A branch of a certain bank in New York City has six ATMs. Let $X$ represent the number of machines in use at a particular time of day. The cdf of $X$ is as follows:
$f (x) = \{\begin{cases} \begin{matrix} 0 x < 1 \\ . 60 1 \leq x < 1 \\ \begin{matrix} . 19 1 \leq x < 2 \\ . 39 2 \leq x < 3 \\ \begin{matrix} . 67 3 \leq x < 4 \\ . 92 4 \leq x < 5 \\ . 97 5 \leq x < 6 \end{matrix} \end{matrix} \end{matrix} \\ 1 6 \leq x \end{cases}$
$P (2 \leq X \leq 5)$ .

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Solution

Finding the probabilities of $P (2 \leq X \leq 5)$ from the given function:
$\Rightarrow P (2 \leq X \leq 5) = P (X \leq 5) - P (X \geq 2)$
$\Rightarrow P (2 \leq X \leq 5) = 0.97 - 0.39$
$= 0.58$
Hence, the answer is $0.58$ .

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A branch of a certain bank in New York City has six ATMs. Let $X$ represent the number of machines in use at a particular time of day. The cdf of $X$ is as follows:
$f (x) = \{\begin{cases} \begin{matrix} 0 x < 1 \\ . 60 1 \leq x < 1 \\ \begin{matrix} . 19 1 \leq x < 2 \\ . 39 2 \leq x < 3 \\ \begin{matrix} . 67 3 \leq x < 4 \\ . 92 4 \leq x < 5 \\ . 97 5 \leq x < 6 \end{matrix} \end{matrix} \end{matrix} \\ 1 6 \leq x \end{cases}$
Calculate the following probabilities directly from the cdf:
$P (X > 3)$

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A branch of a certain bank in New York City has six ATMs. Let X represent the number of machines in use at a particular time of day. The cdf of X is as follows:f(x)=0x<1.601≤x<1.191≤x<2.392≤x<3.673≤x<4.924≤x<5.975≤x<616≤xP(2≤X≤5).

Finding the probabilities of P(2≤X≤5) from the given function:⇒P(2≤X≤5)=P(X≤5)-P(X≥2)⇒P(2≤X≤5)=0.97-0.39 =0.58Hence, the answer is 0.58.

Finding the probabilities of $P (2 \leq X \leq 5)$ from the given function:
$\Rightarrow P (2 \leq X \leq 5) = P (X \leq 5) - P (X \geq 2)$
$\Rightarrow P (2 \leq X \leq 5) = 0.97 - 0.39$
$= 0.58$
Hence, the answer is $0.58$ .