Sure! Let's break down the problem step by step.

Part (a): Find and plot the cumulative distribution function (CDF) ( F_X(x) )

The cumulative distribution function (CDF) ( F_X(x) ) is defined as follows:

[ F_X(x) = P(X \leq x) ]

Given the probability mass function (PMF):

[ P_X(k) = \frac{1}{2^k} \text{ for } k = 1, 2, 3, \ldots ]

We can say that the CDF can be computed as:

[ F_X(x) = \sum_{k=1}^{\lfloor x \rfloor} P_X(k) = \sum_{k=1}^{\lfloor x \rfloor} \frac{1}{2^k} ]

Where ( \lfloor x \rfloor ) is the greatest integer less than or equal to ( x ).

This summation is a geometric series with a common ratio ( r = \frac{1}{2} ). The sum of the first ( n ) terms of a geometric series can be calculated using the formula:

[ S_n = a \frac{1 - r^n}{1 - r} ]

where ( a ) is the first term and ( r ) is the common ratio. Here ( a = \frac{1}{2} ) and ( r = \frac{1}{2} ), so we have:

[ F_X(\lfloor x \rfloor) = \frac{1/2}{1 - 1/2} \left(1 - \left(\frac{1}{2}\right)^{\lfloor x \rfloor}\right) = 1 - \left(\frac{1}{2}\right)^{\lfloor x \rfloor} ]

Now, put the overall expression together:

[ F_X(x) = \begin{cases} 0 & \text{if } x < 1 \ 1 - \left(\frac{1}{2}\right)^{\lfloor x \rfloor} & \text{if } x \geq 1 \end{cases} ]

Plotting the CDF

1. For ( x < 1 ), ( F_X(x) = 0 ).
2. For ( x = 1 ), ( F_X(1) = 1 - (1/2)^1 = 1/2 ).
3. As ( x ) increases, the CDF approaches 1.

You can visualize this by creating points for integer values ( x = 1, 2, 3, \ldots ) and plotting them:

| x | F_X(x) | |---|----------------| | 0 | 0 | | 1 | 0.5 | | 2 | 0.75 | | 3 | 0.875 | | 4 | 0.9375 | | 5 | 0.96875 | | ... | 1 |

The graph will show a step function that rises and levels off, approaching 1.

Part (b): Find ( P(2 \leq X \leq 5) )

This probability can be computed as follows:

[ P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) ]

Calculating each term:

[ P(X = k) = \frac{1}{2^k} ]

Thus,

[ P(X = 2) = \frac{1}{2^2} = \frac{1}{4} ] [ P(X = 3) = \frac{1}{2^3} = \frac{1}{8} ] [ P(X = 4) = \frac{1}{2^4} = \frac{1}{16} ] [ P(X = 5) = \frac{1}{2^5} = \frac{1}{32} ]

Now, adding them together:

[ P(2 \leq X \leq 5) = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} ]

Finding a common denominator (32):

[ = \frac{8}{32} + \frac{4}{32} + \frac{2}{32} + \frac{1}{32} = \frac{15}{32} ]

Part (c): Find ( P(X \geq 4) )

To find ( P(X \geq 4) ), we can use the hint provided:

[ P(X \geq a) = 1 - P(X < a) ]

Thus,

[ P(X \geq 4) = 1 - P(X < 4) = 1 - P(X = 1) - P(X = 2) - P(X = 3) ]

We already calculated:

[ P(X = 1) = \frac{1}{2^1} = \frac{1}{2} ] [ P(X = 2) = \frac{1}{4} ] [ P(X = 3) = \frac{1}{8} ]

Now adding these:

[ P(X < 4) = P(X = 1) + P(X = 2) + P(X = 3) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} ]

Common denominator for addition (8):

[ = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{7}{8} ]

Therefore:

[ P(X \geq 4) = 1 - P(X < 4) = 1 - \frac{7}{8} = \frac{1}{8} ]

Summary:

• The CDF ( F_X(x) ) is defined piecewise and involves summation of a geometric series.
• The probability ( P(2 \leq X \leq 5) = \frac{15}{32} ).
• The probability ( P(X \geq 4) = \frac{1}{8} ).

If you have any further questions or need additional elaboration, feel free to ask!