A fair four-sided spinner has edges numbered 1, 2, 2, 3. A fair three-sided spinner has edges numbered $$$-2,-1,1$$$. Each spinner is spun and the number on the edge on which it comes to rest is noted. The random variable $$$X$$$ is the sum of the two numbers that have been noted. Find $$$\operatorname{Var}(X)$$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_s20_qp_53 Year：2020 Question No：4(b)

Answer:

$\mathrm{E}(\mathrm{X})=\frac{-1+0+3+4+6+4}{12}=\frac{16}{12}=\frac{4}{3}$
$\operatorname{Var}(\mathrm{X})=\frac{1+0+3+8+18+16}{12}-\left(\frac{4}{3}\right)^{2}$
$\frac{37}{18}(=2.06)$

Knowledge points:

5.4.1 draw up a probability distribution table relating to a given situation involving a discrete random variable X, and calculate E(X) and Var(X)

Answer:

Knowledge points:

Solution:

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