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07-02: Exercises — The Normal Distribution and Z-Scores

Notes reference: 07-02: The Normal Distribution and Z-Scores

Q1: Standard normal areas

Find each area. Sketch the curve first.

  1. P(Z < 1.25)
  2. P(Z > 1.25)
  3. P(Z < −0.85)
  4. P(−1.5 < Z < 2.0)
  5. P(Z > −1.96)
  6. P(|Z| > 1.96)

Solution

1.  P(Z < 1.25)          = 0.8944                       (direct table lookup)
2.  P(Z > 1.25)          = 1 − 0.8944 = 0.1056
3.  P(Z < −0.85)         = 0.1977
4.  P(−1.5 < Z < 2.0)    = 0.9772 − 0.0668 = 0.9104
5.  P(Z > −1.96)         = 1 − 0.0250 = 0.9750
6.  P(|Z| > 1.96)        = 2 × 0.0250 = 0.0500          ← the famous 5%

=NORM.S.DIST(1.25, TRUE)                     ' 0.894350
=1-NORM.S.DIST(1.25, TRUE)                   ' 0.105650
=NORM.S.DIST(-0.85, TRUE)                    ' 0.197663
=NORM.S.DIST(2,TRUE)-NORM.S.DIST(-1.5,TRUE)  ' 0.910444
=2*NORM.S.DIST(-1.96, TRUE)                  ' 0.049979

pnorm(1.25); pnorm(1.25, lower.tail = FALSE); pnorm(-0.85)
pnorm(2) - pnorm(-1.5)
2 * pnorm(-1.96)

Q2: Inverse standard normal

Find the z value for each.

  1. Area 0.95 to the left
  2. Area 0.05 to the right
  3. Area 0.025 in each tail (two values)
  4. The 90th percentile
  5. The value with 1% above it

Solution

1.  z = 1.6449
2.  z = 1.6449          (same — 0.05 right = 0.95 left)
3.  z = ±1.9600
4.  z = 1.2816
5.  z = 2.3263

=NORM.S.INV(0.95)       ' 1.644854
=NORM.S.INV(0.975)      ' 1.959964
=NORM.S.INV(0.90)       ' 1.281552
=NORM.S.INV(0.99)       ' 2.326348

qnorm(c(0.95, 0.975, 0.90, 0.99))

Memorize the middle two — 1.645 and 1.960 appear in every confidence interval in Chapter 09.

Q3: Value → probability

Adult systolic blood pressure is normal with μ = 120 and σ = 12.

  1. P(BP < 135)
  2. P(BP > 140)
  3. P(110 < BP < 130)
  4. What percent are classified hypertensive (BP ≥ 140)?

Solution

1.  z = (135 − 120)/12 = 1.25
    P(Z < 1.25) = 0.8944                          →  89.44%

2.  z = (140 − 120)/12 = 1.6667
    P(Z > 1.67) = 1 − 0.9525 = 0.0475             →  4.75%

3.  z₁ = (110 − 120)/12 = −0.8333
    z₂ = (130 − 120)/12 = +0.8333
    P = 0.7977 − 0.2023 = 0.5954                  →  59.54%

4.  Same as #2:  about 4.8% of adults

=NORM.DIST(135, 120, 12, TRUE)                        ' 0.894350
=1-NORM.DIST(140, 120, 12, TRUE)                      ' 0.047790
=NORM.DIST(130,120,12,TRUE)-NORM.DIST(110,120,12,TRUE)' 0.595347

pnorm(135, 120, 12)
pnorm(140, 120, 12, lower.tail = FALSE)
pnorm(130, 120, 12) - pnorm(110, 120, 12)

Q4: Probability → value (inverse)

Same blood-pressure distribution.

  1. The 95th percentile
  2. The value below which 10% fall
  3. The middle 80% of readings
  4. The cutoff for the top 2.5%

Solution

1.  z = 1.6449  →  x = 120 + 1.6449(12) = 120 + 19.74 = 139.74

2.  z = −1.2816 →  x = 120 − 1.2816(12) = 120 − 15.38 = 104.62

3.  Middle 80% leaves 10% in each tail:  z = ±1.2816
    Lower = 120 − 1.2816(12) = 104.62
    Upper = 120 + 1.2816(12) = 135.38
    →  (104.62, 135.38)

4.  Top 2.5% →  z = 1.9600
    x = 120 + 1.96(12) = 120 + 23.52 = 143.52

=NORM.INV(0.95, 120, 12)          ' 139.7383
=NORM.INV(0.10, 120, 12)          ' 104.6214
=NORM.INV(0.90, 120, 12)          ' 135.3786
=NORM.INV(0.975, 120, 12)         ' 143.5196

qnorm(c(0.95, 0.10, 0.90, 0.975), mean = 120, sd = 12)

Which function? If the problem gives you a value and asks for a probability, use NORM.DIST / pnorm. If it gives you a probability and asks for a value, use NORM.INV / qnorm.

Q5: Working backwards to a parameter

A machine fills bottles with a standard deviation of 4 ml. The label says 500 ml, and the company wants only 1% of bottles to fall below the label.

