Chapter 8. 7 multiple choice and 8 worked questions, all with solutions.
These are the Checkpoint 2 practice questions for this topic. Work them with the answers covered.
1. 20% of parcels are flagged for inspection. In 15 processed independently, the probability exactly 3 are flagged is closest to
(A) 0.2501 (B) 0.2001 (C) 0.3101 (D) 0.2000
2. For those 15 parcels, the expected number flagged and the standard deviation are
(A) 3.0 and 2.4 (B) 2.4 and 1.5492 (C) 3.0 and 1.5492 (D) 3.0 and 1.7321
3. Each of 10 invoices is independently queried with probability 0.12. The probability at least one is queried is closest to
(A) 0.2785 (B) 0.7215 (C) 0.1200 (D) 0.8800
4. Of those 15 parcels, the probability that at most one is flagged is closest to
(A) 0.8329 (B) 0.0836 (C) 0.0352 (D) 0.1671
5. An auditor samples 6 invoices from a folder of 30 and counts errors. The binomial is inappropriate here because
(A) sampling without replacement changes $p$ between draws (B) there are more than two outcomes (C) the number sampled is not fixed (D) the invoices are not random
6. A call centre answers 6% of calls within 10 seconds. For a sample of 60 calls, a normal approximation to the count is
(A) inappropriate, because $np < 5$ (B) appropriate, because $n > 30$ (C) appropriate, because $n(1-p) > 5$ (D) inappropriate, because $p$ is small
7. A seed supplier claims half of all seeds germinate. In a tray of $n$ seeds the standard deviation of the number germinating is 4. Then $n$ is
(A) 16 (B) 32 (C) 8 (D) 64
1. 20% of parcels are flagged. 15 are processed independently. (a) State and check the four binomial conditions. (b) Find $\mathrm{P}(X=3)$. (c) Find $E(X)$ and $\sigma$.
2. Explain why ‘at least one’ questions are easier through the complement, using 8 independent trials each succeeding with probability 0.1.
3. A quality scheme rejects a batch if 2 or more of 12 sampled items are defective. If the true defect rate is 0.1, find the probability a batch is rejected.
4. A binomial has $E(X) = 12$ and $V(X) = 8.4$. Find $n$ and $p$.
5. Explain, with reference to the Bernoulli trial, why $E(X) = np$ for a binomial.
6. For $X \sim \mathrm{Bin}(25, 0.4)$, check whether the normal approximation applies, and if so give the approximating normal.
7. Lecture 6 showed a binomial at $p = 0.3$ approaching the normal as $n$ grew from 5 to 60. State the condition for using that approximation and check it at $n = 12$ and at $n = 60$.
8. A batch of 20 items has each independently defective with probability 0.15. (a) Find the expected number and the standard deviation. (b) Find the probability that none is defective. (c) Explain why $\mathrm{P}(X \ge 1)$ is easier through (b).
1. A — $\binom{15}{3}(0.2)^3(0.8)^{12}$.
2. C — $np$ and $\sqrt{np(1-p)}$. The first distractor reports the variance as the standard deviation.
3. B — $1 - (0.88)^{10}$. 0.2785 is the probability that none is.
4. D — $\mathrm{P}(0)+\mathrm{P}(1)$. 0.8329 is the complement.
5. A — 6 of 30 is a large fraction of the folder, so the probability shifts materially as invoices are removed.
6. A — $np = 3.6$. Both $np$ and $n(1-p)$ must clear 5.
7. D — $0.25n = 16$. 16 stops at the variance.
1. (a) Fixed $n = 15$; two outcomes; constant $p = 0.2$; independent. (b) $\binom{15}{3}(0.2)^3(0.8)^{12} = 0.2501$. (c) $E(X) = 3$, $\sigma = 1.5492$.
2. Directly it needs the eight terms $X = 1, \ldots, 8$. The complement needs one: $1 - (0.9)^8 = 0.5695$.
3. $1 - \mathrm{P}(0) - \mathrm{P}(1) = 1 - (0.9)^{12} - 12(0.1)(0.9)^{11} = 0.3410$.
4. $np = 12$ and $np(1-p) = 8.4$, so $1-p = 8.4/12 = 0.7$, giving $p = 0.3$ and $n = 40$.
5. A binomial is the sum of $n$ independent Bernoulli trials, each with expectation $p$. Expectation is linear, so the sum of $n$ copies of $p$ is $np$.
6. $np = 10$ and $n(1-p) = 15$, both above 5, so it applies. The approximation is $N(10, 6)$ — mean $np = 10$ and variance $np(1-p) = 6$, so $\sigma = 2.4495$.
7. The condition is $np \ge 5$ and $n(1-p) \ge 5$. At $n = 12$: $np = 3.6$, which fails, so use the exact binomial. At $n = 60$: $np = 18$ and $n(1-p) = 42$, both clear, so the approximation is sound.
8. (a) $E(X) = 3$, $\sigma = \sqrt{20(0.15)(0.85)} = 1.5969$. (b) $(0.85)^{20} = 0.0388$. (c) ‘At least one’ spans twenty separate terms; its complement is the single case $X = 0$, so one calculation replaces twenty.