Normal Distribution: Z-scores, Percentiles, and Z-table Techniques

Key ideas and setup

  • Any normally distributed data can be modeled by a normal distribution with some mean μ and standard deviation σ.
  • We standardize observations to the standard normal distribution with mean 0 and sd 1 via the z-score:
    z=x−μσz = \frac{x - \mu}{\sigma}
  • The standard normal distribution is used to compare different distributions on a common scale and to translate observations into percentiles.

Percentiles and z-scores

  • A percentile is the observed value in the distribution such that c percent of the distribution is below that value. In other words, the percentile is the cumulative proportion below the value.
  • Example: a z-score of about 0.7 corresponds to roughly the 75th percentile (since z=0.7 is about 75–76% of the distribution below it).
  • The 50th percentile corresponds to z = 0 (the median).
  • Roman’s example (as described): convert scores to z-scores to compare “how far above the mean” each score is.
    • MCAT score: 5 12? (typo in transcript) but z ≈ 1.0 standard deviations above the MCAT mean, which is about the 85.31st percentile.
    • LSAT score: z ≈ 1.89 above mean, which is about the 97.06th percentile.
  • Conclusion from Roman example: the higher z-score (relative to its distribution) indicates better relative performance; Roman did better on LSAT by this metric.

Z-table intuition and two key directions

  • The z-table gives the cumulative probability up to a z-score:
    P(Z≤z)P(Z \le z)
    (i.e., the area under the standard normal curve to the left of z).
  • Two directions when using the table:
    1) Forward direction: given an observed value x, compute its z-score and then look up P(Z≤z)P(Z \le z) to get the percentile.
    2) Backward direction: given a percentile p, find the z-score such that P(Z≤z)≈pP(Z \le z) \approx p, then unstandardize to get x via x=μ+zσx = \mu + z\sigma.
  • Rounding note: when looking up z-scores in the table, round to two decimal places since the table’s precision is two decimals.

Empirical rule (the 68–95–99.7 rule)

  • Central region rule of thumb for normal distributions (empirical rule):
    • About 68% of observations lie within one standard deviation of the mean: P(∣Z∣≤1)≈0.68P(|Z| \le 1) \approx 0.68
    • About 95% lie within two standard deviations: P(∣Z∣≤2)≈0.95P(|Z| \le 2) \approx 0.95
    • About 99.7% lie within three standard deviations: P(∣Z∣≤3)≈0.997P(|Z| \le 3) \approx 0.997
  • In percentile terms, the central 95% corresponds to z in [−1.96, 1.96], since the exact tails are 2.5% each: P(Z≤−1.96)≈0.025,P(Z≤1.96)≈0.975.P(Z \le -1.96) \approx 0.025, \quad P(Z \le 1.96) \approx 0.975.
  • The empirical rule is approximate; for exact probabilities, use the z-table.
  • Example: the central 95% is between z = −1.96 and z = 1.96 (not exactly ±2). The exact area within ±1.96 is 0.95.

Using the z-table to find exact central percentages

  • To capture the central 95% exactly, use the two symmetric z-scores around the mean: ±z, where z is chosen so that the area between −z* and z* is 0.95.
  • Since the tails are 0.025 each, the exact tails occur at z ≈ ±1.96. Use the table to verify: P(Z ≤ 1.96) ≈ 0.975 and P(Z ≤ −1.96) ≈ 0.025.
  • Strategy described in the lecture: use the table in reverse to locate the z-score for a given central percentage (e.g., central 95% → find z such that P(Z ≤ z) ≈ 0.975; then take the negative for the lower bound).
  • Practical takeaway: for a central interval with probability p, find z-values that bound the central region, using the symmetric tails around 0.

Forward problem example (observed value to percentile)

  • Example: normal model with μ = 22, σ = 6. If x = 14, convert to z:
    z=14−226=−86=−1.33z = \frac{14 - 22}{6} = -\frac{8}{6} = -1.33
  • Round z to two decimals: −1.33. Look up P(Z ≤ −1.33) from the z-table: ≈ 0.0918.
  • Therefore, the percentile (percent below x = 14) is about 9.18%.
  • Note: This is a standard way to interpret an observed value within any normal distribution by converting to z and using the table.

Illustrative problem: Hershey's Kisses weights (toy example)

  • Given a normal model for weight (grams): mean μ = 4.76 g, σ = 0.07 g.
  • Step 1: sketch the curve and mark 1 SD boundaries: μ ± σ → 4.69, 4.83; 2 SD → 4.62, 4.90; 3 SD → 4.55, 4.97.
  • Step 2: compute proportion weighing ≤ 4.65 g:
    • z for x = 4.65: z=4.65−4.760.07≈−1.57z = \frac{4.65 - 4.76}{0.07} \approx -1.57
    • Look up P(Z ≤ −1.57) ≈ 0.0582 → about 5.82%.
  • Step 3: compute proportion weighing > 4.81 g:
    • z for x = 4.81: z=4.81−4.760.07≈0.71z = \frac{4.81 - 4.76}{0.07} \approx 0.71
    • P(Z ≤ 0.71) ≈ 0.7611. So P(X > 4.81) = 1 − 0.7611 ≈ 0.2389 → about 23.89% weigh more.
  • Percentile interpretation: a weight of 4.81 g sits near the 76th percentile (since about 76% weigh less).
  • Important note from the lecture: percentile means below that value; if you need the percentile below, you report the below-value probability; if you need the percentile above, use 1 − below probability.

