INTRODUCTION TO APPLIED STATISTICS

Tutorial Sheet 2

DATE : October 26, 2016

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Question 1: Define the following terms

(i) Parameter: A numerical value that describes a characteristic of a population. For example, the population mean (denoted as μ).
(ii) Statistic: A numerical value that describes a characteristic of a sample. For example, the sample mean (denoted as $\bar{x}$).
(iii) Sample space: The set of all possible outcomes of a random experiment, denoted as $S$. For a coin flip, $S = \{H, T\}$.
(iv) An Event: A subset of the sample space. It represents one or more outcomes of an experiment. For example, getting a head in a coin flip is an event.
(v) A random Variable: A variable whose values are numerical outcomes of a random phenomenon. It can be discrete (e.g., number of heads) or continuous (e.g., height).

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Question 2

a) A fair coin is to be flipped four times. Determine the sample space $S$.

• Each flip has 2 outcomes (H = Head, T = Tail).
• Total outcomes: $2^4 = 16$.
• $S = \{$HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT$\}$.

b) Find the probability of observing each event (fair coin, $P(H) = P(T) = 0.5$:
(i) Exactly 1 head:

• Favorable outcomes: HTTT, THTT, TTHT, TTTH (4 outcomes).
• $P = \frac{4}{16} = \frac{1}{4} = 0.25$.
  (ii) At least 1 head:
• Complement: no heads (all tails) = {TTTT} (1 outcome).
• $P = 1 - \frac{1}{16} = \frac{15}{16} = 0.9375$.
  (iii) No more than 2 heads:
• 0 heads: {TTTT} (1), 1 head: 4 outcomes, 2 heads: C(4,2) = 6 outcomes (e.g., HHTT).
• Total favorable: $1 + 4 + 6 = 11$.
• $P = \frac{11}{16} = 0.6875$.
  (iv) Either no tail or no head:
• No tail = all heads {HHHH}, no head = all tails {TTTT}.
• $P = \frac{2}{16} = \frac{1}{8} = 0.125$.
  (v) At least both a head and a tail:
• Exclude all heads and all tails: $16 - 2 = 14$ outcomes.
• $P = \frac{14}{16} = \frac{7}{8} = 0.875$.
  (vi) More than 4 heads:
• Maximum heads is 4.
• $P = 0$.
  (vii) At most 4 tails:
• Maximum tails is 4 (covers all outcomes).
• $P = \frac{16}{16} = 1$.

c) Biased coin ($P(H) = \frac{2}{5} = 0.4$, $P(T) = \frac{3}{5} = 0.6$):

• Use binomial distribution: $X \sim \text{Binomial}(n=4, p=0.4)$, where $X =$ number of heads.
• $P(X = k) = \binom{4}{k} (0.4)^k (0.6)^{4-k}$.
  (i) Exactly 1 head: $$P(X=1) = \binom{4}{1} (0.4)^1 (0.6)^3 = 4 \times 0.4 \times 0.216 = 0.3456.$$

(ii) At least 1 head:

$$P(X \geq 1) = 1 - P(X=0) = 1 - (0.6)^4 = 1 - 0.1296 = 0.8704.$$

(iii) No more than 2 heads:

$$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = (0.6)^4 + 4 \times 0.4 \times (0.6)^3 + \binom{4}{2} (0.4)^2 (0.6)^2 = 0.1296 + 0.3456 + 0.3456 = 0.8208.$$

(iv) Either no tail or no head:

$$P(\text{all heads}) = (0.4)^4 = 0.0256, \quad P(\text{all tails}) = (0.6)^4 = 0.1296, \quad P = 0.0256 + 0.1296 = 0.1552.$$

(v) At least both a head and a tail:

$$P = 1 - (0.0256 + 0.1296) = 1 - 0.1552 = 0.8448.$$

(vi) More than 4 heads: $P = 0$.
(vii) At most 4 tails: $P = 1$.

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Question 3: A fair die is tossed twice. Find the probability of observing:

• Sample space: $6 \times 6 = 36$ equally likely outcomes.
  (i) Odd number on first toss:
  • Odd numbers: 1,3,5 (3 outcomes).
  • $P = \frac{3 \times 6}{36} = \frac{18}{36} = 0.5$.
    (ii) Prime number on both tosses:
  • Primes: 2,3,5 (3 outcomes per toss).
  • $P = \frac{3 \times 3}{36} = \frac{9}{36} = 0.25$.
    (iii) A 2 and a 5:
  • Outcomes: (2,5), (5,2).
  • $P = \frac{2}{36} = \frac{1}{18} \approx 0.0556$.
    (iv) A 3 or 6 (interpreted as on the first toss):
  • First toss: 3 or 6 (2 outcomes).
  • $P = \frac{2 \times 6}{36} = \frac{12}{36} = \frac{1}{3} \approx 0.3333$.
    (v) Even on first toss and 1 on second:
  • Even: 2,4,6 (3 outcomes), second: 1 (1 outcome).
  • $P = \frac{3 \times 1}{36} = \frac{3}{36} = \frac{1}{12} \approx 0.0833$.
    (vi) Numbers whose product is 12:
  • Outcomes: (2,6), (3,4), (4,3), (6,2).
  • $P = \frac{4}{36} = \frac{1}{9} \approx 0.1111$.

