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Report No. 2
Mr. Roderico Y. Dumaug, Jr.
TOPIC OUTLINE
 The Normal Distribution
   1) Introduction
   2) Definition of Terms and Statistical Symbols Used
   3) How To Find Areas Under the Normal Curve
   4) Finding the Unknown Z represented by Zo
   5) Examples
 Hypothesis Testing
The Normal Distribution
 Introduction


     Before exploring the complicated Standard
 Normal Distribution, we must examine how the
 concept of Probability Distribution changes when the
 Random Variable is Continuous.
The Normal Distribution
 Introduction

       A Probability Distribution will give us a Value of P(x) = P(X=x)
  to each possible outcome of x. For the values to make a Probability
  Distribution, we needed two things to happen:

       1. P (x) = P (X = x)
       2. 0 ≤ P (x) ≤ 1

      For a Continuous Random Variable, a Probability Distribution
  must be what is called a Density Curve. This means:

       1. The Area under the Curve is 1.
       2. 0 ≤ P (x) for all outcomes x.
The Normal Distribution
 Introduction: Example:
          Suppose the temperature of a piece of metal is always between 0°F
   and 10°F. Furthermore, suppose that it is equally likely to be any
   temperature in that range. Then the graph of the probability distribution
   for the value of the temperature would look like the one below:
                                0.10
                                                                   Uniform Distribution
  P (X < 5)                            ValuesAREA
                                              are spread
                                       uniformly across
                    Probability 0.05             0.8
                                       the range 0 to 10
                                       0.5     4
P (3 < X ≤ 7)                             Probability:
                                           Area: 80%              P (X > 2)
                                           (4)(0.1) =
                                0.00   Finding the Area Under the Curve
                                              0.4
7 - 3     =     4
                                       2 3 5
                                           5             7
                                                             10
                                                    X
Illustration of the fundamental fact about DENSITY CURVE
The Normal Distribution: Definition
   of Terms and Symbols Used
 Normal Distribution Definition:

   1)   A continuous variable X having the symmetrical, bell shaped distribution
        is called a Normal Random Variable.
   2)   The normal probability distribution (Gaussian distribution) is a
        continuous distribution which is regarded by many as the most significant
        probability distribution in statistics particularly in the field of statistical
        inference.
 Symbols Used:
    “z” – z-scores or the standard scores. The table that transforms every normal
     distribution to a distribution with mean 0 and standard deviation 1. This
     distribution is called the standard normal distribution or simply
     standard distribution and the individual values are called standard
     scores or the z-scores.
    “µ” – the Greek letter “mu,” which is the Mean, and
    “σ” – the Greek letter “sigma,” which is the Standard Deviation
The Normal Distribution: Definition
 of Terms and Symbols Used
 Characteristics of Normal Distribution:
   1) It is “Bell-Shaped” and has a single peak at the center of the
      distribution,
   2) The arithmetic Mean, Median and Mode are equal.
   3) The total area under the curve is 1.00; half the area under the
      normal curve is to the right of this center point and the other
      half to the left of it,
   4) It is Symmetrical about the mean,
   5) It is Asymptotic: The curve gets closer and closer to the X –
      axis but never actually touches it. To put it another way, the
      tails of the curve extend indefinitely in both directions.
   6) The location of a normal distribution is determined by the
      Mean, µ, the Dispersion or spread of the distribution is
      determined by the Standard Deviation, σ.
The Normal Distribution: Graphically
                   Normal Curve is Symmetrical
                      Two halves identical




     Theoretically, curve                  Theoretically, curve
     extends to - ∞                        extends to + ∞

                            Mean, Median
                            and Mode are
                               equal.
AREA UNDER THE NORMAL CURVE
                     P(x1<X<x2)
      P(X<x1)             P(X>x2)




                x1   x2
AREA UNDER THE NORMAL CURVE
 Let us consider a variable X which is normally distributed with a mean of 100 and
 a standard deviation of 10. We assume that among the values of this variable are
 x1= 110 and x2 = 85.
                        110        100                 85       100
                   z1                    1 . 00   z2                   1 . 50
                              10                            10

                                                                 0.7745
                  0.4332
                                                                      0.3413




                                    -1.5                    1
The Standard Normal Probability
Distribution
  The Standard Normal Distribution is a Normal
   Distribution with a Mean of 0 and a Standard Deviation
   of 1.
  It is also called the z distribution
  A z –value is the distance between a selected value
   , designated X, and the population Mean µ, divided by
   the Population Standard Deviation, σ.
  The formula is :
Areas Under the Normal Curve
       z          0          1        2        3        4        5        6        7        8        9

