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Part 1
22​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is​ (a) exactly​ two, (b) more than​ two,
and​ (c) between two and five inclusive. If​ convenient, use technology to find the probabilities.
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Part 1
​(a) ​P(2)equals

0.298 ​(Round to the nearest thousandth as​ needed.)
Part 2
​(b) ​P(xgreater than​2)equals

0.383 ​(Round to the nearest thousandth as​ needed.)
Part 3
​(c) ​P(2less than or equalsxless than or equals​5)equals

enter your response here ​(Round to the nearest thousandth as​ needed.)

This problem involves a **binomial distribution**, as we're dealing with a fixed number of independent trials (10 students), and each student has two possible outcomes: they either use credit cards for the rewards program or they don't. The probability of success (using credit cards for the rewards program) is 22%, or \( p = 0.22 \).

We use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}
\]

Where:
- \( n = 10 \) (the number of students),
- \( k \) is the number of students who say they use credit cards for rewards,
- \( p = 0.22 \) is the probability of success,
- \( 1 - p = 0.78 \) is the probability of failure.

### Part (a) \( P(X = 2) \)
For exactly 2 students, we calculate:

\[
P(X = 2) = \binom{10}{2} (0.22)^2 (0.78)^8
\]

The problem already states that \( P(2) \approx 0.298 \), so this part is confirmed.

### Part (b) \( P(X > 2) \)
To calculate the probability that more than 2 students say they use credit cards for the rewards program, we sum the probabilities for all values of \( X \) greater than 2 (i.e., \( P(X = 3) + P(X = 4) + \dots + P(X = 10) \)).

Alternatively, we can compute the complement:
\[
P(X > 2) = 1 - P(X \leq 2)
\]
where:
\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]
Given that \( P(X > 2) \approx 0.383 \) (from the problem), we can verify that \( P(X \leq 2) \approx 1 - 0.383 = 0.617 \).

### Part (c) \( P(2 \leq X \leq 5) \)
To find the probability that the number of students is between 2 and 5 (inclusive), we sum the probabilities for \( X = 2 \), \( X = 3 \), \( X = 4 \), and \( X = 5 \):
\[
P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\]

Let me calculate this for you:The probability that the number of college students who say they use credit cards because of the rewards program is between 2 and 5 (inclusive) is approximately \( P(2 \leq X \leq 5) = 0.671 \) (rounded to the nearest thousandth).

Would you like further details on any part of the solution or clarification?

Here are 5 related questions to expand on this concept:
1. How does changing the probability of success affect the binomial distribution?
2. What is the expected value and variance for this binomial distribution?
3. How would the probability change if we sampled 20 students instead of 10?
4. What is the probability of exactly 0 students choosing credit cards for rewards?
5. How can we approximate the binomial distribution with a normal distribution?

**Tip:** For binomial distributions with large \( n \) and when \( p \) is not too close to 0 or 1, the normal approximation can be a useful shortcut.

22% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is (a) exactly two, (b) more than two, and (c) between two and five inclusive.

Calculate the probability that the number of college students using credit cards for rewards is exactly two, more than two, or between two and five. Explore binomial distribution concepts with a step-by-step solution. Ideal for college-level statistics students.

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