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07-01: Exercises — Continuous, Uniform and Exponential Distributions

Notes reference: 07-01: Continuous, Uniform and Exponential Distributions

Q1: Discrete or continuous — and why it changes the arithmetic

For each, say whether P(X = 5) is zero or positive, and explain.

  1. X = number of emails received in an hour
  2. X = time in minutes until the next email
  3. X = number of heads in 10 flips
  4. X = weight of a randomly chosen apple in kg

Solution

1. DISCRETE   → P(X = 5) is POSITIVE (a specific, attainable count)
2. CONTINUOUS → P(X = 5) = 0.  Ask instead for P(4.5 < X < 5.5).
3. DISCRETE   → P(X = 5) is POSITIVE
4. CONTINUOUS → P(X = 5) = 0

Consequence for continuous variables:
    P(X < a) = P(X ≤ a)      and      P(a < X < b) = P(a ≤ X ≤ b)

That is NOT true for discrete variables, where the endpoints carry
positive probability. Getting this backwards is what the continuity
correction in 07-02 fixes.

Q2: Uniform distribution

A subway train arrives every 12 minutes. You arrive at a random time, so your wait X is uniform on [0, 12].

  1. The density f(x)
  2. P(wait < 4 minutes)
  3. P(wait between 5 and 9 minutes)
  4. P(wait > 10 minutes)
  5. μ and σ
  6. The 90th percentile of the wait

Solution

a = 0,  b = 12

1.  f(x) = 1/(12 − 0) = 1/12 = 0.08333   for 0 ≤ x ≤ 12,  and 0 elsewhere

2.  P(X < 4)     = (4 − 0)/12  = 4/12  = 0.3333
3.  P(5 < X < 9) = (9 − 5)/12  = 4/12  = 0.3333
4.  P(X > 10)    = (12 − 10)/12 = 2/12 = 0.1667

5.  μ = (0 + 12)/2 = 6.0 minutes
    σ = √((12 − 0)²/12) = √12 = 3.4641 minutes

6.  P90:  0.90 = (x − 0)/12  →  x = 10.8 minutes

Note that #2 and #3 give the same answer — for a uniform distribution only the length of the interval matters, not where it sits.

=1/(12-0)                   ' density  -> 0.08333
=(4-0)/(12-0)               ' P(X<4)   -> 0.33333
=(12-0)/2                   ' mean     -> 6
=SQRT((12-0)^2/12)          ' sd       -> 3.4641
=0+0.90*(12-0)              ' P90      -> 10.8

punif(4, 0, 12)                              # 0.3333
punif(9, 0, 12) - punif(5, 0, 12)            # 0.3333
punif(10, 0, 12, lower.tail = FALSE)         # 0.1667
qunif(0.90, 0, 12)                           # 10.8

U = stats.uniform(loc=0, scale=12)
U.cdf(4), U.cdf(9) - U.cdf(5), U.sf(10), U.ppf(0.90), U.mean(), U.std()

Q3: Uniform on a non-zero start

Delivery times are uniform between 20 and 50 minutes.

  1. P(delivery in under 30 minutes)
  2. P(delivery takes 35 to 45 minutes)
  3. μ and σ
  4. If the company promises "under 45 minutes or it's free", what fraction of orders are free?

Solution

a = 20,  b = 50,  b − a = 30

1.  P(X < 30)      = (30 − 20)/30 = 10/30 = 0.3333
2.  P(35 < X < 45) = (45 − 35)/30 = 10/30 = 0.3333
3.  μ = (20 + 50)/2 = 35 minutes
    σ = √(30²/12) = √75 = 8.6603 minutes
4.  P(X > 45) = (50 − 45)/30 = 5/30 = 0.1667   →  16.7% of orders are free

Business note: giving away one order in six is expensive. Promising "under 48 minutes" would cut it to 2/30 = 6.7%.

Q4: Exponential — waiting time

A call centre receives calls at 6 per hour, so waiting time between calls is exponential with λ = 6 per hour.

  1. Mean and standard deviation of the wait, in minutes
  2. P(next call within 5 minutes)
  3. P(next call takes more than 20 minutes)
  4. P(wait between 5 and 20 minutes)
  5. The 95th percentile of the wait

Solution

λ = 6 per hour.  Work in hours, convert at the end.

1.  μ = 1/λ = 1/6 hour = 10 minutes
    σ = 1/λ = 10 minutes                (mean = SD for the exponential)

2.  5 minutes = 1/12 hour
    P(X ≤ 1/12) = 1 − e^(−6/12) = 1 − e^(−0.5) = 1 − 0.606531 = 0.393469
                                                              →  0.3935

3.  20 minutes = 1/3 hour
    P(X > 1/3) = e^(−6/3) = e^(−2) = 0.135335                 →  0.1353

4.  P(5 < X < 20 min) = P(X ≤ 1/3) − P(X ≤ 1/12)
                      = (1 − 0.135335) − 0.393469
                      = 0.864665 − 0.393469 = 0.471196        →  0.4712

5.  P95:  1 − e^(−6x) = 0.95  →  e^(−6x) = 0.05
          −6x = ln(0.05) = −2.995732
          x = 0.499289 hours = 29.96 minutes