At what mean should the machine be set?

Solution

Want P(X < 500) = 0.01

z for a left-tail area of 0.01 is −2.3263

        500 − μ
−2.3263 = ─────────
            4

500 − μ = −9.3054
      μ = 509.31 ml

Setting the mean to 509.31 ml wastes about 9.3 ml per bottle.
Reducing σ from 4 to 2 would let the mean drop to
   μ = 500 + 2.3263(2) = 504.65 ml
— halving the giveaway. Process CONSISTENCY is worth more than
process CENTERING here.

=500+2.3263*4         ' 509.31
=500-NORM.S.INV(0.01)*4    ' same, since NORM.S.INV(0.01) = -2.3263

Q6: Normal approximation to the binomial

A survey finds 55% of adults own a pet. In a random sample of 200 adults, find P(at least 120 own a pet).

Solution

n = 200,  p = 0.55

CHECK:  np = 110 ≥ 5  ✓     nq = 90 ≥ 5  ✓     → approximation is valid

μ = np      = 110
σ = √(npq)  = √(200 × 0.55 × 0.45) = √49.5 = 7.0356

CONTINUITY CORRECTION:  P(X ≥ 120)  →  P(X > 119.5)

z = (119.5 − 110) / 7.0356 = 9.5 / 7.0356 = 1.3502

P(Z > 1.35) = 1 − 0.9115 = 0.0885                →  8.85%

Exact binomial:  1 − BINOM.DIST(119, 200, 0.55, TRUE) = 0.0885
Without the correction: z = 10/7.0356 = 1.4213 → 0.0776 — off by 12%.

=1-NORM.DIST(119.5, 110, SQRT(200*0.55*0.45), TRUE)   ' 0.08849  approx
=1-BINOM.DIST(119, 200, 0.55, TRUE)                   ' 0.08847  exact

pnorm(119.5, 110, sqrt(49.5), lower.tail = FALSE)     # 0.088487
pbinom(119, 200, 0.55, lower.tail = FALSE)            # 0.088474

Q7: Continuity corrections

Write the normal interval for each binomial question.

Binomial Normal
P(X = 25) ?
P(X ≤ 25) ?
P(X < 25) ?
P(X ≥ 25) ?
P(X > 25) ?
P(20 ≤ X ≤ 25) ?

Solution

Binomial Normal
P(X = 25) P(24.5 < X < 25.5)
P(X ≤ 25) P(X < 25.5)
P(X < 25) P(X < 24.5)
P(X ≥ 25) P(X > 24.5)
P(X > 25) P(X > 25.5)
P(20 ≤ X ≤ 25) P(19.5 < X < 25.5)

The rule: expand the interval by 0.5 in the direction that includes the endpoint; shrink by 0.5 where the endpoint is excluded.

Q8: Is the data normal?

You have 60 observations. Describe four checks, and what each would show if the data were right-skewed.

Solution

1. HISTOGRAM
   Normal:      symmetric bell
   Right-skew:  a long tail stretching right, peak pushed left

2. BOXPLOT
   Normal:      median centred in the box, whiskers similar, few outliers
   Right-skew:  median left of centre, long right whisker, outliers above

3. NORMAL Q-Q PLOT  (the most sensitive visual check)
   Normal:      points hug the straight reference line
   Right-skew:  points curve UPWARD at the right end (above the line)

4. SKEWNESS / KURTOSIS  and  SHAPIRO-WILK
   Normal:      skew ≈ 0, excess kurtosis ≈ 0, Shapiro p > 0.05
   Right-skew:  skew > 1, Shapiro p < 0.05

x <- rexp(60, 0.2)                       # deliberately right-skewed

par(mfrow = c(1, 3))
hist(x, breaks = 12, col = "#8A5FBF", border = "white")
boxplot(x, col = "#0FA3A3")
qqnorm(x, pch = 19); qqline(x, col = "#0FA3A3", lwd = 2)
par(mfrow = c(1, 1))

psych::skew(x)          # well above 1
shapiro.test(x)         # small p  →  reject normality

fig, axes = plt.subplots(1, 3, figsize=(12, 3))
axes[0].hist(x, bins=12, color="#8A5FBF", edgecolor="white")
axes[1].boxplot(x)
stats.probplot(x, dist="norm", plot=axes[2])
stats.skew(x), stats.shapiro(x)

=SKEW(A2:A61)          ' well above 1 for right-skewed data
=KURT(A2:A61)          ' Excel reports EXCESS kurtosis; ≈ 0 if normal
' Q-Q plot by hand:
'   1. sort the data into column B
'   2. C2: =NORM.S.INV((ROW()-1.5)/60)     expected z-score
'   3. scatter B against C; a straight line means normal

Tip

Failing a normality test does not doom your analysis. The Central Limit Theorem (08-02) makes the sample mean approximately normal even when the raw data is not — which is what most tests actually require.

⬅️ Previous: 07-01: Exercises — Continuous, Uniform and Exponential Distributions ➡️ Next: 08-01: Exercises — Sampling Methods and Bias