CLASS 9

Chapter 5: Normal Distribution (Part II)

Introduction to Standardization Method

  • The situation does not always assume $ = 0$ and $ = 1$.

  • The standard normal distribution is uncommon but simplifies the process due to the existence of z-tables.

Understanding Z-Scores

  • Definition: A z-score (or standardized value) signifies how many standard deviations a value $x$ is either above or below the mean.

  • Formula for Calculating Z-Scores:

    • Sample: z=x−xˉsz = \frac{x - \bar{x}}{s}

    • Population: z=x−μσz = \frac{x - \mu}{\sigma}

  • Rounding: Z-scores should be rounded to two decimal places (e.g., $2.31$).

Properties of Z-Scores

  1. Z-scores are unitless quantities.

  2. If a particular value $x$ is lower than the mean, then $z < 0$. Conversely, if $x$ surpasses the mean, then $z > 0$.

Using the Standard Normal Distribution

  • Z-scores allow for the comparison of data points across different distributions, facilitating probability calculations.

  • Probability Relation: The relationship between raw scores and z-scores means that:

    • P(X < x) = P(Z < z)

    • Where z=x−μσz = \frac{x - \mu}{\sigma}

Examples of Calculating Z-Scores and Probabilities

Example 1: Height Calculation

  • Problem Statement: Find the z-score for a height of $182.6$ cm, given a mean height of $175.1$ cm and a standard deviation of $4$ cm.

  • Calculation of z-score:

    • z=182.6−175.14=1.875 ≈1.88z = \frac{182.6 - 175.1}{4} = 1.875 \ \approx 1.88

Example 2: Area Calculation

  • Given:

    • Mean ($\mu$) = $46$, Standard Deviation ($\sigma$) = $9$, X value = $37$.

  • Calculation of z-score:

    • z=37−469=−1z = \frac{37 - 46}{9} = -1

  • Using the z-table, find the probability:

    • P(z < -1) = 0.1587

  • Therefore, the area of the shaded region = $0.1587$.

Conversion Between X-Scores and Z-Scores

  • To convert from an x-score to a z-score:

    • z=x−μσz = \frac{x - \mu}{\sigma}

  • To convert a z-score back to an x-score:

    • x=μ+σzx = \mu + \sigma z

Example 3: Finding the X-Score

  • Problem Statement: Given a mean of $6$ and a standard deviation of $4.2$, find the x-score corresponding to $z = -0.44$.

  • Application of the formula:

    • x=6+(−0.44)(4.2)=6−1.848=4.2x = 6 + (-0.44)(4.2) = 6 - 1.848 = 4.2

Additional Example on Heart Rate

  • Adult males have resting heart rates normally distributed:

    • Mean ($\mu$) = $78$ bpm, Standard Deviation ($\sigma$) = $13.5$ bpm.

    • Probability that a randomly selected adult male has a resting heart rate less than $110$ bpm:

    • Calculation:

      • P(X < 110) = P\left(z < \frac{110 - 78}{13.5}\right) = P(z < 2.37) \

    • This probability is $99.11%$.

Example 4: Higher Heart Rate Probability

  • Problem Statement: Find the probability that an adult male's resting heart rate is greater than $93$ bpm:

    • Calculation:

    • P(X > 93) = P\left(z > \frac{93 - 78}{13.5}\right) = P(z > 1.11)

    • Result: $13.35%$ chance that a randomly selected adult male has a heart rate greater than $93$ bpm.

Probability of Events Over an Interval

  • Finding probability for an interval (e.g., between $51$ and $105$ bpm):

    • Calculate z-values:

    • z=51−7813.5=−2z = \frac{51 - 78}{13.5} = -2

    • z=105−7813.5$</p></li></ul></li><li><p>Combineprobabilities:</p><ul><li><p>z = \frac{105 - 78}{13.5} \$</p></li></ul></li><li><p>Combine probabilities:</p><ul><li><p>P(51 < X < 105) = P(-2 < z < 2)

    • Resulting probability $
      ightarrow 95.42%$.

Sample Mean Probabilities

  • For a sample mean with size $n$:

    • The formula P(\bar{X} < x) = P\left(z < \frac{x - \mu}{\sigma/\sqrt{n}}\right)

Example of Sample Mean:

  • Problem: Probability that the mean resting heart rate of $40$ adult males is less than $73$ bpm:

  • Finding z-value:

    • z = \frac{73 - 78}{\frac{13.5}{\sqrt{40}}} = -2.34

    • Result probability approximately $0.96%$.

Problem Set and Practice

  • Key Calculation: 1) Find the probability a randomly selected value is between $1.5$ and $7$. For normal distribution, mean is $4.8$, and s.d. is $3$. Z-scores must be calculated to find respective probabilities.

    • Probability for $z < -1.1$ and $z < 0.7333$ should be found next.

  • Quartiles Calculation:

    • First Quartile $Q1$ (25th percentile) and Third Quartile $Q3$ (75th percentile) should be calculated for a data set with $\mu = 26$.

Advanced Example on Pregnancy Duration

  1. Giving Birth After 298 Days:

    • For mean $=268$ days, standard deviation $=15$ days, calculate:

    • z = \frac{298-268}{15} \$$

    • Find probability using z-tables.

  2. Pregnancy Duration Less than 1%:

    • Identify z-value for $1\%$ and find corresponding x-value.

Practical Examples in Occupational Settings

  1. Downloading Time Normal Distribution: Mean of $7$ seconds, s.d. of $4$ seconds.

    • Determine the value such that only $30\%$ exceed it.

  2. Significant Values in Specific Heights for Possible Applications:

    • Calculate z-values, and corresponding heights separating significant from non-significant heights in a dataset with $ = 22$ inch, $ = 1.2$ inch.

Summary of Findings on Gender Height Percentiles

  • Military height requirements can be computed based on women’s height data, using mean of $63.8$ inches and standard deviation of $2.2$ inches to find eligible candidates.

Conclusion

  • The z-score method allows for the comparison of any normally distributed data, facilitating to estimate probabilities and necessary areas under the curve effectively, highlighting its application across various fields such as healthcare, psychology, and occupational standards for design and evaluations.