4.3 Distributions

Topic	4.3 Distributions
Tag
Source	N16-TZ0-P1-2(HL)
Question Text	The faces of a fair six - sided die are numbered 1,3,3,5,5,7. Let $X$ be the discrete random variable that models the score obtained when this die is rolled. By completing the probability distribution table for $X$ find the expected value of $X$ .

\begin{aligned} E(X)=1 \times \frac{1}{6} & +3 \times \frac{1}{3}+5 \times \frac{1}{3}+7 \times \frac{1}{6} \\ & =\frac{24}{6}=4 \end{aligned}

Answer: 4

Topic	4.3 Distributions
Tag
Source	M16-TZ2-P1-5(HL)
Question Text	A biased coin is tossed six times. The probability of obtaining a head in any one throw is $p$ . Let $X$ be the number of heads obtained. (a) Find, in terms of $p$ , an expression for $P(X=5)$ . [multiple choice] (a) $p(1-p)^5$ (b) $6 p(1-p)^5$ (c) $p^5(1-p)$ (d) $6 p^5(1-p)$
Total Mark	2
Correct Answer	c
Explanation	na
Mark Scheme	$\begin{gathered} X \sim B(6, p) \\ P(X=4)=(6 C 5) p^5(1-p)^1 \\ 6 p^5(1-p)^1 \end{gathered}$ Answer: C
Question Text	(i) Determine the value of p for which P(X = 4) is a maximum. (a) 1/2 (b) 5/9 (c) 2/3 (d) 5/6 Total mark : 3 Correct Answer : d Explanation : na Mark Scheme : $\begin{gathered} \frac{d}{d p}\left(6 p^5-6 p^6\right)=30 p^4-36 p^5 \\ 6 p^4(5-6 p)=0 \\ p=\frac{5}{6} \end{gathered}$ Answer: D (ii) For this value of p , determine the expected number of heads. Total Mark : 3 Correct Answer : 5 Explanation : na Mark Scheme : E(X) = np = 6 x (5/6) = 5 Answer: 5

Topic	4.3 Distributions
Tag
Source	N19/5/MATHL/HP1/ENG/TZ1/XX/1
Question Text	The probability distribution of a discrete random variable, $X$ , is given by the following table, where $N$ and $p$ are constants. (a) Find the value of $p$. [2] [multiple choice] Options: (A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4

Question Text	(a) Find the value of $p$. [2] [multiple choice] Options: (A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4
Total Mark	2
Correct Answer	c
Explanation	na
Mark Scheme	p = 1 - 0.2 - 0.4 - 0.1 = 0.3 Answer: C
Question Text	(b) Given that E(X) = 8, find the value of N.
Total Mark	3
Correct Answer	16
Explanation	na
Mark Scheme	attempt to find E(X) 0.2 + 5 x 0.4 + 10 x 0.1 + N x 0.3 = 8 Rearrange, N=4.8/0.3=16 Answer: 16

Topic	4.3 Distributions
Tag
Source	M19/5/MATHL/HP1/ENG/TZ1/XX/6
Question Text	Let X be a random variable which follows a normal distribution with mean µ. Given that P(X < μ - 3) = 0.3, find (a) P(X > μ + 3); (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4
Total Mark	2
Correct Answer	c
Explanation	na
Mark Scheme	Due to symmetry, P(X > μ + 3) = P(X < μ - 3) = 0.3 Answer: C
Question Text	(b) P(X < μ + 3 ∣ X > μ - 3) can be written as $\frac{a}{b}$ where a and b are positive integers in lowest terms. Find the value of a + b.
Total Mark	5
Correct Answer	11
Explanation	na
Mark Scheme	$\begin{gathered} P(X<\mu+3 \mid X>\mu-3)=\frac{P(X>\mu-3 \cap X<\mu+3)}{P(X>\mu-3)} \\ \frac{P(\mu-3<X<\mu+3)}{P(X>\mu-3)}=\frac{0 . .4}{0.7}=\frac{4}{7} \end{gathered}$ Answer: 11