Standard Normal Distribution and Z-Tables

Empirical Rule Review and Limitations

  • Empirical Rule Application: The empirical rule can be used to find proportions or percentages for intervals that correspond to exact standard deviation increments from the mean.
    • Specifically, if an interval is 1,2,1, 2,, or 33 standard deviations away from the mean, the empirical rule is applicable.
  • Example from Previous Lecture: Given a variable x\text{x} follows a normal distribution with mean μ=50\mu = 50 and standard deviation σ=5\sigma = 5.
    • Interval 4545 to 5555: This interval is one standard deviation away from the mean (50−5=4550 - 5 = 45, 50+5=5550 + 5 = 55). According to the empirical rule, the proportion of data in this interval is 0.680.68.
    • Interval 4949 to 5353: This interval does not match ,or, or3 standard deviations from the mean.
      • Limitation: The empirical rule cannot be used for this interval because it does not involve 1, 2,,or, or3 standard deviations from the mean.

Introduction to Standard Normal Distribution

  • Problem Addressed: To find proportions for intervals where the empirical rule is not applicable, the standard normal distribution is introduced.
  • Varying Normal Distributions: The shape of a normal distribution's density curve changes based on its population mean \muandpopulationstandarddeviationand population standard deviation\sigma.
    • Example: If \mu = 5andand\sigma = 2,thecurvehasacertainspread.If, the curve has a certain spread. If\mu = 5andand\sigma = 1,thecurveisnarrower(lessvariation).If, the curve is narrower (less variation). If\mu = 0andand\sigma = 1, this is the standard normal distribution.
  • Definition of Standard Normal Distribution: A specific normal distribution where the mean is \mu = 0andthestandarddeviationisand the standard deviation is\sigma = 1.
    • Variable Notation: The variable for a standard normal distribution is typically denoted as \text{z}.
  • Purpose: Any normal distribution can be converted into a standard normal distribution. Once converted, a standard normal table (Table A) can be used to find proportions (areas) for any given point.

Calculating Normal Proportions Using Table A

  • No Direct Formula: There is no simple algebraic formula to analytically calculate normal proportions.
  • Methods: Proportions are obtained using statistical software or tables.
    • Course Approach: In this course, students will use Table A from the textbook.
  • Cumulative Proportions: Table A provides cumulative proportions.
    • Definition: The cumulative proportion for a value \text{x}(or(or\text{z}inthestandardnormalcase)istheproportionofobservationsinthedistributionthatare<strong>lessthanorequalto</strong>in the standard normal case) is the proportion of observations in the distribution that are <strong>less than or equal to</strong>\text{x}.Thiscorrespondstotheareaunderthedensitycurvetotheleftof. This corresponds to the area under the density curve to the left of\text{x}.
    • Importance: Understanding how to use this table is crucial for subsequent chapters where proportions and z-values are frequently used.
  • Interpreting Table A: The table gives the area for \text{P}(\text{Z} < \text{z}_{1}).
    • If you locate \text{z}{1}inthetable,itdirectlygivestheproportionofin the table, it directly gives the proportion of\text{z}valueslessthanvalues less than\text{z}{1}(theareaunderthecurvetotheleftof(the area under the curve to the left of\text{z}_{1}).
  • Reading Z-Values from the Table:
    • Z-values are split into two parts for table lookup:
      1. First whole number and first decimal number: Located in the first column of the table.
      2. Second decimal number: Located in the first row of the table.
    • Example 1: For \text{z} = 0.12
      • First part: 0.1 (from the first column).
      • Second part: 0.02 (from the first row).
      • The intersection gives \text{P}(\text{Z} < 0.12).
    • Example 2: For \text{z} = -0.21
      • First part: -0.2 (from the first column).
      • Second part: 0.01 (from the first row).
      • The intersection gives \text{P}(\text{Z} < -0.21).

Solving Proportion Problems with Standard Normal Distribution

  • Problem A: Find \text{P}(\text{Z} < 2.85)
    • Graphically: This is the area to the left of \text{z} = 2.85inastandardnormalcurve(in a standard normal curve (\mu = 0).
    • Table Lookup: Divide 2.85intointo2.8(column)and(column) and0.05 (row).
    • Result: From Table A, \text{P}(\text{Z} < 2.85) = 0.9978.
  • Problem B: Find \text{P}(\text{Z} > 2.85)
    • Graphically: This is the area to the right of \text{z} = 2.85.
    • Strategy: Since Table A gives cumulative proportions (area to the left), use the fact that the total area under the curve is 1.
      • P(Z > 2.85) = 1 - P(Z < 2.85)
    • Result: 1 - 0.9978 = 0.0022.
  • Problem C: Find \text{P}(\text{Z} < -1.66)
    • Table Lookup: Divide -1.66intointo-1.6(column)and(column) and0.06 (row).
    • Result: From Table A, \text{P}(\text{Z} < -1.66) = 0.0485.
  • Problem D: Find \text{P}(-1.66 < \text{Z} < 2.85)
    • Graphically: This is the area between \text{z} = -1.66andand\text{z} = 2.85.
    • Strategy: Subtract the cumulative proportion of the lower bound from the cumulative proportion of the upper bound.
      • P(-1.66 < Z < 2.85) = P(Z < 2.85) - P(Z < -1.66)
    • Result: 0.9978 - 0.0485 = 0.9493.

Converting Any Normal Distribution to Standard Normal

  • Real-World Scenario: Most real-world normal distributions (e.g., student heights) are not standard normal distributions (\mu \neq 0oror\sigma \neq 1).
  • Conversion Formula: To convert any normal variable \text{X}(withmean(with mean\muandstandarddeviationand standard deviation\sigma)intoastandardnormalvariable) into a standard normal variable\text{Z}, use the formula:
    z = \frac{x - \mu}{\sigma}
  • Process: This conversion allows us to take any observation \text{x},transformitintoa, transform it into a\text{z}−score,andthenuseTableAtofindproportionsorareasassociatedwith-score, and then use Table A to find proportions or areas associated with\text{x}.
  • Conservation of Proportion: The proportion (area) associated with \text{x}valuesintheoriginalnormaldistributionis<strong>equal</strong>totheproportionassociatedwiththecorrespondingvalues in the original normal distribution is <strong>equal</strong> to the proportion associated with the corresponding\text{z} values in the standard normal distribution.
    • Example: If we want to find \text{P}(\text{X} < \text{a}), we convert the inequality:
      P(X < a) = P\left(\frac{X - \mu}{\sigma} < \frac{a - \mu}{\sigma}\right) = P\left(Z < \frac{a - \mu}{\sigma}\right)
  • Numerical Example: Given \text{X}followsanormaldistributionwithfollows a normal distribution with\mu = 10andand\sigma = 3.Find. Find\text{P}(\text{X} < 14.5).
    • Convert \text{X}toto\text{Z}:
      P(X < 14.5) = P\left(Z < \frac{14.5 - 10}{3}\right)
      P(X < 14.5) = P\left(Z < \frac{4.5}{3}\right)
      P(X < 14.5) = P(Z < 1.5)
    • Next step (not performed in transcript): Use Table A to find \text{P}(\text{Z} < 1.5)$$