Ch7 crv00 by Mohamed Metwalli

Chapter 6,7,8 Continuous Probability Distributions

Continuous Probability Distributions • A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values)  thickness of an item  time required to complete a task  temperature of a solution  height • These can potentially take on any value, depending only on the ability to measure precisely and accurately.

Continuous Probability Distributions

The Uniform Distribution â&#x20AC;˘ The uniform distribution is a probability distribution that has equal probabilities for all possible outcomes of the random variable â&#x20AC;˘ Because of its shape it is also called a rectangular distribution

The Uniform Distribution The Continuous Uniform Distribution: 1 ba

if a  X  b

otherwise

f(X) =

where f(X) = value of the density function at any X value a = minimum value of X b = maximum value of X

a  b μ  2 2 (σ -1 b )2 a

The Uniform Distribution

• The mean of a uniform distribution is:

• The standard deviation is:

a  b 2  6 μ    4 2 (σb1 -2 a)2(61 -2)2 1.547

The Uniform Distribution

Example: Uniform probability distribution over the range 2 ≤ X ≤ 6: 1 f(X) = 6 - 2 = .25 for 2 ≤ X ≤ 6 f(X) .25 2

Example

The Exponential Distribution • Used to model the length of time between two occurrences of an event (the time between arrivals)  Examples: • Time between trucks arriving at an unloading dock • Time between transactions at an ATM Machine • Time between phone calls to the main operator

The Exponential Distribution

 Defined by a single parameter, its mean λ (lambda)  The probability that an arrival time is less than some

specified time X is

P(arrival time  X)  1  e  λX where

e = mathematical constant approximated by 2.71828 λ = the population mean number of arrivals per unit X = any value of the continuous variable where 0 < X < 

The Exponential Distribution Example: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes? 

The mean number of arrivals per hour is 15, so λ = 15



Three minutes is .05 hours



P(arrival time < .05) = 1 – e-λX = 1 – e-(15)(.05) = .5276



So there is a 52.76% probability that the arrival time between successive customers is less than three minutes

Introduction to Normal Distributions

Properties of a Normal Distribution Continuous random variable â&#x20AC;˘ Has an infinite number of possible values that can be represented by an interval on the number line. Hours spent studying in a day 0

The time spent studying can be any number between 0 and 24.

Continuous probability distribution â&#x20AC;˘ The probability distribution of a continuous random variable. 17

Properties of Normal Distributions Normal distribution â&#x20AC;˘ A continuous probability distribution for a random variable, x. â&#x20AC;˘ The most important continuous probability distribution in statistics. â&#x20AC;˘ The graph of a normal distribution is called the normal curve.

x 18

Properties of Normal Distributions 1. The mean, median, and mode are equal. 2. The normal curve is bell-shaped and symmetric about the mean. 3. The total area under the curve is equal to one. 4. The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean. Total area = 1

Îź

x 19

Properties of Normal Distributions 5. Between μ – σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ – σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points. Inflection points

μ  3σ

μ  2σ

μσ

μ+σ

μ + 2σ

μ + 3σ

x 20

Means and Standard Deviations • A normal distribution can have any mean and any positive standard deviation. • The mean gives the location of the line of symmetry. • The standard deviation describes the spread of the data.

μ = 3.5 σ = 1.5 Larson/Farber 4th ed

μ = 3.5 σ = 0.7

μ = 1.5 σ = 0.7 28

Example: Understanding Mean and Standard Deviation 1. Which curve has the greater mean?

Solution: Curve A has the greater mean (The line of symmetry of curve A occurs at x = 15. The line of symmetry of curve B occurs at x = 12.) Larson/Farber 4th ed

Example: Understanding Mean and Standard Deviation 2. Which curve has the greater standard deviation?

