Science:Math Exam Resources/Courses/MATH105/April 2011/Question 09 (f)

MATH105 April 2011

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Question 09 (f)
Each of the short-answer questions below is worth 5 points. Put your answer in the box provided and show your work. No credit will be given for the answer without the correct accompanying work. A discrete random variable takes only two values, $0$ and $1.$ Find $p=P_{r}(X=1)$ if the variance of $X$ is $1/4.$

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Hint
The variance for a discrete random variable is defined to be $\operatorname {Var} (X)=\sum _{x_{i}}(x_{i}-\mu )^{2}P(X=x_{i})$ where the values $x_{i}$, are the possible outcomes for the random variable $X$ and $\mu$ is the mean: $\mu =\sum _{x_{i}}x_{i}P(X=x_{i}).$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The variance for a discrete random variable is defined to be $\operatorname {Var} (X)=\sum _{x_{i}}(x_{i}-\mu )^{2}P(X=x_{i})$ where the values $x_{i}$, are the possible outcomes for the random variable $X$ and $\mu$ is the mean: $\mu =\sum _{x_{i}}x_{i}P(X=x_{i}).$ In this case, there are only two possible outcomes $X=0$ or $X=1$. If we define the probability $P(X=1)$ with $p$ then by the law that the sum of the probabilities for all events must sum to one, we know that $P(X=0)=1-P(X=1)=1-p$. Thus, $\,\mu =0P(X=0)+1P(X=1)=p.$ Now we can use the definition of variance and get an equation for $p$. ${\begin{aligned}\operatorname {Var} (X)=(0-p)^{2}\underbrace {P(X=0)} _{=1-p}+(1-p)^{2}P(X=1)&=1/4\\-p^{2}(1-p)+(1-p)^{2}p&=1/4\\-p^{2}+p-1/4&=0\\p^{2}-p+1/4&=0\\(p-1/2)^{2}&=0\end{aligned}}$ Thus, the value of $p$ such that Var(X) = 1/4 is $p=1/2$.

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