             0          0     0.004    0.008    0.012    0.016   0.0199   0.0239   0.0279   0.0319   0.0359

            0.1   0.0398     0.0438   0.0478   0.0517   0.0557   0.0596   0.0636   0.0675   0.0714   0.0754

            1.4   0.4192     0.4207   0.4222   0.4236   0.4251   0.4265   0.4279   0.4292   0.4306   0.4319


           1.5 0.4332        0.4345   0.4357    0.437   0.4382   0.4394   0.4406   0.4418   0.4429   0.4441

            1.6   0.4452     0.4463   0.4474   0.4484   0.4495   0.4505   0.4515   0.4525   0.4535   0.4545




              0.4332


      µ = 283          µ = 285.4          Grams
        0              1.50               z Values
How to Find Areas Under the
Normal Curve
Let       Z be a standardized random variable
          P stands for Probability
          Φ(z) indicates the area covered
                    under the Normal Curve.
      a.) P (0 ≤ Z≤≤ZZ≤≤2.45) Φ(-0.35) – Φ (0.91)
      b.) (-0.35 1.53) = Φ Φ (2.45)
      c.) (0.91          0) = (1.53)
                               = 0.4929 - 0.3186
                           = 0.1368
                              0.4370
                               = 0.1743



                                                      This shaded area, from
                                                    This shaded area, from
                                                       This shaded area, from
                                                    Z Z = -0.35Z = 1.53, = 2.45,
                                                      = 0 and until Z = 0,
                                                       Z – 0.91 until Z
                                                      represents the probability
                                                    represents the probability
                                                       represents the probability
                                                    value ofof0.1368
                                                      value of0.4370.
                                                       value 0.1743
How to Find Areas Under the
Normal Curve
Let      Z be a standardized random variable
         P stands for Probability
         Φ(z) indicates the area covered
                   under the Normal Curve.

d.) P (-2.0 ≤ Z ≤ 0.95) = Φ (-2.0) + Φ (0.95)
                        =0.4772 + 0.3289
                        = 0.8061



                                                This shaded area, from
                                                Z = -2.0 until Z = 0.95
                                                represents the probability
                                                value of 0.8061.
How to Find Areas Under the
Normal Curve
Let       Z be a standardized random variable
          P stands for Probability
          Φ(z) indicates the area covered
                    under the Normal Curve.

 e.) P (-1.5 ≤ Z ≤ -0.5) = Φ (-1.5) – Φ (-0.5)
                         =0.4332 - 0.1915
                         = 0.2417


                                                 This shaded area, from
                                                 Z = -1.5 until Z = -0.5
                                                 represents the probability
                                                 value of 0.2417.
How to Find Areas Under the
Normal Curve
Let     Z be a standardized random variable
        P stands for Probability
        Φ(z) indicates the area covered
                  under the Normal Curve.

 f.) P(Z ≥ 2.0) = 0.5 – Φ (2.0)               This shaded area, from
                = 0.5 – 0.4772                Z = 2.0 until beyond Z = 3
                                              represents the probability
                = 0.0228
                                              value of 0.0228.
How to Find Areas Under the
Normal Curve
Let       Z be a standardized random variable
          P stands for Probability
          Φ(z) indicates the area covered
                    under the Normal Curve.

 g.) P (Z ≤ 1.5) = 0.5 + Φ (1.5)
                 = 0.5 + 0.4332
                = 0.9332



                                                This shaded area, from
                                                Z = 1.5 until beyond Z = -3
                                                represents the probability
                                                value of 0.9332
Finding the unknown Z represented
by Zo
Finding the unknown Z represented
by Zo

   P(Z ≤ Z0) = 0.8461
      0.5 + X = 0.8461
      X = 0.8461 – 0.5
       X = 0.3461
       Z0 = 1.02 ans.
  • P(-1.72 ≤ Z ≤ Z0) = 0.9345
    •   Φ (-1.72) + X = 0.9345
    •   X = 0.9345 – 0.4573
    •   X = 0.4772
    •   Z0 = 2.0
Finding the unknown Z represented
by Zo

               CASE         MNEMONICS
            0 ≤ Z ≤ Zo          Φ(Zo )
          (-Zo ≤ Z ≤ 0)         Φ(Zo )
            Z1 ≤ Z ≤ Z2     Φ(Z2 ) – Φ(Z1)
          (-Z1 ≤ Z ≤ Z2)    Φ(Z1) + Φ(Z2 )
         (-Z1 ≤ Z ≤ - Z2)   Φ(Z1) - Φ(Z2 )
              Z ≥ Zo         0.5 – Φ(Zo )
              Z ≤ Zo         0.5 + Φ(Zo )
              Z ≤ -Zo        0.5 – Φ(Zo )
              Z ≥ -Zo        0.5 + Φ(Zo )
The event X has a normal distribution with mean µ = 10 and
Variance = 9. Find the probability that it will fall:

a.) between 10 and 11
b.) between 12 and 19
c.) above 13
d.) at x = 11
e.) between 8 amd 12
x         10       10
a .) Z                              0
                          3
           x        11    10        1
   Z                                        0 . 33
                          3         3