Solution: Curve B has the greater standard deviation (Curve B is more spread out than curve A.) Larson/Farber 4th ed

Example: Interpreting Graphs The heights of fully grown white oak trees are normally distributed. The curve represents the distribution. What is the mean height of a fully grown white oak tree? Estimate the standard deviation. Solution: Îź = 90 (A normal curve is symmetric about the mean)

Larson/Farber 4th ed

Ď&#x192; = 3.5 (The inflection points are one standard deviation away from the mean)

The Standard Normal Distribution Standard normal distribution • A normal distribution with a mean of 0 and a standard deviation of 1. Area = 1 3

2

1

• Any x-value can be transformed into a z-score by using the formula Value - Mean x-m z  Standard deviation s

Larson/Farber 4th ed

The Standard Normal Distribution • If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution. Normal Distribution

x-m z s

s m

Standard Normal Distribution

s m

• Use the Standard Normal Table to find the cumulative area under the standard normal curve. Larson/Farber 4th ed

Properties of the Standard Normal Distribution 1. The cumulative area is close to 0 for z-scores close to z = 3.49. 2. The cumulative area increases as the z-scores increase.

Area is close to 0 z = 3.49

z 3

Larson/Farber 4th ed

2

1

Properties of the Standard Normal Distribution 3. The cumulative area for z = 0 is 0.5000. 4. The cumulative area is close to 1 for z-scores close to z = 3.49.

Area is close to 1 z 3

2

Larson/Farber 4th ed

1

z=0 Area is 0.5000

z = 3.49

Example: Using The Standard Normal Table Find the cumulative area that corresponds to a z-score of 1.15.

Solution: Find 1.1 in the left hand column. Move across the row to the column under 0.05

The area to the left of z = 1.15 is 0.8749. Larson/Farber 4th ed

Example: Using The Standard Normal Table Find the cumulative area that corresponds to a z-score of -0.24.

Solution: Find -0.2 in the left hand column. Move across the row to the column under 0.04

The area to the left of z = -0.24 is 0.4052.

Larson/Farber 4th ed

Finding Areas Under the Standard Normal Curve 1. Sketch the standard normal curve and shade the appropriate area under the curve. 2. Find the area by following the directions for each case shown. a. To find the area to the left of z, find the area that corresponds to z in the Standard Normal Table. 2.

The area to the left of z = 1.23 is 0.8907

1. Use the table to find the area for the z-score

Larson/Farber 4th ed

Finding Areas Under the Standard Normal Curve b. To find the area to the right of z, use the Standard Normal Table to find the area that corresponds to z. Then subtract the area from 1. 2. The area to the left of z = 1.23 is 0.8907.

3. Subtract to find the area to the right of z = 1.23: 1 ď&#x20AC; 0.8907 = 0.1093.

1. Use the table to find the area for the z-score. Larson/Farber 4th ed

Finding Areas Under the Standard Normal Curve c. To find the area between two z-scores, find the area corresponding to each z-score in the Standard Normal Table. Then subtract the smaller area from the larger area. 2. The area to the left of z = 1.23 is 0.8907. 3. The area to the left of z = ď&#x20AC;0.75 is 0.2266. 1. Use the table to find the area for the z-scores.

Larson/Farber 4th ed

4. Subtract to find the area of the region between the two z-scores: 0.8907 ď&#x20AC; 0.2266 = 0.6641.

Example: Finding Area Under the Standard Normal Curve Find the area under the standard normal curve to the left of z = -0.99. Solution:

0.1611 ď&#x20AC;0.99

From the Standard Normal Table, the area is equal to 0.1611. Larson/Farber 4th ed

Example: Finding Area Under the Standard Normal Curve Find the area under the standard normal curve to the right of z = 1.06. Solution: 1 ď&#x20AC; 0.8554 = 0.1446

0.8554

1.06

From the Standard Normal Table, the area is equal to 0.1446. Larson/Farber 4th ed

Example: Finding Area Under the Standard Normal Curve Find the area under the standard normal curve between z = 1.5 and z = 1.25. Solution:

0.8944 0.0668 = 0.8276 0.8944

0.0668 1.50

1.25

From the Standard Normal Table, the area is equal to 0.8276. Larson/Farber 4th ed

Summary â&#x20AC;˘ Interpreted graphs of normal probability distributions â&#x20AC;˘ Found areas under the standard normal curve

