We appreciate your visit to According to a report tex 79 7 tex of murders are committed with a firearm a If 200 murders are randomly selected how many would. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
- Calculate the expected number of murders committed with a firearm: $\mu = 200 \times 0.797 = 159.4$.
- Calculate the standard deviation: $\sigma = \sqrt{200 \times 0.797 \times 0.203} \approx 5.688$.
- Calculate the interval $\mu \pm 2\sigma$: $159.4 \pm 2(5.688) = [148.024, 170.776]$.
- Since 173 is greater than $170.776$, it is unusual. The answer is $\boxed{159.4}$ and A.
### Explanation
1. Understand the problem
We are given that $79.7\%$ of murders are committed with a firearm. We are asked to find the expected number of murders committed with a firearm in a random sample of 200 murders, and to determine if observing 173 murders by firearm is unusual.
2. Define variables
Let $p = 0.797$ be the probability of a murder being committed with a firearm. Let $n = 200$ be the sample size. Let $X$ be the number of murders committed with a firearm in the sample. $X$ follows a binomial distribution with parameters $n$ and $p$.
3. Calculate the expected value
The expected number of murders committed with a firearm is $\mu = np = 200 \times 0.797 = 159.4$.
4. Calculate the standard deviation
The standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.797 \times (1-0.797)} = \sqrt{200 \times 0.797 \times 0.203} = \sqrt{32.3362} \approx 5.688$.
5. Calculate the interval
To determine if observing 173 murders by firearm is unusual, we calculate the interval $\mu \pm 2\sigma = 159.4 \pm 2(5.688) = 159.4 \pm 11.376$. The lower bound is $159.4 - 11.376 = 148.024$ and the upper bound is $159.4 + 11.376 = 170.776$.
6. Determine if 173 is unusual
Since 173 is greater than 170.776, 173 is greater than $\mu + 2\sigma$. Therefore, it would be unusual to observe 173 murders by firearm in a random sample of 200 murders.
7. Final Answer
The expected number of murders committed with a firearm is $159.4$. It would be unusual to observe 173 murders by firearm because 173 is greater than $\mu + 2\sigma$.
### Examples
This type of calculation is useful in criminology and public health to understand trends in violent crime and to assess whether specific interventions are having an impact. For example, if a city implements a gun control policy, one could analyze whether the observed number of firearm murders decreases significantly compared to the expected number based on historical data. This helps policymakers evaluate the effectiveness of their strategies.
- Calculate the standard deviation: $\sigma = \sqrt{200 \times 0.797 \times 0.203} \approx 5.688$.
- Calculate the interval $\mu \pm 2\sigma$: $159.4 \pm 2(5.688) = [148.024, 170.776]$.
- Since 173 is greater than $170.776$, it is unusual. The answer is $\boxed{159.4}$ and A.
### Explanation
1. Understand the problem
We are given that $79.7\%$ of murders are committed with a firearm. We are asked to find the expected number of murders committed with a firearm in a random sample of 200 murders, and to determine if observing 173 murders by firearm is unusual.
2. Define variables
Let $p = 0.797$ be the probability of a murder being committed with a firearm. Let $n = 200$ be the sample size. Let $X$ be the number of murders committed with a firearm in the sample. $X$ follows a binomial distribution with parameters $n$ and $p$.
3. Calculate the expected value
The expected number of murders committed with a firearm is $\mu = np = 200 \times 0.797 = 159.4$.
4. Calculate the standard deviation
The standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.797 \times (1-0.797)} = \sqrt{200 \times 0.797 \times 0.203} = \sqrt{32.3362} \approx 5.688$.
5. Calculate the interval
To determine if observing 173 murders by firearm is unusual, we calculate the interval $\mu \pm 2\sigma = 159.4 \pm 2(5.688) = 159.4 \pm 11.376$. The lower bound is $159.4 - 11.376 = 148.024$ and the upper bound is $159.4 + 11.376 = 170.776$.
6. Determine if 173 is unusual
Since 173 is greater than 170.776, 173 is greater than $\mu + 2\sigma$. Therefore, it would be unusual to observe 173 murders by firearm in a random sample of 200 murders.
7. Final Answer
The expected number of murders committed with a firearm is $159.4$. It would be unusual to observe 173 murders by firearm because 173 is greater than $\mu + 2\sigma$.
### Examples
This type of calculation is useful in criminology and public health to understand trends in violent crime and to assess whether specific interventions are having an impact. For example, if a city implements a gun control policy, one could analyze whether the observed number of firearm murders decreases significantly compared to the expected number based on historical data. This helps policymakers evaluate the effectiveness of their strategies.
Thanks for taking the time to read According to a report tex 79 7 tex of murders are committed with a firearm a If 200 murders are randomly selected how many would. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
