A survey showed that 35​% of human resource professionals are at companies that rejected job candidates because of information found on their social media. If 24 human resource professionals are randomly​ selected, would 13 be a significantly high number to be at companies that rejected job candidates because of information found on their social​ media? Why or why​ not?


To determine if 13 is a significantly high number, we can model the number of human resource (HR) professionals at companies that rejected candidates based on social media information as a binomial random variable. Here's the process:

### Step 1: Define the parameters
1. **Sample size (\( n \))**: 24 HR professionals are selected.
2. **Probability of success (\( p \))**: The probability that a randomly selected HR professional is from a company that rejected candidates based on social media information is 35%, or \( p = 0.35 \).
3. **Number of successes (\( X \))**: We want to see if 13 is significantly high for the number of HR professionals from companies with these rejections.

### Step 2: Determine the mean and standard deviation of the distribution
For a binomial distribution, the mean \( \mu \) and standard deviation \( \sigma \) are given by:
\[
\mu = n \cdot p
\]
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

Substitute the values:
\[
\mu = 24 \cdot 0.35 = 8.4
\]
\[
\sigma = \sqrt{24 \cdot 0.35 \cdot 0.65} \approx 2.34
\]

### Step 3: Calculate the z-score for \( X = 13 \)
The z-score tells us how many standard deviations 13 is from the mean. The formula for the z-score is:
\[
z = \frac{X - \mu}{\sigma}
\]
Substitute \( X = 13 \), \( \mu = 8.4 \), and \( \sigma \approx 2.34 \):
\[
z = \frac{13 - 8.4}{2.34} \approx 1.97
\]

### Step 4: Interpret the z-score
A z-score of 1.97 corresponds to a probability of about 0.9750, meaning that 13 is in the upper 2.5% of outcomes. This suggests that 13 is indeed a significantly high number, as it falls outside of the typical range we’d expect for this binomial distribution.

### Conclusion
Yes, 13 HR professionals is a significantly high number for those at companies rejecting candidates due to social media information, as it is quite far above the mean for this scenario.

Would you like more details or have any questions? 

Here are some related questions:
1. What if the percentage were 50% instead of 35%?
2. How would we test if a number is significantly low rather than high?
3. Could this approach be used for continuous distributions?
4. How does increasing the sample size affect the probability?
5. Why is the z-score useful in identifying significant outcomes?

**Tip:** Always calculate both the mean and standard deviation before deciding if an observed outcome is "significantly high" or "significantly low."

A survey showed that 35% of human resource professionals are at companies that rejected job candidates because of information found on their social media. If 24 human resource professionals are randomly selected, would 13 be a significantly high number to be at companies that rejected job candidates because of information found on their social media? Why or why not?

Explore whether 13 out of 24 HR professionals represent a significantly high number for companies that reject candidates due to social media findings. This problem uses the binomial distribution model, z-score, and normal approximation, suitable for students in Grades 11-12.

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