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Question 4: A small clinic has only 8 bed spaces. If 14 patients are brought in, in how many possible ways can these patients be accommodated (one per bed)?

• Select 8 patients out of 14 and assign to distinct beds: $$\binom{14}{8} \times 8! = \frac{14!}{6!} = 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 = 121{,}080{,}960.$$

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Question 5: In how many ways can a committee of 7 be selected from 30 individuals?

• Order does not matter (combination): $$\binom{30}{7} = \frac{30!}{7! \times 23!} = \frac{30 \times 29 \times 28 \times 27 \times 26 \times 25 \times 24}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 2{,}035{,}800.$$

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Question 6

a) Patients with symptoms: 25% strep throat (S), 40% allergy (A), 10% both.
(i) $P(S) = 0.25$.
(ii) $P(A) = 0.40$.
(iii) $P(S \text{ or } A) = P(S) + P(A) - P(S \text{ and } A) = 0.25 + 0.40 - 0.10 = 0.55$.
b) Independence and mutual exclusivity:

• Independent? $P(S \text{ and } A) = 0.10$, $P(S) \times P(A) = 0.25 \times 0.40 = 0.10$. Yes, independent.
• Mutually exclusive? $P(S \text{ and } A) = 0.10 \neq 0$. No, not mutually exclusive.

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Question 7: Select 5 children from 11 (4 males, 7 females).

a) Different selections:

$$\binom{11}{5} = \frac{11!}{5! \times 6!} = 462.$$

b) 2 males and 3 females:

$$\binom{4}{2} \times \binom{7}{3} = 6 \times 35 = 210.$$

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Question 8: Virus present in 1.4% of population. Test accuracy: $P(\text{positive} \mid \text{virus}) = 0.995$, $P(\text{positive} \mid \text{no virus}) = 0.01$.

Let $V$ = has virus, $T^+$ = test positive.

• $P(V) = 0.014$, $P(V^c) = 0.986$.
  a) $P(\text{no virus}) = P(V^c) = 0.986$.
  b) $P(\text{test negative} \mid \text{individual has virus}) = 1 - 0.995 = 0.005$.
  c) $P(\text{test negative} \mid \text{individual no virus}) = 1 - 0.01 = 0.99$.
  d) $P(\text{test positive})$:

$$P(T^+) = P(T^+ \mid V)P(V) + P(T^+ \mid V^c)P(V^c) = (0.995 \times 0.014) + (0.01 \times 0.986) = 0.01393 + 0.00986 = 0.02379.$$

e) If test positive:
(i) $P(V \mid T^+) = \frac{P(T^+ \mid V)P(V)}{P(T^+)} = \frac{0.01393}{0.02379} \approx 0.5855$.
(ii) $P(V^c \mid T^+) = 1 - 0.5855 = 0.4145$.
f) If test negative:
(i) $P(V \mid T^-) = \frac{P(T^- \mid V)P(V)}{P(T^-)} = \frac{0.005 \times 0.014}{1 - 0.02379} = \frac{0.00007}{0.97621} \approx 0.0000717$.
(ii) $P(V^c \mid T^-) = 1 - 0.0000717 \approx 0.9999283$.

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Question 9: Same as Question 4

• Answer: $121{,}080{,}960$ ways.

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Question 10: Same as Question 5

• Answer: $2{,}035{,}800$ ways.

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Question 11: Probability of male birth = 0.52. Family has 12 children.

• $X \sim \text{Binomial}(n=12, p=0.52)$, $X =$ number of boys.
  a)
  (i) $P(X=3) = \binom{12}{3} (0.52)^3 (0.48)^9 \approx 0.0418$.
  (ii) $P(X=10) = \binom{12}{10} (0.52)^{10} (0.48)^2 \approx 0.0220$.
  (iii) $P(X \geq 10) = P(X=10) + P(X=11) + P(X=12) \approx 0.0220 + 0.0043 + 0.0004 = 0.0267$.
  (iv) More than half girls (girls > 6, i.e., boys $\leq 5$): $$P(X \leq 5) \approx 0.3343 \quad (\text{sum of } P(X=0) \text{ to } P(X=5)).$$

(v) At most 7 girls (boys $\geq 5$):

$$P(X \geq 5) \approx 0.8425.$$

(vi) At least 5 boys: $P(X \geq 5) \approx 0.8425$.
(vii) At most 9 girls (boys $\geq 3$):

$$P(X \geq 3) \approx 0.9863.$$

(viii) All girls: $P(X=0) = (0.48)^{12} \approx 0.00015$.
b) Expected number of sons:

$$E[X] = n p = 12 \times 0.52 = 6.24.$$

• Interpretation: Long-run average number of sons per family with 12 children.

