A branch of a certain bank 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*) = 

0
  
*x* < 0

0.07
  
0 ≤ *x* < 1

0.22
  
1 ≤ *x* < 2

0.32
  
2 ≤ *x* < 3

0.54
  
3 ≤ *x* < 4

0.89
  
4 ≤ *x* < 5

0.96
  
5 ≤ *x* < 6

1
  
6 ≤ *x*

Calculate the following probabilities directly from the cdf:

(a)    *p*(2), that is, 

*P*(*X* = 2)




(b)    *P*(*X* > 3)


(c)    

*P*(2 ≤ *X* ≤ 5)




(d)    

*P*(2 < *X* < 5)


Let's calculate each probability based on the cumulative distribution function (CDF) \( F(x) \).

### (a) \( p(2) = P(X = 2) \)

To find \( P(X = 2) \), we need to find the probability that \( X = 2 \). We can calculate this by finding the difference in the CDF values just before and just after 2:

\[
p(2) = P(X = 2) = F(2) - F(1)
\]

From the given CDF:
- \( F(2) = 0.32 \)
- \( F(1) = 0.22 \)

Thus,
\[
p(2) = 0.32 - 0.22 = 0.10
\]

### (b) \( P(X > 3) \)

To find \( P(X > 3) \), we calculate it as \( 1 - F(3) \) because \( F(3) \) represents \( P(X \leq 3) \):

\[
P(X > 3) = 1 - F(3)
\]

From the CDF:
- \( F(3) = 0.54 \)

Therefore,
\[
P(X > 3) = 1 - 0.54 = 0.46
\]

### (c) \( P(2 \leq X \leq 5) \)

To calculate \( P(2 \leq X \leq 5) \), we find the difference between \( F(5) \) and \( F(2) \):

\[
P(2 \leq X \leq 5) = F(5) - F(2)
\]

From the CDF:
- \( F(5) = 0.96 \)
- \( F(2) = 0.32 \)

Thus,
\[
P(2 \leq X \leq 5) = 0.96 - 0.32 = 0.64
\]

### (d) \( P(2 < X < 5) \)

To find \( P(2 < X < 5) \), we need to find the probability between 2 and 5, excluding both endpoints. This can be calculated as:

\[
P(2 < X < 5) = F(4) - F(2)
\]

From the CDF:
- \( F(4) = 0.89 \)
- \( F(2) = 0.32 \)

Therefore,
\[
P(2 < X < 5) = 0.89 - 0.32 = 0.57
\]

### Summary of Results
- **(a)** \( p(2) = 0.10 \)
- **(b)** \( P(X > 3) = 0.46 \)
- **(c)** \( P(2 \leq X \leq 5) = 0.64 \)
- **(d)** \( P(2 < X < 5) = 0.57 \)

Would you like more details on these calculations?

---

1. How is a cumulative distribution function (CDF) used to calculate probabilities?
2. What is the difference between \( P(X \geq x) \) and \( P(X > x) \) when using a CDF?
3. Why does the probability of a specific value, \( P(X = x) \), depend on the CDF increments?
4. How does a CDF differ from a probability density function (PDF)?
5. Can the probability mass function (PMF) be derived from the CDF directly?

**Tip:** Remember that for discrete random variables, probabilities at specific values are found by the difference in CDF values at neighboring points.

A branch of a certain bank 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) =

0       x < 0

0.07    0 ≤ x < 1

0.22    1 ≤ x < 2

0.32    2 ≤ x < 3

0.54    3 ≤ x < 4

0.89    4 ≤ x < 5

0.96    5 ≤ x < 6

1       6 ≤ x

Calculate the following probabilities directly from the cdf:

(a) p(2), that is, P(X = 2)

(b) P(X > 3)

(c) P(2 ≤ X ≤ 5)

(d) P(2 < X < 5)

Explore probability calculations based on the cumulative distribution function (CDF) for ATM usage at a bank branch. This solution provides a step-by-step approach to find probabilities like P(X = 2), P(X > 3), and interval probabilities, using core probability concepts and CDF properties for discrete variables.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation