Question #229286

A discrete random variable 𝑌 has a cumulative probability distribution as shown below.


𝑌 = 𝑦 0 1 2 3

𝑃(𝑌 ≤ 𝑦) 5𝑘 10𝑘 11𝑘 12𝑘


(a) Determine the value of 𝑘.

(b) Calculate expected value of 𝑌.

(c) Calculate variance and standard deviation of 𝑌.


1
Expert's answer
2022-02-22T13:47:09-0500

(a) Find the value of k from the condition

pi=1\sum {{p_i}} = 1

Then

5k+10k+11k+12k=138k=1k=1385k + 10k + 11k + 12k = 1 \Rightarrow 38k = 1 \Rightarrow k = \frac{1}{{38}}

The distribution series has the form

y0123p538103811381238\begin{matrix} y&0&1&2&3\\ p&{\frac{5}{{38}}}&{\frac{{10}}{{38}}}&{\frac{{11}}{{38}}}&{\frac{{12}}{{38}}} \end{matrix}

Answer: k=138k = \frac{1}{{38}}

(b) E(Y)=yipi=05+110+211+31238=6838=3419E(Y) = \sum {{y_i}} {p_i} = \frac{{0 \cdot 5 + 1 \cdot 10 + 2 \cdot 11 + 3 \cdot 12}}{{38}} = \frac{{68}}{{38}} = \frac{{34}}{{19}}

Answer: E(Y)=3419E(Y) = \frac{{34}}{{19}}

(c) Let's find the variance:

D(Y)=M(Y2)M2(Y)=05+110+411+91238(3419)2=383361D(Y) = M({Y^2}) - {M^2}\left( Y \right) = \frac{{0 \cdot 5 + 1 \cdot 10 + 4 \cdot 11 + 9 \cdot 12}}{{38}} - {\left( {\frac{{34}}{{19}}} \right)^2} = \frac{{383}}{{361}}

Then standard deviation is

σ(Y)=D(Y)=383361=38319\sigma \left( Y \right) = \sqrt {D(Y)} = \sqrt {\frac{{383}}{{361}}} = \frac{{\sqrt {383} }}{{19}}

Answer: D(Y)=383361D(Y) = \frac{{383}}{{361}} ; σ(Y)=38319\sigma \left( Y \right) = \frac{{\sqrt {383} }}{{19}}


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