 - 99.74%

µµ-1σµ+1σ

µ-2σ µ+2σµ+3σµ-3σ

EXAMPLE 2

The daily water usage per person in Surabaya is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in Surabaya will use how many gallons of water?

About 68% of the daily water usage will lie between 15 and 25 gallons.

EXAMPLE 3

What is the probability that a person from Surabaya selected at random will use between 20 and 24 gallons per day?

00.05

2020

X

z

80.05

2024

X

z

P(20<X<24)=P[(20-20)/5 < (X-20)/5 < (24-20)/5 ]=P[ 0<Z<0.8 ]

The Normal Approximation to the Binomial

The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.

The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.

The Normal Approximation continued

Recall for the binomial experiment: There are only two mutually exclusive

outcomes (success or failure) on each trial.

A binomial distribution results from counting the number of successes.

Each trial is independent. The probability is fixed from trial to

trial, and the number of trials n is also fixed.

The Normal Approximation

normal

binomial

Continuity Correction Factor

The value 0.5 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 (the normal distribution).

Four cases may arise: For the P at least X occur, use the area above (X – 0,5) For the P that more than X occur, use the area above (X

+ 0,5) For the P that X or fewer occur, use the area below (X

+ 0,5) For the P that fewer than X occur, use the area below (X

– 0,5)

Continuity Correction Factor

Because the normal distribution can take all real numbers (is continuous) but the binomial distribution can only take integer values (is discrete), a normal approximation to the binomial should identify the binomial event "8" with the normal interval "(7.5, 8.5)" (and similarly for other integer values). The figure below shows that for P(X > 7) we want the magenta region which starts at 7.5.

n=20 and p=0.25 P(X ≥ 8)=? First step: determine mean () & standard deviation

()

Second step: determine the z-value

Third step: check the area below normal curve

(20)(0.25) 5n

(1 ) (5)(1 0.25) 3.57 1.94n

1. Without correction factor of 0.5:

check in table-Z: P(X ≥ 8)=0.5-0.4394=0.0606

2. With correction factor of 0.5: P(X ≥ 8) P(X ≥ 8-0.5=7.5)

check in table-Z: P(X ≥ 7.5)=0.5-0.4015=0.0985

The exact solution from binomial distribution function is 0.1019.

The continuity correct factor is important for the accuracy of the normal approximation of binomial.

The approximation is quite good.

8 51.55

1.94

Xz

7.5 51.29

1.94

Xz

EXAMPLE 5

A recent study by a marketing research firm showed that 15% of American households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras?

30)200)(15(. n

What is the variance?

5.25)15.1)(30()1(2 n

0498.55.25

What is the standard deviation?

What is the mean?

What is the probability that less than 40 homes in the sample have video cameras?

We use the correction factor, so X is 39.5. The value of z is 1.88.

88.10498.5

0.305.39

X

z

EXAMPLE 5 continued

Example 5 continued

From Appendix D the area between 0 and 1.88 on the z scale is .4699.

So the area to the left of 1.88 is .5000 + .4699 = .9699.

The likelihood that less than 40 of the 200 homes have a video camera is about 97%.

EXAMPLE 5

0 1 2 3 4

P(z<1.88)=.5000+.4699=.9699

z=1.88

z

Soal

KPS MP melakukan polling dengan menyebarkan 60 kuisioner. Probabilitas kembalinya kuisioner adalah 80%. Hitung probabilitas :50 kuisioner kembaliAntara 45-55 kuisioner kembaliKurang dari 55 kuisioner kembali20 atau lebih kuisioner kembali

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Chapter SevenThe Normal Probability DistributionThe Normal Probability Distribution

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Statistik 1 6 distribusi probabilitas normal