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Hi, the problem is attached. Thanks

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    What you have tried? – 2017-01-22
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    You have first to compute the height of the $\delta$ peakn(which is equivalent to an area): use for that the fact the total probability is 1. Could you say what you obtain ? – 2017-01-22
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    I know that the area under probability density function should be 1, but from what I see it's equal to 0.6, so I am confused then how is that possible? for question a, I compute the area as 2*0.2 and for question b I suppose the probability is zero, since its a point. – 2017-01-22

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Your confusion is totally valid because your $f_{\bar{x}}$ is not what we call a probability density function; this terminology is reserved only for continuous random variables. On the other hand, the usual arrow notation (as you see in Figure 1) denotes point mass. So $\bar{x}$ is a mixture of continuous random variable and discrete random variable. In this sense the problem is severely misleading readers.

Anyway, forgetting about the term 'probability density function' and accepting that Figure somehow represents a distribution $F_{\bar{x}}$ such that

  • $F_{\bar{x}}$ has uniform density of $0.2$ over the interval $[-2,1]$, and
  • $F_{\bar{x}}$ has probability mass at $2$, and
  • The total mass of $F_{\bar{x}}$ is of course $1$.

Then it follows that

$$ \operatorname{Pr}(\bar{x} = 2) = 1 - \operatorname{Pr}(\bar{x} \neq 2) = 1 - \int_{-2}^{1} 0.2 \, dx = 1 - 0.6 = 0.4. $$

Now you have all the required ingredients. Let us for instance solve [a].

$$ \operatorname{Pr}(\bar{x} \geq -1) = \operatorname{Pr}(\bar{x} \in [-1,1]) + \operatorname{Pr}(\bar{x} = 2) = \int_{-1}^{1} 0.2 \, dx + 0.4 = 0.8. $$

Alternatively,

$$ \operatorname{Pr}(\bar{x} \geq -1) = 1 - \operatorname{Pr}(\bar{x} < -1) = \int_{-2}^{-1} 0.2 \, dx = 0.2. $$

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    Hey thank you for your clarification and time. So as far as I understand the probability of Pr{x=0} then will be equal to the probability of Pr{x=2} which equals to 0.4. Is this correct? – 2017-01-22
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    @KyanaShayan, I'm afraid that I am not following your comment. Notice that there is no probability mass at $0$. Only uniform density $0.2$ is given near $0$, which leads to $$ \operatorname{Pr}(\bar{x} = 0) = \int_{0}^{0} 0.2 \, dx = 0. $$ Just recall that this kind of behavior is what we expect for continuous RVs. Although $\bar{x}$ is not exactly a continuous RV, it behaves the same as continuous RV when restricted on $[-2, 1]$. – 2017-01-22