Larson/Farber 4th ed

Normal Distributions: Finding Probabilities

Larson/Farber 4th ed

Objectives â&#x20AC;˘ Find probabilities for normally distributed variables

Larson/Farber 4th ed

Probability and Normal Distributions • If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval. μ = 500 σ = 100

P(x < 600) = Area

μ =500 600

Larson/Farber 4th ed

Probability and Normal Distributions Normal Distribution μ = 500 σ = 100 P(x < 600)

Standard Normal Distribution μ=0 σ=1

x  m 600  500 z  1 s 100

P(z < 1) z

μ =500 600

μ=0 1

Same Area P(x < 500) = P(z < 1) Larson/Farber 4th ed

Example: Finding Probabilities for Normal Distributions A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed.

Solution: Finding Probabilities for Normal Distributions Normal Distribution μ = 2.4 σ = 0.5

Standard Normal Distribution μ=0 σ=1

x  m 2  2.4 z   0.80 s 0.5

P(x < 2)

P(z < -0.80)

0.2119

2 2.4

-0.80 0

P(x < 2) = P(z < -0.80) = 0.2119 50

Example: Finding Probabilities for Normal Distributions A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes.

Larson/Farber 4th ed

Solution: Finding Probabilities for Normal Distributions Normal Distribution μ = 45 σ = 12

Standard Normal Distribution μ=0 σ=1

x - m 24 - 45   -1.75 s 12 x - m 54 - 45 z2    0.75 s 12 z1 

P(24 < x < 54)

P(-1.75 < z < 0.75)

0.7734 0.0401 x

45 54

-1.75

0 0.75

P(24 < x < 54) = P(-1.75 < z < 0.75) = 0.7734 – 0.0401 = 0.7333 Larson/Farber 4th ed

Example: Finding Probabilities for Normal Distributions Find the probability that the shopper will be in the store more than 39 minutes. (Recall Îź = 45 minutes and Ď&#x192; = 12 minutes)

Larson/Farber 4th ed

Solution: Finding Probabilities for Normal Distributions Normal Distribution μ = 45 σ = 12 z

P(x > 39)

Standard Normal Distribution μ=0 σ=1 x - m 39 - 45   -0.50 s 12

P(z > -0.50)

0.3085 x

39 45

-0.50 0

P(x > 39) = P(z > -0.50) = 1– 0.3085 = 0.6915 Larson/Farber 4th ed

Example: Finding Probabilities for Normal Distributions If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes? Solution: Recall P(x > 39) = 0.6915 200(0.6915) =138.3 (or about 138) shoppers

Larson/Farber 4th ed

Example: Using Technology to find Normal Probabilities Assume that cholesterol levels of men in the United States are normally distributed, with a mean of 215 milligrams per deciliter and a standard deviation of 25 milligrams per deciliter. You randomly select a man from the United States. What is the probability that his cholesterol level is less than 175? Use a technology tool to find the probability.

Larson/Farber 4th ed

Solution: Using Technology to find Normal Probabilities Must specify the mean, standard deviation, and the xvalue(s) that determine the interval.

Larson/Farber 4th ed

Summary â&#x20AC;˘ Found probabilities for normally distributed variables

Larson/Farber 4th ed

Normal Distributions: Finding Values

Larson/Farber 4th ed

Objectives • Find a z-score given the area under the normal curve • Transform a z-score to an x-value • Find a specific data value of a normal distribution given the probability

Larson/Farber 4th ed

Finding values Given a Probability â&#x20AC;˘ In section 5.2 we were given a normally distributed random variable x and we were asked to find a probability. â&#x20AC;˘ In this section, we will be given a probability and we will be asked to find the value of the random variable x. 5.2 x

probability

5.3 Larson/Farber 4th ed

Example: Finding a z-Score Given an Area Find the z-score that corresponds to a cumulative area of 0.3632.

Solution: 0.3632

z 0

Larson/Farber 4th ed

Solution: Finding a z-Score Given an Area â&#x20AC;˘ Locate 0.3632 in the body of the Standard Normal Table.

