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.

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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.

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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.

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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.

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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.

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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

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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.