  P ( 10       X   11 )       P(0       Z      0 . 33 )
                              ( 0 . 33 )      0 . 1293
x         12        10      2
b .) Z                                       0 . 67
                           3           3
           x          19        10      9
     Z                                         3
                           3            3

  P ( 12       X   19 )        P ( 0 . 67     Z       3)
                                ( 3)        ( 0 . 67 )     0 . 4987   0 . 2486   0 . 251
x          13       10        3
c .) Z                                        1
                          3             3

   P( X       13 )   P( Z          1)       0 .5       ( 1)

                     0 .5      0 . 3413           0 . 1587
d .) P ( X         11 )        0
           x               8           10          2
e .) Z                                                    0 . 67
                                   3              3
    P( 8       X          12 )         P ( 0 . 67         Z        0 . 67 )
                                            ( 0 . 33 )   0 . 1293
2. A random variable X has a normal distribution
        with mean 5 and variance 16.

        a.) Find an interval (b,c) so that the probability of X
        lying in the interval is 0.95.
        b.) Find d so that the probability that X ≥ d is 0.05.


                                                   1
Solution: A.
   P (b ≤ X ≤ c) = P (Z b ≤ Z ≤ Zc)                2


                 = P (-1.96 (4) ≤X -5 ≤ 1.96 (4)
                 = P (-7.84 + 5 ≤ X ≤ 7.84 + 5)
   P (b ≤ X ≤ c) = P (-2.84 ≤ X ≤ 12.84 )

               thus: b = -2.84 and c = 12.84
2. A random variable X has a normal distribution
        with mean 5 and variance 16.

        a.) Find an interval (b,c) so that the probability of X
        lying in the interval is 0.95.
        b.) Find d so that the probability that X ≥ d is 0.05.


                                           1   0.5 – 0.05 = 0.45
Solution B:
   P ( X ≥ d ) = P (Z ≥ Zd) = 0.05≥            from the table

                                               0.45    Z = 1.64
               = P (X -5 ≥ 6.56)
               = P (X ≥ 6.56 + 5)
   P ( X ≥ d ) = P (X ≥ 11.56)

              thus: d = 11.56
3. A certain type of storage battery last on the
     average 3.0 years, with a standard deviation σ of
     0.5 year. Assuming that the battery are normally
     distributed, find the probability that a given battery
     will last less than 2.3 years.

Solution:                                      X   2 .3     3    0 .7
                                           z                            1.4
                                                     0 .5       0 .5
            P (X < 2.3) = P (Z < -1.4)
                        = 0.5 – Φ (-1.4)
                        = 0.5 – 0.4192
            P (X < 2.3) = 0.0808

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STATISTICS: Normal Distribution

