This is a binomial probability problem since each selected American either lives in a city with a population greater than 100,000 (success) or does not (failure).

Given:

• $n = 50$ (total selections)
• $p = 0.65$ (probability of success)
• $q = 1 - p = 0.35$ (probability of failure)

The binomial probability formula is:

$P(X = k) = \binom{n}{k} p^k q^{n-k}$

where $\binom{n}{k}$ is the binomial coefficient.

Approximating with the Normal Distribution

Since $n$ is large, we use the normal approximation to the binomial distribution:

$X \sim N(\mu, \sigma^2)$

where
$\mu = np = 50(0.65) = 32.5$ $\sigma = \sqrt{npq} = \sqrt{50(0.65)(0.35)} \approx 3.63$

Using this, we standardize to the normal $Z$-score:

$Z = \frac{X - \mu}{\sigma}$

Now, let’s compute the probabilities.

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(a) $P(X = 35)$

Using the binomial formula directly is cumbersome, so we approximate using the normal distribution with a continuity correction ($X = 35$ is approximated as $P(34.5 \leq X \leq 35.5)$):

$Z_1 = \frac{34.5 - 32.5}{3.63} = \frac{2}{3.63} \approx 0.55$

$Z_2 = \frac{35.5 - 32.5}{3.63} = \frac{3}{3.63} \approx 0.83$

Using the standard normal table:

$P(Z \leq 0.83) \approx 0.7967, \quad P(Z \leq 0.55) \approx 0.7088$

$P(34.5 \leq X \leq 35.5) = 0.7967 - 0.7088 = 0.0879$

Answer: $P(X = 35) \approx 0.0879$

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(b) $P(X \leq 31)$ (At Most 31)

Using continuity correction: $P(X \leq 31) = P(X \leq 31.5)$

$Z = \frac{31.5 - 32.5}{3.63} = \frac{-1}{3.63} \approx -0.28$

$P(Z \leq -0.28) \approx 0.3897$

Answer: $P(X \leq 31) \approx 0.3897$

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(c) $P(X \geq 30)$ (At Least 30)

$P(X \geq 30) = 1 - P(X \leq 29.5)$

$Z = \frac{29.5 - 32.5}{3.63} = \frac{-3}{3.63} \approx -0.83$

$P(Z \leq -0.83) \approx 0.2033$

$P(X \geq 30) = 1 - 0.2033 = 0.7967$

Answer: $P(X \geq 30) \approx 0.7967$

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(d) $P(31 \leq X \leq 39)$ (Between 31 and 39, inclusive)

Using continuity correction: $P(30.5 \leq X \leq 39.5)$

$Z_1 = \frac{30.5 - 32.5}{3.63} = \frac{-2}{3.63} \approx -0.55$

$Z_2 = \frac{39.5 - 32.5}{3.63} = \frac{7}{3.63} \approx 1.93$

From the normal table:

$P(Z \leq 1.93) \approx 0.9732, \quad P(Z \leq -0.55) \approx 0.2912$

$P(31 \leq X \leq 39) = 0.9732 - 0.2912 = 0.6820$

Answer: $P(31 \leq X \leq 39) \approx 0.6820$

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Final Answers:

(a) $0.0879$
(b) $0.3897$
(c) $0.7967$
(d) $0.6820$

Would you like any additional clarification or step-by-step calculations?

Related Questions:

1. How would the probabilities change if the percentage of Americans living in large cities increased to 70%?
2. What is the probability that fewer than 25 people live in large cities?
3. How can we use the Poisson approximation instead of the normal approximation?
4. What is the probability that exactly half of the selected people live in large cities?
5. If we change the sample size to 100, how does it affect the standard deviation?

Tip:

Always check if a binomial distribution problem meets the conditions for a normal approximation: $np \geq 5$ and $nq \geq 5$. If these conditions fail, use the exact binomial probability formula.