Inverse problem (percentile to value) with Hershey's Kisses

  • If a candy is at the 62nd percentile (P = 0.62): find the corresponding z such that P(Z ≤ z) ≈ 0.62.
  • Approximate z ≈ 0.31 (since P(Z ≤ 0.31) ≈ 0.62–0.63 depending on table; the table usually gives near 0.62).
  • Then unstandardize to get x: x = \mu + z\sigma = 4.76 + 0.31(0.07) \approx 4.7817 \text{ g}
    \approx 4.78 \text{ g}.
  • The lecture notes show a similar process, using the closest table entry to the target percentile and then computing x from μ and σ.

Top 15% problem (upper tail) and its relation to percentiles

  • To be in the top 15% of weights, a candy must weigh at a percentile of 85% or higher (since 15% are above the value).
  • Look up the z corresponding to the 85th percentile (P(Z ≤ z) ≈ 0.85). The z-score is about z≈1.04z \approx 1.04.
  • Translate back to weight: x=μ+zσ=4.76+1.04(0.07)≈4.76+0.0728≈4.8328 g.x = \mu + z\sigma = 4.76 + 1.04(0.07) \approx 4.76 + 0.0728 \approx 4.8328 \text{ g}.
  • The lecture notes indicate using z ≈ 1.04 for the 85th percentile in this context.

Chipotle calories case study (realistic application)

  • Data: mean calories per order ≈ 1070, standard deviation ≈ 300.
  • Question: What percentage of meals weigh between 500 and 990 calories?
  • Steps:
    1) Compute z for the upper bound (990):
    z<em>high=990−1070300=−0.27z<em>{high} = \frac{990 - 1070}{300} = -0.27 P(Z ≤ −0.27) ≈ 0.3936. 2) Compute z for the lower bound (500): z</em>low=500−1070300=−1.9z</em>{low} = \frac{500 - 1070}{300} = -1.9
    P(Z ≤ −1.9) ≈ 0.0287.
    3) Area between = 0.3936 − 0.0287 = 0.3649.
  • Therefore, about 36.49% of meals weigh between 500 and 990 calories.
  • Practical note: to interpret a percentile or proportion for a range, subtract the smaller cumulative probability from the larger one.

Two common pitfalls and clarifications

  • Percentile vs. percentage below vs. above:
    • Percentile of a value is the percentage of observations below that value.
    • To get the percentage above a value, subtract the below-probability from 1.
  • When asked for the percentage in a range, use the difference of the two (larger percentile minus smaller percentile).
  • If the problem asks for the top X%, use the complement to convert to the lower-tail percentile, or directly use the z-score corresponding to that upper-tail percentile (e.g., 85th percentile for top 15%).

Worked recap: procedures you can follow

  • Standardize any observation: z=x−μσz = \frac{x - \mu}{\sigma}
  • Use the z-table to find the corresponding percentile: P(Z≤z)P(Z \le z)
  • If you need a percentile from a given probability, locate the z-score in the table that yields the nearest cumulative probability.
  • To translate a z back to the original units: x=μ+zσx = \mu + z\sigma
  • For a probability interval [a, b], compute the z-scores for a and b, look up their cumulative probabilities, then subtract: P(a < X < b) = P(Z \le zb) - P(Z \le za)

Final summary notes and study-friendly tips

  • Always start by identifying μ and σ for the distribution you’re working with.
  • Convert to z to compare across different distributions, then translate back as needed.
  • The z-table is your friend for exact percentiles; remember the forward (x → z → percentile) and backward (percentile → z → x) directions.
  • When using the empirical rule, remember it’s approximate; for precision, rely on the z-table.
  • Practice with varied examples (e.g., weight, calories, test scores) to fluently switch between units and interpretations.

Problems to practice (recommended flow)

  • Roman’s LSAT vs MCAT comparison using z-scores and percentiles.
  • Exact central intervals: find z for central 80%, 90%, and 95% using the z-table.
  • Hershey's Kisses: compute probabilities for x = 4.65, 4.81; interpret percentiles.
  • Inverse problems: given a percentile (e.g., 62nd, 85th), find the corresponding x.
  • Range problems: probability mass between two values (e.g., Chipotle calories between 500 and 990).

Clinical nuance from the lecture

  • The exercise demonstrates practical use of normal modeling in decision-making (e.g., “which score did better?”), statistical interpretation, and the importance of understanding tails and percentiles for real-world judgments.

Notes on notational conventions used in this lecture

  • Observed value: x
  • Population parameters: mean μ, standard deviation σ
  • Standardized value: z
  • Standard normal distribution: Z ~ N(0, 1)
  • Probability statements: P(Z≤z)P(Z \le z) or equivalently the percentile of x via the inverse link.

This set of notes captures the major and minor points from the lecture content, including formulas, exact numerical references, examples, and the practical workflow for working with normal distributions, z-scores, and percentiles.