  • 1. Report No. 2 Mr. Roderico Y. Dumaug, Jr.
  • 2. TOPIC OUTLINE  The Normal Distribution 1) Introduction 2) Definition of Terms and Statistical Symbols Used 3) How To Find Areas Under the Normal Curve 4) Finding the Unknown Z represented by Zo 5) Examples  Hypothesis Testing
  • 3. The Normal Distribution  Introduction Before exploring the complicated Standard Normal Distribution, we must examine how the concept of Probability Distribution changes when the Random Variable is Continuous.
  • 4. The Normal Distribution  Introduction A Probability Distribution will give us a Value of P(x) = P(X=x) to each possible outcome of x. For the values to make a Probability Distribution, we needed two things to happen: 1. P (x) = P (X = x) 2. 0 ≤ P (x) ≤ 1 For a Continuous Random Variable, a Probability Distribution must be what is called a Density Curve. This means: 1. The Area under the Curve is 1. 2. 0 ≤ P (x) for all outcomes x.
  • 5. The Normal Distribution  Introduction: Example: Suppose the temperature of a piece of metal is always between 0°F and 10°F. Furthermore, suppose that it is equally likely to be any temperature in that range. Then the graph of the probability distribution for the value of the temperature would look like the one below: 0.10 Uniform Distribution P (X < 5) ValuesAREA are spread uniformly across Probability 0.05 0.8 the range 0 to 10 0.5 4 P (3 < X ≤ 7) Probability: Area: 80% P (X > 2) (4)(0.1) = 0.00 Finding the Area Under the Curve 0.4 7 - 3 = 4 2 3 5 5 7 10 X Illustration of the fundamental fact about DENSITY CURVE
  • 6. The Normal Distribution: Definition of Terms and Symbols Used  Normal Distribution Definition: 1) A continuous variable X having the symmetrical, bell shaped distribution is called a Normal Random Variable. 2) The normal probability distribution (Gaussian distribution) is a continuous distribution which is regarded by many as the most significant probability distribution in statistics particularly in the field of statistical inference.  Symbols Used:  “z” – z-scores or the standard scores. The table that transforms every normal distribution to a distribution with mean 0 and standard deviation 1. This distribution is called the standard normal distribution or simply standard distribution and the individual values are called standard scores or the z-scores.  “µ” – the Greek letter “mu,” which is the Mean, and  “σ” – the Greek letter “sigma,” which is the Standard Deviation
  • 7. The Normal Distribution: Definition of Terms and Symbols Used  Characteristics of Normal Distribution: 1) It is “Bell-Shaped” and has a single peak at the center of the distribution, 2) The arithmetic Mean, Median and Mode are equal. 3) The total area under the curve is 1.00; half the area under the normal curve is to the right of this center point and the other half to the left of it, 4) It is Symmetrical about the mean, 5) It is Asymptotic: The curve gets closer and closer to the X – axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. 6) The location of a normal distribution is determined by the Mean, µ, the Dispersion or spread of the distribution is determined by the Standard Deviation, σ.
  • 8. The Normal Distribution: Graphically Normal Curve is Symmetrical Two halves identical Theoretically, curve Theoretically, curve extends to - ∞ extends to + ∞ Mean, Median and Mode are equal.
  • 9. AREA UNDER THE NORMAL CURVE P(x1<X<x2) P(X<x1) P(X>x2) x1 x2
  • 10. AREA UNDER THE NORMAL CURVE Let us consider a variable X which is normally distributed with a mean of 100 and a standard deviation of 10. We assume that among the values of this variable are x1= 110 and x2 = 85. 110 100 85 100 z1 1 . 00 z2 1 . 50 10 10 0.7745 0.4332 0.3413 -1.5 1
  • 11. The Standard Normal Probability Distribution  The Standard Normal Distribution is a Normal Distribution with a Mean of 0 and a Standard Deviation of 1.  It is also called the z distribution  A z –value is the distance between a selected value , designated X, and the population Mean µ, divided by the Population Standard Deviation, σ.  The formula is :
  • 12. Areas Under the Normal Curve z 0 1 2 3 4 5 6 7 8 9 0 0 0.004 0.008 0.012 0.016 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0754 1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 1.5 0.4332 0.4345 0.4357 0.437 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 1.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 0.4332 µ = 283 µ = 285.4 Grams 0 1.50 z Values
