Chapters 8 to 10 — the binomial, continuous distributions, and $t$, chi-squared and $F$.
Fifteen problems. Work them with the solutions covered — reading a worked answer and producing one are different skills, and only the second is any use with a clock running.
Keep four decimal places unless a question says otherwise.
1. 20% of parcels are flagged for inspection. 14 are processed independently.
(a) State the four binomial conditions and confirm each. (b) Find $\mathrm{P}(X = 4)$. (c) Find $E(X)$ and $\sigma$.
2. For the same binomial, find $\mathrm{P}(X \le 2)$, and explain why this is not $1 - \mathrm{P}(X \ge 2)$.
3. For each, say whether the binomial applies; if not, name the failing condition.
(a) Number of sixes in 20 rolls of a fair die. (b) Number of red cards in 10 drawn from a deck without replacement. (c) Number of emails arriving between 2 and 3 pm. (d) Number of defective bulbs in 40 drawn from a shipment of 50,000.
4. Monthly spend is normal with $\mu = 340$ and $\sigma = 45$.
(a) Find the fraction above 400. (b) Find the fraction between 280 and 400. (c) Above what level do the top 2.5\% fall?
5. Processing times are uniform between 8 and 32 minutes.
(a) Find the mean and standard deviation. (b) Find $\mathrm{P}(X < 20)$. (c) Find $\mathrm{P}(15 < X < 25)$.
6. Time between machine faults is exponential with a mean of 6 days.
(a) Find the probability of more than 9 days without a fault. (b) Find the probability a fault occurs within 3 days. (c) State the standard deviation.
7. A classmate computes $\mathrm{P}(X = 25) = 0.031$ for a continuous variable.
(a) What is wrong? (b) What have they most likely computed? (c) What would a sensible question have asked?
8. Using Stata functions or tables, find:
(a) $t_{0.025,\,19}$ (b) $t_{0.05,\,29}$ (c) $\chi^2{0.05,\,12}$ (d) $F{0.05,\,4,\,15}$
9. Explain, in two sentences and without formulas, why the $t$ distribution has fatter tails than the normal.
10. A sample of $n = 20$ gives $\bar{x} = 46$ and $s = 10$, against a claimed mean of 40.
(a) Should you use $z$ or $t$, and why? (b) How many degrees of freedom? (c) Compute the test statistic.
11. Why can a chi-squared value never be negative, and why is the distribution right-skewed? Answer from the construction, not from a table.
12. A classmate computes $F_{0.05,\,12,\,4}$ when the question asked for $F_{0.05,\,4,\,12}$.
(a) Does the order matter? (b) Give both values. (c) Why are they different?
13. A binomial has $n = 50$ and $p = 0.06$.
(a) Check whether the normal approximation is appropriate. (b) State the rule you applied. (c) What should you use instead if it fails?
14. $X \sim \text{Bin}(n, 0.4)$ has a standard deviation of 3.0984.
(a) Find $n$. (b) Find $E(X)$. (c) Find $\mathrm{P}(X = 0)$.
15. Match each to the distribution you would use, and say why in a few words:
(a) The number of defective items in 25 independent draws. (b) The time until the next customer arrives. (c) A measurement that is equally likely anywhere in a range. (d) The sum of many small independent effects.
1. (a) Fixed $n = 14$; two outcomes (flagged or not); constant $p = 0.2$; independent processing. (b) $\binom{14}{4}(0.2)^4(0.8)^{10} = 1001 \times 0.0016 \times 0.1074 = 0.1720$. (c) $E(X) = 2.8$ and $\sigma = \sqrt{14(0.2)(0.8)} = 1.4967$.
2. $\mathrm{P}(X \le 2) = \mathrm{P}(0) + \mathrm{P}(1) + \mathrm{P}(2) = 0.4481$.
It is not $1 - \mathrm{P}(X \ge 2)$ because both events include $X = 2$, so they overlap rather than partition. The correct complement of $X \le 2$ is $X \ge 3$.
3. (a) Yes — all four hold, with $p = 1/6$. (b) No — drawing without replacement makes $p$ change between draws, and 10 of 52 is too large a fraction to ignore. (c) No — there is no fixed number of trials. (d) Yes, effectively. Strictly without replacement, but 40 from 50,000 leaves $p$ constant to any accuracy that matters.
4. (a) $z = (400-340)/45 = 1.3333$, so $\mathrm{P} = 0.0912$. (b) $z = -1.3333$ and $z = 1.3333$, giving $0.8176$. (c) $z = 1.959964$, so $x = 340 + 1.959964(45) = 428.20$.
5. (a) $\mu = (8+32)/2 = 20$; $\sigma = 24/\sqrt{12} = 6.9282$. (b) $(20-8)/24 = 0.5000$. (c) $(25-15)/24 = 0.4167$.
6. (a) $e^{-9/6} = e^{-1.5} = 0.2231$. (b) $1 - e^{-3/6} = 1 - 0.6065 = 0.3935$. (c) 6 days — for an exponential the standard deviation equals the mean.
7. (a) For a continuous variable the probability of any exact value is zero. (b) The density at 25, which is a height, not a probability — and heights can even exceed 1. (c) An interval: $\mathrm{P}(24.5 < X < 25.5)$, or $\mathrm{P}(X < 25)$.
8. (a) 2.0930 — invt(19, 0.975)
(b) 1.6991 — invt(29, 0.95)
(c) 21.0261 — invchi2(12, 0.95)
(d) 3.0556 — invF(4, 15, 0.95)
9. The $t$ divides by $s$ rather than $\sigma$, and $s$ is itself random — it varies from sample to sample instead of being a fixed number. When $s$ happens to come out small the whole ratio is pushed far from zero, so extreme values arise more often than the normal predicts.
10. (a) $t$ — $\sigma$ is unknown and $s$ was estimated from the same 20 observations. (b) $\nu = 19$. (c) SE $= 10/\sqrt{20} = 2.2361$, so $t = (46-40)/2.2361 = 2.6833$.
11. A $\chi^2_\nu$ is the sum of $\nu$ squared standard normals. Every term is a square, so every term is at least zero and so is the sum. It is right-skewed because squaring stretches large deviations far more than small ones, so the upper tail is pulled out while the lower end is bounded at zero.
12. (a) Yes, very much. (b) $F_{0.05,4,12} = 3.2592$ and $F_{0.05,12,4} = 5.9117$. (c) $F$ is a ratio of two chi-squareds, each divided by its own degrees of freedom. Swapping them swaps which group is in the numerator, which is a different question entirely.
13. (a) $np = 50(0.06) = 3$, which is below 5, so the approximation is not
appropriate. ($n(1-p) = 47$ passes, but both must.)
(b) The approximation needs $np \ge 5$ and $n(1-p) \ge 5$.
(c) The exact binomial — binomial() in Stata, or the formula directly.
14. (a) $\sigma^2 = n(0.4)(0.6) = 0.24n = 3.0984^2 = 9.6$, so $n = 40$. (b) $E(X) = 40(0.4) = 16$. (c) $(0.6)^{40} = 1.33 \times 10^{-9}$ — effectively zero.
15. (a) Binomial — a fixed number of independent two-outcome trials. (b) Exponential — waiting time until the next event at a constant rate. (c) Uniform — a flat density over the range. (d) Normal — by the Central Limit Theorem, sums of many small independent effects tend to a bell whatever the pieces look like.