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Question 12: Probability drug reaches market = 0.12. Firm has 15 compounds.

• $X \sim \text{Binomial}(n=15, p=0.12)$.
  a)
  (i) None reach market: $P(X=0) = (0.88)^{15} \approx 0.1470$.
  (ii) One or more: $P(X \geq 1) = 1 - 0.1470 = 0.8530$.
  (iii) At most 10: $P(X \leq 10) \approx 1$ (since mean $= 15 \times 0.12 = 1.8$, right tail negligible).
  (iv) Exactly half (7.5): Impossible, $P = 0$.
  (v) Between 8 and 13: $P(8 \leq X \leq 13) \approx 0$ (negligible).
  b) Expected value and variance:

$$E[X] = n p = 15 \times 0.12 = 1.8, \quad \text{Var}(X) = n p (1-p) = 15 \times 0.12 \times 0.88 = 1.584.$$

• Comment: Expected 1.8 drugs reach market, with moderate variability.

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Question 13: $Z \sim N(0,1)$.

a) Probabilities:
(i) $P(Z \leq 1.49) = \Phi(1.49) \approx 0.9319$.
(ii) $P(Z < -1.49) = \Phi(-1.49) \approx 0.0681$.
(iii) $P(Z \geq 2) = 1 - \Phi(2) \approx 0.0228$.
(iv) $P(-2.5 \leq Z) = \Phi(2.5) \approx 0.9938$ (since symmetric).
(iv) $P(|Z| < 1.2) = 2\Phi(1.2) - 1 \approx 2 \times 0.8849 - 1 = 0.7698$.
(v) $P(Z > -3.5) \approx 1$.
(vi) $P(1.5 \leq Z < 3) = \Phi(3) - \Phi(1.5) \approx 0.9987 - 0.9332 = 0.0655$.
(vii) $P(-1 \leq Z < 2.5) = \Phi(2.5) - \Phi(-1) \approx 0.9938 - 0.1587 = 0.8351$.
(viii) $P(-1.8 \leq Z < 1.8) = 2\Phi(1.8) - 1 \approx 2 \times 0.9641 - 1 = 0.9282$.
b) Find $z$:
(i) $P(Z < z) = 0.1762$: $z \approx -0.93$.
(ii) $P(Z > z) = 0.80$ → $P(Z < z) = 0.20$: $z \approx -0.84$.
(iii) $P(-z < Z < z) = 0.954$ → $P(Z < z) = 0.977$: $z = 2.00$.
(iv) $P(-6 \leq Z < z) = 0.5$ → $z \approx 0$ (since $P(Z < -6) \approx 0$).

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Question 14: $X \sim N(60, 4)$ (mean 60, variance 4, std dev 2).

a) Probabilities:
(i) $P(55 \leq X \leq 63) = P(-2.5 \leq Z \leq 1.5) \approx 0.9270$.
(ii) $P(X < 70) = P(Z < 5) \approx 1$.
(iii) $P(X \geq 61) = P(Z \geq 0.5) \approx 0.3085$.
(iv) $P(X \leq 60.8) = P(Z \leq 0.4) \approx 0.6554$.
(iv) $P(57.8 < X) = P(Z > -1.1) \approx 0.8643$.
b) Find $x$ (1 decimal place):
(i) $P(X < x) = 0.84$ → $z = 0.99$, $x = 60 + 0.99 \times 2 = 62.0$.
(ii) $P(X > x) = 0.8664$ → $P(X < x) = 0.1336$, $z \approx -1.11$, $x = 60 - 1.11 \times 2 = 57.8$.
(iii) $P(53 < X \leq x) = 0.9$ → $x \approx 62.6$.
(iv) $P(x \leq X < 66) = 0.5$ → $x \approx 60.0$.

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Question 15: Heights of 2-year-old boys $\sim N(34.5, 1.4^2)$.

(i) Between 32.5 and 36.5:

$$P(32.5 < X < 36.5) = P(-1.4286 < Z < 1.4286) \approx 0.8468.$$

(ii) Less than 31:

$$P(X < 31) = P(Z < -2.5) \approx 0.0062.$$

(iii) More than 37.5:

$$P(X > 37.5) = P(Z > 2.1429) \approx 0.0162.$$

(iv) Between 31.5 and 35:

$$P(31.5 < X < 35) = P(-2.1429 < Z < 0.3571) \approx 0.6234.$$

@Dr. Microbiota

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End of Solutions