  • 13. How to Find Areas Under the Normal Curve Let Z be a standardized random variable P stands for Probability Φ(z) indicates the area covered under the Normal Curve. a.) P (0 ≤ Z≤≤ZZ≤≤2.45) Φ(-0.35) – Φ (0.91) b.) (-0.35 1.53) = Φ Φ (2.45) c.) (0.91 0) = (1.53) = 0.4929 - 0.3186 = 0.1368 0.4370 = 0.1743 This shaded area, from This shaded area, from This shaded area, from Z Z = -0.35Z = 1.53, = 2.45, = 0 and until Z = 0, Z – 0.91 until Z represents the probability represents the probability represents the probability value ofof0.1368 value of0.4370. value 0.1743
  • 14. How to Find Areas Under the Normal Curve Let Z be a standardized random variable P stands for Probability Φ(z) indicates the area covered under the Normal Curve. d.) P (-2.0 ≤ Z ≤ 0.95) = Φ (-2.0) + Φ (0.95) =0.4772 + 0.3289 = 0.8061 This shaded area, from Z = -2.0 until Z = 0.95 represents the probability value of 0.8061.
  • 15. How to Find Areas Under the Normal Curve Let Z be a standardized random variable P stands for Probability Φ(z) indicates the area covered under the Normal Curve. e.) P (-1.5 ≤ Z ≤ -0.5) = Φ (-1.5) – Φ (-0.5) =0.4332 - 0.1915 = 0.2417 This shaded area, from Z = -1.5 until Z = -0.5 represents the probability value of 0.2417.
  • 16. How to Find Areas Under the Normal Curve Let Z be a standardized random variable P stands for Probability Φ(z) indicates the area covered under the Normal Curve. f.) P(Z ≥ 2.0) = 0.5 – Φ (2.0) This shaded area, from = 0.5 – 0.4772 Z = 2.0 until beyond Z = 3 represents the probability = 0.0228 value of 0.0228.
  • 17. How to Find Areas Under the Normal Curve Let Z be a standardized random variable P stands for Probability Φ(z) indicates the area covered under the Normal Curve. g.) P (Z ≤ 1.5) = 0.5 + Φ (1.5) = 0.5 + 0.4332 = 0.9332 This shaded area, from Z = 1.5 until beyond Z = -3 represents the probability value of 0.9332
  • 18. Finding the unknown Z represented by Zo
  • 19. Finding the unknown Z represented by Zo  P(Z ≤ Z0) = 0.8461 0.5 + X = 0.8461 X = 0.8461 – 0.5 X = 0.3461 Z0 = 1.02 ans. • P(-1.72 ≤ Z ≤ Z0) = 0.9345 • Φ (-1.72) + X = 0.9345 • X = 0.9345 – 0.4573 • X = 0.4772 • Z0 = 2.0
  • 20. Finding the unknown Z represented by Zo CASE MNEMONICS 0 ≤ Z ≤ Zo Φ(Zo ) (-Zo ≤ Z ≤ 0) Φ(Zo ) Z1 ≤ Z ≤ Z2 Φ(Z2 ) – Φ(Z1) (-Z1 ≤ Z ≤ Z2) Φ(Z1) + Φ(Z2 ) (-Z1 ≤ Z ≤ - Z2) Φ(Z1) - Φ(Z2 ) Z ≥ Zo 0.5 – Φ(Zo ) Z ≤ Zo 0.5 + Φ(Zo ) Z ≤ -Zo 0.5 – Φ(Zo ) Z ≥ -Zo 0.5 + Φ(Zo )
  • 21. The event X has a normal distribution with mean µ = 10 and Variance = 9. Find the probability that it will fall: a.) between 10 and 11 b.) between 12 and 19 c.) above 13 d.) at x = 11 e.) between 8 amd 12
  • 22. x 10 10 a .) Z 0 3 x 11 10 1 Z 0 . 33 3 3 P ( 10 X 11 ) P(0 Z 0 . 33 ) ( 0 . 33 ) 0 . 1293
  • 23. x 12 10 2 b .) Z 0 . 67 3 3 x 19 10 9 Z 3 3 3 P ( 12 X 19 ) P ( 0 . 67 Z 3) ( 3) ( 0 . 67 ) 0 . 4987 0 . 2486 0 . 251
  • 24. x 13 10 3 c .) Z 1 3 3 P( X 13 ) P( Z 1) 0 .5 ( 1) 0 .5 0 . 3413 0 . 1587
  • 25. d .) P ( X 11 ) 0 x 8 10 2 e .) Z 0 . 67 3 3 P( 8 X 12 ) P ( 0 . 67 Z 0 . 67 ) ( 0 . 33 ) 0 . 1293
  • 26. 2. A random variable X has a normal distribution with mean 5 and variance 16. a.) Find an interval (b,c) so that the probability of X lying in the interval is 0.95. b.) Find d so that the probability that X ≥ d is 0.05. 1 Solution: A. P (b ≤ X ≤ c) = P (Z b ≤ Z ≤ Zc) 2 = P (-1.96 (4) ≤X -5 ≤ 1.96 (4) = P (-7.84 + 5 ≤ X ≤ 7.84 + 5) P (b ≤ X ≤ c) = P (-2.84 ≤ X ≤ 12.84 ) thus: b = -2.84 and c = 12.84
  • 27. 2. A random variable X has a normal distribution with mean 5 and variance 16. a.) Find an interval (b,c) so that the probability of X lying in the interval is 0.95. b.) Find d so that the probability that X ≥ d is 0.05. 1 0.5 – 0.05 = 0.45 Solution B: P ( X ≥ d ) = P (Z ≥ Zd) = 0.05≥ from the table 0.45 Z = 1.64 = P (X -5 ≥ 6.56) = P (X ≥ 6.56 + 5) P ( X ≥ d ) = P (X ≥ 11.56) thus: d = 11.56
  • 28. 3. A certain type of storage battery last on the average 3.0 years, with a standard deviation σ of 0.5 year. Assuming that the battery are normally distributed, find the probability that a given battery will last less than 2.3 years. Solution: X 2 .3 3 0 .7 z 1.4 0 .5 0 .5 P (X < 2.3) = P (Z < -1.4) = 0.5 – Φ (-1.4) = 0.5 – 0.4192 P (X < 2.3) = 0.0808