We will solve this problem using the binomial probability distribution:

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

where:

• $n = 30$ (total job listings),
• $p = 0.74$ (probability that a listing conducts social media screening),
• $q = 1 - p = 0.26$.

However, since calculating individual binomial probabilities for multiple values of $k$ can be cumbersome, we will approximate it using the normal approximation to the binomial distribution:

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

where:

• Mean: $\mu = np = 30(0.74) = 22.2$,
• Standard deviation:

$\sigma = \sqrt{np(1 - p)} = \sqrt{30(0.74)(0.26)} \approx 2.35.$

We will use this normal approximation with the continuity correction to calculate each probability.

(a) $P(X \geq 19)$

Using continuity correction:

$P(X \geq 19) \approx P\left( Z \geq \frac{18.5 - 22.2}{2.35} \right)$

(b) $P(X < 17)$

Using continuity correction:

$P(X < 17) \approx P\left( Z < \frac{16.5 - 22.2}{2.35} \right)$

(c) $P(19 \leq X \leq 23)$

Using continuity correction:

$P(19 \leq X \leq 23) \approx P\left( \frac{18.5 - 22.2}{2.35} \leq Z \leq \frac{23.5 - 22.2}{2.35} \right)$

Now, we will compute these probabilities. scipy.stats as stats

Given data

n = 30 p = 0.74 q = 1 - p

Mean and standard deviation

mu = n * p sigma = (n * p * q) ** 0.5

(a) P(X >= 19) using continuity correction P(X >= 18.5)

z_a = (18.5 - mu) / sigma p_a = 1 - stats.norm.cdf(z_a)

(b) P(X < 17) using continuity correction P(X < 16.5)

z_b = (16.5 - mu) / sigma p_b = stats.norm.cdf(z_b)

(c) P(19 <= X <= 23) using continuity correction P(18.5 <= X <= 23.5)

z_c1 = (18.5 - mu) / sigma z_c2 = (23.5 - mu) / sigma p_c = stats.norm.cdf(z_c2) - stats.norm.cdf(z_c1)

Round results to three decimal places

p_a_rounded = round(p_a, 3) p_b_rounded = round(p_b, 3) p_c_rounded = round(p_c, 3)

p_a_rounded, p_b_rounded, p_c_roundedThe probabilities are:

(a) $P(X \geq 19) \approx 0.938$
(b) $P(X < 17) \approx 0.009$
(c) $P(19 \leq X \leq 23) \approx 0.644$

Would you like any further explanation or a different approach?

Here are five related questions to expand on this topic:

1. How does increasing the sample size affect the normal approximation to the binomial distribution?
2. What are some real-world examples where the normal approximation to the binomial distribution is useful?
3. How would the probabilities change if the probability of social media screening was 80% instead of 74%?
4. Why do we use a continuity correction when approximating binomial probabilities with a normal distribution?
5. What are the conditions under which the normal approximation to the binomial distribution is appropriate?

Tip: The normal approximation works best when $np$ and $nq$ are both greater than 5, ensuring that the binomial distribution is not too skewed.