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This problem deals with binomial probability because we're given a fixed number of trials (44 students), a binary outcome (whether a student needs to take another math class or not), and a constant probability (60% of students still need to take another math class).

Let's break down the steps to solve each part:

### Given:
- Probability that a student needs to take another math class, \( p = 0.60 \)
- Number of students, \( n = 44 \)
- We are asked to compute probabilities for various outcomes.

We will use the **binomial probability formula**:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
Where:
- \( X \) is the number of students who need to take another math class.
- \( k \) is the desired number of students.
- \( \binom{n}{k} \) is the binomial coefficient.

For the cumulative probabilities, we'll sum probabilities for different values.

### a. Probability that exactly 28 of them need to take another math class.

We calculate \( P(X = 28) \).

### b. Probability that at most 27 of them need to take another math class.

This is the cumulative probability:
\[
P(X \leq 27) = P(X = 0) + P(X = 1) + \dots + P(X = 27)
\]

### c. Probability that at least 27 of them need to take another math class.

This is the complementary probability:
\[
P(X \geq 27) = 1 - P(X \leq 26)
\]

### d. Probability that between 25 and 30 of them (inclusive) need to take another math class.

We sum the probabilities for \( P(25 \leq X \leq 30) \):
\[
P(25 \leq X \leq 30) = P(X = 25) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30)
\]

I'll compute these probabilities now.Here are the calculated probabilities, rounded to four decimal places as required:

- **a.** Probability that exactly 28 of them need to take another math class: \( 0.1099 \)
- **b.** Probability that at most 27 of them need to take another math class: \( 0.6288 \)
- **c.** Probability that at least 27 of them need to take another math class: \( 0.4919 \)
- **d.** Probability that between 25 and 30 (inclusive) of them need to take another math class: \( 0.6206 \)

Would you like further explanation on any part of these calculations?

Here are 5 related questions you might find useful:
1. How does the binomial distribution relate to the normal distribution for large \( n \)?
2. What is the difference between cumulative and point probabilities in binomial distributions?
3. How would the probabilities change if 70% of students needed another math class instead of 60%?
4. How do you calculate the binomial coefficient \( \binom{n}{k} \) manually?
5. What is the expected number of students who need to take another math class?

**Tip**: For large binomial problems, approximating the distribution with a normal distribution can simplify the calculations.

60% of all students at a college still need to take another math class. Round your answers to four decimal places. If 44 students are randomly selected, find the probability that: a) Exactly 28 of them need to take another math class. b) At most 27 of them need to take another math class. c) At least 27 of them need to take another math class. d) Between 25 and 30 (including 25 and 30) of them need to take another math class.

Solve a probability problem involving binomial distribution where 60% of students need another math class. Learn how to calculate the probabilities for exact, at most, at least, and between scenarios for 44 students, with answers rounded to four decimal places.

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