Answer :
Sure, let's break this down step-by-step:
### Part (a)
Question: If 200 murders are randomly selected, how many would we expect to be committed with a firearm?
Solution:
1. Total number of murders: [tex]\( 200 \)[/tex]
2. Percentage of murders committed with a firearm: [tex]\( 60.8\% \)[/tex]
To find the expected number of murders committed with a firearm, we multiply the total number of murders by the percentage (in decimal form):
[tex]\[ \text{Expected number of murders with a firearm} = 200 \times \left(\frac{60.8}{100}\right) \][/tex]
[tex]\[ \text{Expected number of murders with a firearm} = 200 \times 0.608 \][/tex]
[tex]\[ \text{Expected number of murders with a firearm} = 121.6 \][/tex]
Therefore, we would expect approximately 121.6 murders to be committed with a firearm.
### Part (b)
Question: Would it be unusual to observe 142 murders by firearm in a random sample of 200 murders? Why?
Solution:
1. We previously calculated the expected number of murders with a firearm to be 121.6.
2. To determine if observing 142 murders by firearm is unusual, we need to consider the expected variability in the number of murders. We do this by calculating the z-score and the corresponding probability.
First, let’s find the mean and standard deviation for the binomial distribution:
- Mean (μ): We already know that the expected number of murders with a firearm (mean) is 121.6.
- Standard Deviation (σ): For a binomial distribution, the standard deviation can be calculated using the formula:
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} \][/tex]
Where [tex]\( n \)[/tex] is the total number of trials (murders), and [tex]\( p \)[/tex] is the probability of success (percentage of murders committed with a firearm).
[tex]\[ \sigma = \sqrt{200 \times 0.608 \times (1 - 0.608)} \][/tex]
[tex]\[ \sigma = \sqrt{200 \times 0.608 \times 0.392} \][/tex]
[tex]\[ \sigma = \sqrt{200 \times 0.238336} \][/tex]
[tex]\[ \sigma = \sqrt{47.6672} \][/tex]
[tex]\[ \sigma \approx 6.9 \][/tex]
Now, calculate the z-score for observing 142 murders by firearm:
[tex]\[ \text{z-score} = \frac{\text{observed value} - \text{mean}}{\text{standard deviation}} \][/tex]
[tex]\[ \text{z-score} = \frac{142 - 121.6}{6.9} \][/tex]
[tex]\[ \text{z-score} \approx \frac{20.4}{6.9} \][/tex]
[tex]\[ \text{z-score} \approx 2.95 \][/tex]
Next, we find the probability associated with this z-score (using standard normal distribution tables or a calculator). A z-score of 2.95 corresponds to a very high probability (close to 1). Specifically, it corresponds to a cumulative probability of approximately [tex]\( 0.9984 \)[/tex].
In other words, the probability that 142 or more murders are committed with a firearm, given our binomial distribution, is very high.
Conclusion:
Since the cumulative probability (0.9984) is greater than 0.95, and this indicates that the occurrence is very rare or extreme in the right tail of the distribution, it would indeed be unusual to observe 142 murders by firearm in a random sample of 200 murders.
Therefore, it would be unusual to observe 142 murders by firearm in a random sample of 200 murders. The high z-score and corresponding probability indicate that such an outcome is on the extreme end of the expected range of variation.
### Part (a)
Question: If 200 murders are randomly selected, how many would we expect to be committed with a firearm?
Solution:
1. Total number of murders: [tex]\( 200 \)[/tex]
2. Percentage of murders committed with a firearm: [tex]\( 60.8\% \)[/tex]
To find the expected number of murders committed with a firearm, we multiply the total number of murders by the percentage (in decimal form):
[tex]\[ \text{Expected number of murders with a firearm} = 200 \times \left(\frac{60.8}{100}\right) \][/tex]
[tex]\[ \text{Expected number of murders with a firearm} = 200 \times 0.608 \][/tex]
[tex]\[ \text{Expected number of murders with a firearm} = 121.6 \][/tex]
Therefore, we would expect approximately 121.6 murders to be committed with a firearm.
### Part (b)
Question: Would it be unusual to observe 142 murders by firearm in a random sample of 200 murders? Why?
Solution:
1. We previously calculated the expected number of murders with a firearm to be 121.6.
2. To determine if observing 142 murders by firearm is unusual, we need to consider the expected variability in the number of murders. We do this by calculating the z-score and the corresponding probability.
First, let’s find the mean and standard deviation for the binomial distribution:
- Mean (μ): We already know that the expected number of murders with a firearm (mean) is 121.6.
- Standard Deviation (σ): For a binomial distribution, the standard deviation can be calculated using the formula:
[tex]\[ \sigma = \sqrt{n \times p \times (1 - p)} \][/tex]
Where [tex]\( n \)[/tex] is the total number of trials (murders), and [tex]\( p \)[/tex] is the probability of success (percentage of murders committed with a firearm).
[tex]\[ \sigma = \sqrt{200 \times 0.608 \times (1 - 0.608)} \][/tex]
[tex]\[ \sigma = \sqrt{200 \times 0.608 \times 0.392} \][/tex]
[tex]\[ \sigma = \sqrt{200 \times 0.238336} \][/tex]
[tex]\[ \sigma = \sqrt{47.6672} \][/tex]
[tex]\[ \sigma \approx 6.9 \][/tex]
Now, calculate the z-score for observing 142 murders by firearm:
[tex]\[ \text{z-score} = \frac{\text{observed value} - \text{mean}}{\text{standard deviation}} \][/tex]
[tex]\[ \text{z-score} = \frac{142 - 121.6}{6.9} \][/tex]
[tex]\[ \text{z-score} \approx \frac{20.4}{6.9} \][/tex]
[tex]\[ \text{z-score} \approx 2.95 \][/tex]
Next, we find the probability associated with this z-score (using standard normal distribution tables or a calculator). A z-score of 2.95 corresponds to a very high probability (close to 1). Specifically, it corresponds to a cumulative probability of approximately [tex]\( 0.9984 \)[/tex].
In other words, the probability that 142 or more murders are committed with a firearm, given our binomial distribution, is very high.
Conclusion:
Since the cumulative probability (0.9984) is greater than 0.95, and this indicates that the occurrence is very rare or extreme in the right tail of the distribution, it would indeed be unusual to observe 142 murders by firearm in a random sample of 200 murders.
Therefore, it would be unusual to observe 142 murders by firearm in a random sample of 200 murders. The high z-score and corresponding probability indicate that such an outcome is on the extreme end of the expected range of variation.
