To solve this, we need to determine if the probability of selecting 13 or more human resource professionals from a sample of 24 who are at companies that reject candidates based on their social media is significantly high. This can be done using a binomial distribution.

Given Information:

• The probability that a human resource professional is at a company that rejects job candidates because of their social media is $p = 0.31$.
• The sample size is $n = 24$.
• We are interested in the probability of 13 or more human resource professionals out of 24 (i.e., $X = 13$).

We use the binomial distribution $X \sim B(n, p)$, where $n$ is the number of trials (24) and $p$ is the probability of success (0.31).

For a binomial distribution, the mean $\mu$ and standard deviation $\sigma$ are calculated as:

• Mean: $\mu = n \cdot p = 24 \cdot 0.31 = 7.44$
• Standard deviation: $\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{24 \cdot 0.31 \cdot 0.69} \approx 2.73$

Now, we calculate the z-score for $X = 13$ using the normal approximation to the binomial distribution (since $n$ is large enough): $z = \frac{X - \mu}{\sigma} = \frac{13 - 7.44}{2.73} \approx \frac{5.56}{2.73} \approx 2.04$

Next, we look up the probability corresponding to this z-score. A z-score of 2.04 corresponds to a cumulative probability of about 0.9793. This means the probability of getting 13 or more successes (since it's a right tail) is: $P(X \geq 13) = 1 - P(Z \leq 2.04) = 1 - 0.9793 = 0.0207$

Interpretation:

Since the probability of observing 13 or more professionals at companies rejecting candidates due to social media is only 0.0207, which is a very low probability (typically, if the probability is less than 0.05, we consider it to be significantly high), we conclude that 13 is significantly high.

Answer:

A. Yes, 13 would be significantly high because the probability of 13 or more is 0.0207, which is low.

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Would you like more details or have any questions about this solution?

Here are some follow-up questions:

1. How would the result change if the probability $p$ were different, say 0.4?
2. What would happen if the sample size were larger, say 50?
3. How do you calculate the normal approximation for a binomial distribution?
4. How would you apply this method to a non-binary outcome (e.g., survey responses with more than two categories)?
5. What are the assumptions for using the normal approximation to a binomial distribution?

Tip: The normal approximation works well when $n$ is large enough and both $n \cdot p$ and $n \cdot (1 - p)$ are greater than 5.