The z-score is -0.35. â&#x20AC;˘ The values at the beginning of the corresponding row and at the top of the column give the z-score. Larson/Farber 4th ed

Example: Finding a z-Score Given an Area Find the z-score that has 10.75% of the distributionâ&#x20AC;&#x2122;s area to its right.

Solution: 1 â&#x20AC;&#x201C; 0.1075 = 0.8925 0

0.1075

Because the area to the right is 0.1075, the cumulative area is 0.8925. Larson/Farber 4th ed

Solution: Finding a z-Score Given an Area â&#x20AC;˘ Locate 0.8925 in the body of the Standard Normal Table.

The z-score is 1.24. â&#x20AC;˘ The values at the beginning of the corresponding row and at the top of the column give the z-score. Larson/Farber 4th ed

Example: Finding a z-Score Given a Percentile Find the z-score that corresponds to P5. Solution: The z-score that corresponds to P5 is the same z-score that corresponds to an area of 0.05. 0.05 z

The areas closest to 0.05 in the table are 0.0495 (z = -1.65) and 0.0505 (z = -1.64). Because 0.05 is halfway between the two areas in the table, use the z-score that is halfway between -1.64 and -1.65. The z-score is -1.645. Larson/Farber 4th ed

Transforming a z-Score to an x-Score To transform a standard z-score to a data value x in a given population, use the formula x = Îź + zĎ&#x192;

Larson/Farber 4th ed

Example: Finding an x-Value The speeds of vehicles along a stretch of highway are normally distributed, with a mean of 67 miles per hour and a standard deviation of 4 miles per hour. Find the speeds x corresponding to z-sores of 1.96, -2.33, and 0. Solution: Use the formula x = μ + zσ •z = 1.96:

x = 67 + 1.96(4) = 74.84 miles per hour

•z = -2.33: x = 67 + (-2.33)(4) = 57.68 miles per hour •z = 0:

x = 67 + 0(4) = 67 miles per hour

Notice 74.84 mph is above the mean, 57.68 mph is below the mean, and 67 mph is equal to the mean. Larson/Farber 4th ed

Example: Finding a Specific Data Value Scores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment? Solution: 1 â&#x20AC;&#x201C; 0.05 = 0.95 0 Larson/Farber 4th ed

5% ? ?

z x

An exam score in the top 5% is any score above the 95th percentile. Find the z-score that corresponds to a cumulative area of 0.95. 69

Solution: Finding a Specific Data Value From the Standard Normal Table, the areas closest to 0.95 are 0.9495 (z = 1.64) and 0.9505 (z = 1.65). Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. That is, z = 1.645.

5% 0 75 Larson/Farber 4th ed

1.645 ?

z x 70

Solution: Finding a Specific Data Value Using the equation x = μ + zσ x = 75 + 1.645(6.5) ≈ 85.69 5% 0

1.645

75 85.69

z x

The lowest score you can earn and still be eligible for employment is 86. Larson/Farber 4th ed

Summary • Found a z-score given the area under the normal curve • Transformed a z-score to an x-value • Found a specific data value of a normal distribution given the probability

Larson/Farber 4th ed

Normal Approximations to Binomial Distributions

Larson/Farber 4th ed

Objectives â&#x20AC;˘ Determine when the normal distribution can approximate the binomial distribution â&#x20AC;˘ Find the correction for continuity â&#x20AC;˘ Use the normal distribution to approximate binomial probabilities

Larson/Farber 4th ed

Normal Approximation to a Binomial • The normal distribution is used to approximate the binomial distribution when it would be impractical to use the binomial distribution to find a probability. Normal Approximation to a Binomial Distribution • If np  5 and nq  5, then the binomial random variable x is approximately normally distributed with  mean μ = np  standard deviation σ  npq

Larson/Farber 4th ed

Normal Approximation to a Binomial â&#x20AC;˘ Binomial distribution: p = 0.25

â&#x20AC;˘ As n increases the histogram approaches a normal curve. Larson/Farber 4th ed

Example: Approximating the Binomial Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation.

