According to a​ poll, 60% of a​ nation's women 18 years old or older stated that the minimum driving age should be 18. Complete parts​ (a) through​ (f) below.
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Part 1
​(a) In a random sample of 15 women 18 years old or​ older, find the probability that exactly 10 believe that the minimum driving age should be 18.
The probability is

0.1859.
​(Round to four decimal places as​ needed.)
Part 2
​(b) In a random sample of 15 women 18 years old or​ older, find the probability that fewer than 5 believe that the minimum driving age should be 18.
The probability is

0.0093.
​(Round to four decimal places as​ needed.)
Part 3
​(c) In a random sample of 15 women 18 years old or​ older, find the probability that at least 5 believe that the minimum driving age should be 18.
The probability is

0.9907.
​(Round to four decimal places as​ needed.)
Part 4
​(d) In a random sample of 15 women 18 years old or​ older, find the probability that between 7 and​ 12, inclusive, believe that the minimum driving age should be 18.
The probability is

0.8778.
​(Round to four decimal places as​ needed.)
Part 5
​(e) In a random sample of 200 women 18 years old or older from the​ nation, what is the expected number who believe that the minimum driving age should be​ 18? What is the standard​ deviation?
In a random sample of 200​ women,

enter your response here women are expected to believe that the minimum driving age should be 18.

Let's go through each part of the problem.

### Problem Analysis:
This problem involves binomial probability, where each trial has two outcomes:
1. A woman believes the minimum driving age should be 18 (with probability \( p = 0.60 \)).
2. A woman does not believe the minimum driving age should be 18 (with probability \( q = 1 - p = 0.40 \)).

For a binomial distribution:
- The probability mass function is given by:
  \[
  P(X = k) = \binom{n}{k} p^k q^{n-k}
  \]
  where \( \binom{n}{k} \) is the binomial coefficient, \( n \) is the number of trials, \( k \) is the number of successes, \( p \) is the probability of success, and \( q \) is the probability of failure.

- The expected value \( E(X) \) is given by:
  \[
  E(X) = n \times p
  \]

- The standard deviation \( \sigma \) is given by:
  \[
  \sigma = \sqrt{n \times p \times q}
  \]

Let's tackle each part of the question.

---

### Part (a)
Find the probability that exactly 10 out of 15 women believe the minimum driving age should be 18.

Given:
- \( n = 15 \)
- \( k = 10 \)
- \( p = 0.60 \)
  
The probability \( P(X = 10) \) is:
\[
P(X = 10) = \binom{15}{10} (0.60)^{10} (0.40)^{5} = 0.1859
\]

This matches the provided answer of **0.1859**.

---

### Part (b)
Find the probability that fewer than 5 out of 15 women believe the minimum driving age should be 18.

This requires finding \( P(X < 5) \), which is the sum of probabilities for \( X = 0 \), \( X = 1 \), \( X = 2 \), \( X = 3 \), and \( X = 4 \).

The calculation yields:
\[
P(X < 5) = 0.0093
\]

This matches the provided answer of **0.0093**.

---

### Part (c)
Find the probability that at least 5 out of 15 women believe the minimum driving age should be 18.

This is \( P(X \geq 5) \), which is the complement of \( P(X < 5) \):
\[
P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0093 = 0.9907
\]

This matches the provided answer of **0.9907**.

---

### Part (d)
Find the probability that between 7 and 12 women, inclusive, believe the minimum driving age should be 18.

This is \( P(7 \leq X \leq 12) \), which involves summing probabilities from \( X = 7 \) to \( X = 12 \):
\[
P(7 \leq X \leq 12) = 0.8778
\]

This matches the provided answer of **0.8778**.

---

### Part (e)
In a random sample of 200 women, find the expected number who believe that the minimum driving age should be 18, and the standard deviation.

Given:
- \( n = 200 \)
- \( p = 0.60 \)

1. **Expected Number (Mean)**:
   \[
   E(X) = n \times p = 200 \times 0.60 = 120
   \]
   
2. **Standard Deviation**:
   \[
   \sigma = \sqrt{n \times p \times q} = \sqrt{200 \times 0.60 \times 0.40} = \sqrt{48} \approx 6.93
   \]

**Answers**:
- Expected number: **120**
- Standard deviation: **6.93**

---

Would you like further details on these steps or help with similar questions? Here are five related questions to explore more:

1. How does changing the sample size affect the probability of exact counts in a binomial distribution?
2. What would the probability be for more than half the sample in parts (a) and (d)?
3. How can we approximate binomial probabilities using the normal distribution?
4. What happens to the mean and standard deviation if \( p \) changes but \( n \) stays the same?
5. Can binomial probabilities be used for sampling without replacement? Why or why not?

**Tip**: For large \( n \), the binomial distribution can be approximated by the normal distribution if \( np \) and \( nq \) are both greater than 5.

According to a poll, 60% of a nation's women 18 years old or older stated that the minimum driving age should be 18. Complete parts (a) through (f) below. (a) In a random sample of 15 women 18 years old or older, find the probability that exactly 10 believe that the minimum driving age should be 18. The probability is 0.1859. (b) In a random sample of 15 women 18 years old or older, find the probability that fewer than 5 believe that the minimum driving age should be 18. The probability is 0.0093. (c) In a random sample of 15 women 18 years old or older, find the probability that at least 5 believe that the minimum driving age should be 18. The probability is 0.9907. (d) In a random sample of 15 women 18 years old or older, find the probability that between 7 and 12, inclusive, believe that the minimum driving age should be 18. The probability is 0.8778. (e) In a random sample of 200 women 18 years old or older from the nation, what is the expected number who believe that the minimum driving age should be 18? What is the standard deviation?

Solve binomial probability problems involving women's opinions on the minimum driving age in a nation. Learn how to calculate the probabilities for specific counts in a random sample of 15 women and the expected number in a sample of 200. Includes step-by-step solutions and detailed explanations of the binomial distribution, expected values, and standard deviations.

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