=EXPON.DIST(1/12, 6, TRUE)                        ' 0.393469
=1-EXPON.DIST(1/3, 6, TRUE)                       ' 0.135335
=EXPON.DIST(1/3,6,TRUE)-EXPON.DIST(1/12,6,TRUE)   ' 0.471196
=-LN(1-0.95)/6                                     ' 0.499289 hr = 29.96 min

pexp(1/12, rate = 6)                       # 0.3934693
pexp(1/3, 6, lower.tail = FALSE)           # 0.1353353
qexp(0.95, 6) * 60                         # 29.957 minutes

E = stats.expon(scale=1/6)                 # scipy takes SCALE = 1/lambda
E.cdf(1/12), E.sf(1/3), E.ppf(0.95) * 60

Unit discipline. Keep λ and x in the same time unit throughout, then convert only the final answer.

Using the Q4 call centre (λ = 6 per hour), compute P(no calls in the next 20 minutes) two ways.

Solution

WAY 1 — EXPONENTIAL (waiting time)
    "No call in 20 minutes" = "the wait exceeds 20 minutes"
    P(X > 1/3 hour) = e^(−6 × 1/3) = e^(−2) = 0.135335

WAY 2 — POISSON (count of events)
    In 20 minutes the expected count is λ' = 6 × (1/3) = 2
    P(N = 0) = 2⁰ e^(−2) / 0! = e^(−2) = 0.135335

IDENTICAL — 0.1353.

They are two views of the same process: Poisson counts the events,
exponential measures the gaps between them.

pexp(1/3, 6, lower.tail = FALSE)      # 0.1353353
dpois(0, lambda = 6 * 1/3)            # 0.1353353

Q6: Memorylessness

A component's lifetime is exponential with a mean of 5 years.

  1. P(it lasts more than 3 years)
  2. P(it lasts more than 8 years)
  3. Given that it has already lasted 5 years, P(it lasts 3 more)
  4. Compare #3 with #1 and explain.

Solution

Mean 5 years  →  λ = 1/5 = 0.2 per year

1.  P(X > 3) = e^(−0.2 × 3) = e^(−0.6) = 0.548812

2.  P(X > 8) = e^(−0.2 × 8) = e^(−1.6) = 0.201897

3.  P(X > 8 | X > 5) = P(X > 8) / P(X > 5)
                     = e^(−1.6) / e^(−1.0)
                     = 0.201897 / 0.367879
                     = 0.548812

4.  #3 = #1 EXACTLY.  This is MEMORYLESSNESS:
        P(X > s + t | X > s) = P(X > t)

    A 5-year-old component is, under this model, exactly as likely to
    survive another 3 years as a brand-new one.

WHEN THIS IS WRONG: real components wear out, so their failure rate
RISES with age. The exponential is appropriate for random, age-independent
failures (a lightning strike, a cosmic-ray bit flip), not for mechanical
wear. Use a Weibull distribution when the failure rate changes with age.

Q7: Density is not probability

A uniform distribution on [0, 0.5] has f(x) = 2.

  1. Is a "probability" of 2 legal?
  2. What IS the probability that X falls in [0.1, 0.3]?

Solution

1.  f(x) = 1/(0.5 − 0) = 2.  This is a DENSITY, not a probability.
    Densities may exceed 1 — they are probability PER UNIT of x.
    Only AREAS are probabilities, and areas are still ≤ 1:
        total area = 2 × 0.5 = 1   ✓

2.  P(0.1 ≤ X ≤ 0.3) = width × height = (0.3 − 0.1) × 2 = 0.4

Rule: for a continuous variable, PROBABILITY = AREA UNDER THE CURVE.
The height alone means nothing on its own.

Q8: Simulate and check

Simulate 100,000 exponential waits with λ = 6 per hour and verify the theoretical mean, SD, and P(X > 1/3).

Solution

set.seed(3)
w <- rexp(1e5, rate = 6)

c(theory_mean = 1/6,        sim_mean = mean(w))
c(theory_sd   = 1/6,        sim_sd   = sd(w))
c(theory_p    = exp(-2),    sim_p    = mean(w > 1/3))

hist(w, breaks = 60, col = "#0FA3A3", border = NA, freq = FALSE,
     main = "Exponential(rate = 6)", xlab = "hours")
curve(dexp(x, 6), add = TRUE, col = "#5B2A86", lwd = 2)

rng = np.random.default_rng(3)
w = rng.exponential(scale=1/6, size=100_000)

1/6, w.mean()
1/6, w.std(ddof=1)
np.exp(-2), (w > 1/3).mean()

' A2: =-LN(RAND())/6      — inverse-transform sampling; fill down 10,000 rows
=AVERAGE(A2:A10001)       ' ≈ 0.1667
=STDEV.S(A2:A10001)       ' ≈ 0.1667
=COUNTIF(A2:A10001,">"&1/3)/10000    ' ≈ 0.1353

What to notice: the simulated mean and SD are both close to 1/6 — the exponential's signature property, and a good sanity check that your λ/scale parameterization is right.

⬅️ Previous: 06-02: Exercises — Binomial and Poisson Distributions ➡️ Next: 07-02: Exercises — The Normal Distribution and Z-Scores