1. Fifty-one percent of adults in the U.S. whose New Yearâ&#x20AC;&#x2122;s resolution was to exercise more achieved their resolution. You randomly select 65 adults in the U.S. whose resolution was to exercise more and ask each if he or she achieved that resolution.

Larson/Farber 4th ed

Solution: Approximating the Binomial • You can use the normal approximation n = 65, p = 0.51, q = 0.49 np = (65)(0.51) = 33.15 ≥ 5 nq = (65)(0.49) = 31.85 ≥ 5 • Mean: μ = np = 33.15 • Standard Deviation:

σ  npq  65 × 0.51× 0.49 » 4.03

Larson/Farber 4th ed

Example: Approximating the Binomial Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can find, find the mean and standard deviation.

2. Fifteen percent of adults in the U.S. do not make New Yearâ&#x20AC;&#x2122;s resolutions. You randomly select 15 adults in the U.S. and ask each if he or she made a New Yearâ&#x20AC;&#x2122;s resolution.

Larson/Farber 4th ed

Solution: Approximating the Binomial • You cannot use the normal approximation n = 15, p = 0.15, q = 0.85 np = (15)(0.15) = 2.25 < 5 nq = (15)(0.85) = 12.75 ≥ 5 • Because np < 5, you cannot use the normal distribution to approximate the distribution of x.

Larson/Farber 4th ed

Correction for Continuity â&#x20AC;˘ The binomial distribution is discrete and can be represented by a probability histogram. â&#x20AC;˘ To calculate exact binomial probabilities, the binomial formula is used for each value of x and the results are added. â&#x20AC;˘ Geometrically this corresponds to adding the areas of bars in the probability histogram.

Larson/Farber 4th ed

Correction for Continuity • When you use a continuous normal distribution to approximate a binomial probability, you need to move 0.5 unit to the left and right of the midpoint to include all possible x-values in the interval (correction for continuity). Exact binomial probability P(x = c)

c Larson/Farber 4th ed

Normal approximation P(c – 0.5 < x < c + 0.5)

c– 0.5 c c+ 0.5 82

Example: Using a Correction for Continuity Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

1. The probability of getting between 270 and 310 successes, inclusive. Solution: • The discrete midpoint values are 270, 271, …, 310. • The corresponding interval for the continuous normal distribution is 269.5 < x < 310.5 Larson/Farber 4th ed

Example: Using a Correction for Continuity Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

2. The probability of getting at least 158 successes. Solution: • The discrete midpoint values are 158, 159, 160, …. • The corresponding interval for the continuous normal distribution is x > 157.5 Larson/Farber 4th ed

Example: Using a Correction for Continuity Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

3. The probability of getting less than 63 successes. Solution: • The discrete midpoint values are …,60, 61, 62. • The corresponding interval for the continuous normal distribution is x < 62.5 Larson/Farber 4th ed

Using the Normal Distribution to Approximate Binomial Probabilities In Words 1. Verify that the binomial distribution applies. 2. Determine if you can use the normal distribution to approximate x, the binomial variable. 3. Find the mean m and standard deviations for the distribution. Larson/Farber 4th ed

In Symbols Specify n, p, and q. Is np  5? Is nq  5?

m  np s 

npq

Using the Normal Distribution to Approximate Binomial Probabilities In Words 4. Apply the appropriate continuity correction. Shade the corresponding area under the normal curve. 5. Find the corresponding zscore(s). 6. Find the probability.

