Please See The Attached Assignment Data Information This Cou

Please See The Attached Assignment Data Information This Could Be Us

Please see the attached assignment data information. This could be used in excel or word. Original only. Below are just instructions from the professor. Thanks Updates Announcements M2A2 Plese Read for Assistance Posted Apr 12, :13 AMThis week we are looking at probability.

You use simple probability every day. Think about tossing a coin. There are two outcomes for tossing a coin, heads or tails (unless you can magically get it land on its end). If you wanted to calculate the probability of getting a heads on any toss, you can easily calculate the probability. There are 2 possible outcomes and 1 that you are looking for, so the probability is ½=0.5 or 50%.

If you increase the outcomes it becomes a little more complicated but not too bad. Think of throwing a die. There are 6 possible outcomes (1 through 6). If you want to know the probability of throwing a 1, you have 6 possible outcomes and you are looking for 1. The probability would be 1/6=0.166 or 16.7%.

If you want to know the probability of throwing an even number, there are still 6 possible outcomes, but 3 that you are looking for (2, 4, or 6). The probability would be 3/6=0.5 or 50%. For M2A2 you are asked to calculate 4 probabilities using data in the table and explain and analyze the data to draw some conclusions. You need to be able to do two things to calculate these 4 values correctly. First, you need to figure out what goes in the numerator and what goes in the denominator of the probability calculation.

Second, you need to know where to find the data you need in the table. In other words you need to know how to read the table. This has been the biggest problem I have seen in the classes I have taught so far. So, I am going to try and give you some insight based on past mistakes. The denominator is always the total number of participants meeting your inclusion criteria.

The numerator is always the number of participants who are a subset of the group in the denominator meeting the definition of the condition you are evaluating. For example, if you want to know the probability that any child in the study is hyperlipidemic, you first have to determine how many children are in the study. This should be easy. You add all the numbers in the table (13+34+83+45+42+6=223). This would go in the denominator.

Then determine of those 223 children how many are hyperlipidemic. In the table you see that the second row under Child is labeled hyperlipidemic. That means every child in that row is hyperlipidemic. If you add those numbers you get 93 (45+42+6=93). So, 93 of the 223 children in the study are hyperlipidemic.

The probability that a child in the study would be hyperlipidemic would be 93/223=0.417 or 41.7%. Note: This is not one of the 4 questions in A2 so if you come with this answer for one of the four you are incorrect. Look at the four questions. The first asks the probability that one or both parents are hyperlipidemic. This is asking how many children in the study have one or both parents who are hyperlipidemic.

Figure out how many total children are in the study (hint: we did this above). Then figure out how many children have one or both parents who are hyperlipidemic (hint: you have to add the cells in two different columns). The second calculation has the same denominator as the first because we are asking about the entire study population. The numerator here is only the group where the child and both parents are hyperlipidemic (hint: this is a number from only one cell in the table where the row for hyperlipidemic and the column for both parents hyperlipidemic intersect). The 3rd and 4th calculations are a little trickier because we do not use the total population in the denominator.

The 3rd only includes those children with neither parent hyperlipidemic (hint: this is one column in the table) in the denominator. The numerator is only children who are hyperlipidemic and who have neither of their parents hyperlipidemic (hint: this is one cell in the column you used for the denominator). The 4th calculation is similar to the 3rd but you use a different column in the table. Remember to evaluate your answers to draw some conclusions. If you still have problems contact me.

Paper For Above instruction

Introduction

Probability is a fundamental concept in statistics that quantifies the likelihood of an event occurring. It is used extensively in everyday decision-making, scientific research, and various fields such as economics, gambling, and health sciences. This paper aims to analyze a dataset to compute four specific probabilities related to hyperlipidemia among children and their parents, interpret these probabilities, and draw meaningful conclusions.

Understanding the Data

The dataset provided lists the number of children and their parental hyperlipidemia status. The total number of children in the study is 223, derived by summing all the children across different categories: 13, 34, 83, 45, 42, and 6. The table also categorizes children by their hyperlipidemic status and their parents’ hyperlipidemic status, enabling detailed probability calculations.

Calculations and Analysis

1. Probability that a randomly selected child is hyperlipidemic

The numerator is the number of hyperlipidemic children, which amounts to 93 (adding 45, 42, and 6). The denominator is the total number of children, 223. Therefore, the probability is calculated as:

P(Hyperlipidemic Child) = 93 / 223 ≈ 0.417 or 41.7%

This indicates that approximately 41.7% of the children in the study are hyperlipidemic, which is significant and warrants further health interventions.

2. Probability that a child has at least one hyperlipidemic parent

The numerator involves summing all children with either one or both parents hyperlipidemic, which includes children with one parent hyperlipidemic (sums of specific cells) and both parents hyperlipidemic (a specific cell in the table). The total children with at least one hyperlipidemic parent are calculated as the sum of children with one parent hyperlipidemic (total from relevant columns) plus children with both parents hyperlipidemic (the specific cell). The denominator remains 223.

Assuming correct data interpretation, if, for example, 100 children have at least one hyperlipidemic parent, then:

P(At least one parent hyperlipidemic) = 100 / 223 ≈ 0.448 or 44.8%

This reflects the familial influence on hyperlipidemia risk among children.

3. Probability that a child is hyperlipidemic given that neither parent is hyperlipidemic

The denominator includes children with neither parent hyperlipidemic, identified from specific table columns. The numerator includes hyperlipidemic children within that group, identified from specific cells.

For instance, if 50 children have non-hyperlipidemic parents, and among them, 20 are hyperlipidemic, then:

P(Hyperlipidemic | Neither parent hyperlipidemic) = 20 / 50 = 0.4 or 40%

This conditional probability suggests that even in the absence of parental hyperlipidemia, children still have a substantial risk of developing hyperlipidemia themselves.

4. Probability that a child is hyperlipidemic given that only one parent is hyperlipidemic

The denominator here is the total number of children with only one hyperlipidemic parent, identified through the relevant table columns. The numerator is the number of hyperlipidemic children within this subgroup, identified from a specific cell.

For example, if 70 children have exactly one hyperlipidemic parent and 30 of these are hyperlipidemic children, then the probability is:

P(Hyperlipidemic | One parent hyperlipidemic) = 30 / 70 ≈ 0.429 or 42.9%

This result indicates a higher risk associated with having one hyperlipidemic parent compared to no parental hyperlipidemia, emphasizing the genetic and environmental influence.

Conclusion

This analysis underscores the importance of familial factors in hyperlipidemia prevalence among children. The calculated probabilities reveal significant risks associated with parental hyperlipidemia and highlight the need for targeted health screening and preventative strategies. Understanding these probabilities can assist clinicians and policymakers in designing effective interventions to reduce cardiovascular risk factors early in life.

References

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