Larson/Farber 4th ed

In Symbols Add or subtract 0.5 from endpoints.

x-m zď&#x20AC;˝ s Use the Standard Normal Table. 87

Example: Approximating a Binomial Probability Fifty-one percent of adults in the U. S. whose New Year’s resolution was to exercise more achieved their resolution. You randomly select 65 adults in the U. S. whose resolution was to exercise more and ask each if he or she achieved that resolution. What is the probability that fewer than forty of them respond yes? (Source: Opinion Research Corporation)

Solution: • Can use the normal approximation (see slide 89) μ = 65∙0.51 = 33.15 σ  65 × 0.51 × 0.49 » 4.03 Larson/Farber 4th ed

Solution: Approximating a Binomial Probability • Apply the continuity correction: Fewer than 40 (…37, 38, 39) corresponds to the continuous normal distribution interval x < 39.5 Normal Distribution μ = 33.15 σ = 4.03 P(x < 39.5)

z

Standard Normal μ=0 σ=1

x - m 39.5 - 33.15  » 1.58 s 4.03

μ =33.15

39.5

P(z < 1.58) 0.9429 μ =0

1.58

P(z < 1.58) = 0.9429 Larson/Farber 4th ed

Example: Approximating a Binomial Probability A survey reports that 86% of Internet users use Windows® Internet Explorer ® as their browser. You randomly select 200 Internet users and ask each whether he or she uses Internet Explorer as his or her browser. What is the probability that exactly 176 will say yes? (Source: 0neStat.com)

Solution: • Can use the normal approximation np = (200)(0.86) = 172 ≥ 5 nq = (200)(0.14) = 28 ≥ 5 μ = 200∙0.86 = 172 Larson/Farber 4th ed

σ  200 × 0.86 × 0.14 » 4.91 90

Solution: Approximating a Binomial Probability • Apply the continuity correction: Exactly 176 corresponds to the continuous normal distribution interval 175.5 < x < 176.5 Normal Distribution Standard Normal μ = 172 σ = 4.91 x - m 175.5 - 172 μ=0 σ=1 z   » 0.71 P(175.5 < x < 176.5)

s 4.91 x - m 176.5 - 172 z2   » 0.92 s 4.91 1

μ =172 176.5 175.5

P(0.71 < z < 0.92)

0.8212 0.7611

μ =0 0.92 0.71

P(0.71 < z < 0.92) = 0.8212 – 0.7611 = 0.0601 Larson/Farber 4th ed

‫الحنحدار الخطى البسيط‬ ‫‪Simple linear Regression‬‬ ‫• أس لوب إحص ائي أ و طريق ة رياضي ة تس تخدم لدراس ة وتحلي ل تأثي ر‬ ‫متغير كمي مستقل على متغير كمي آخر تابع ‪ ،‬والتنبؤ بقيمة المتغير‬ ‫التابططع بدللططة المتغيططر المسططتقل بعططد إيجاد معادلططة الحنحدار الخطططي‬ ‫البسيط ‪،‬‬ ‫• متى يسمى الحنحدار خطي بسيط‬ ‫• متى يسمى الحنحدار خطي متعدد)أكثر من متغي مستقل ومتغي تابع واحد (‪.‬‬ ‫~‬ ‫‪y  B0  B1 x1  B2 x2‬‬

‫• مثال‪ :‬معدل أداء العامل يعتمد على عدد سنوات خبرته و ذكائه و مؤهله العلمي وعمره‬

‫• المتغي“ر المؤثــ“ر“ ‪ ” x‬يسمي )المستقل(‬ ‫• المتغي“ر الذي يتأث“ر ” ‪ ”y‬يسمي )التابع(‪,‬‬ ‫• أمثلة‪:‬‬

‫‪ ‬أثر الدخل ” ‪ ”x‬على الحنفاق ” ‪”y‬‬ ‫‪ ‬دراسة أثر سعر سلعة ما ” ‪” x‬علي كمية المبيعات من هذه السلعة ” ‪y‬‬

‫الخلةصة‬ ‫الفرق بين الرتباط والحنحدار‬ ‫الرتباط والحنحدار وجهان لعملة واحدة‬ ‫فالرتباط يقيس درجة أو قوة مدى العلقة بين المتغي“رات بقيمة رقمية‬ ‫محصورة بين ‪1+‬و‪1-‬‬ ‫الحنحدار يعـب“ر عن هذه العلقة بمعادلة رياضية خطية ) أو غي“ر خطية (‬ ‫تفيد في التنبؤ بقيم المتغي“ر التابع بافت“راض قيم معينة للمتغي“ر المستقلة‬