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Sagot :
To determine which statement represents the percent of American adults with total cholesterol scores in the borderline-high range ([tex]$200-239 \text{ mg/dL}$[/tex]), we'll use the properties of the normal distribution. Here's a detailed, step-by-step solution:
### 1. Understanding the Problem
The problem gives us a normal distribution with the following parameters:
- The mean ([tex]\(\mu\)[/tex]) of the distribution: [tex]\(200 \text{ mg/dL}\)[/tex]
- The standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(30 \text{ mg/dL}\)[/tex]
The borderline-high range for cholesterol levels is from [tex]\(200 \text{ mg/dL}\)[/tex] to [tex]\(239 \text{ mg/dL}\)[/tex].
### 2. Standardization Using Z-Scores
The next step is to convert the given cholesterol levels to z-scores using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
For the limits of the borderline-high range:
- Lower limit: [tex]\(200 \text{ mg/dL}\)[/tex]
[tex]\[ z_{lower} = \frac{200 - 200}{30} = 0 \][/tex]
- Upper limit: [tex]\(239 \text{ mg/dL}\)[/tex]
[tex]\[ z_{upper} = \frac{239 - 200}{30} = 1.3 \][/tex]
Thus, the range [tex]\(200 \text{ mg/dL} \leq X \leq 239 \text{ mg/dL}\)[/tex] corresponds to the z-score range [tex]\(0 \leq z \leq 1.3\)[/tex].
### 3. Identifying the Statements
We are given four statements with different z-score ranges:
1. [tex]\(P(-1.3 \leq z \leq 0)\)[/tex]
- This range does not correspond to the borderline-high range.
2. [tex]\(P(-0.77 \leq z \leq 1)\)[/tex]
- This range does not correspond to the borderline-high range.
3. [tex]\(P(0 \leq z \leq 1.3)\)[/tex]
- This matches the z-score range for the borderline-high cholesterol levels.
4. [tex]\(P(0 \leq z \leq 0.77)\)[/tex]
- This range does not encompass the full range of borderline-high cholesterol levels.
### 4. Conclusion
By examining the z-score ranges, it is clear that the statement which represents the percent of American adults with total cholesterol scores in the borderline-high range ([tex]$200-239 \text{ mg/dL}$[/tex]) is:
[tex]\[P(0 \leq z \leq 1.3)\][/tex]
### 1. Understanding the Problem
The problem gives us a normal distribution with the following parameters:
- The mean ([tex]\(\mu\)[/tex]) of the distribution: [tex]\(200 \text{ mg/dL}\)[/tex]
- The standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(30 \text{ mg/dL}\)[/tex]
The borderline-high range for cholesterol levels is from [tex]\(200 \text{ mg/dL}\)[/tex] to [tex]\(239 \text{ mg/dL}\)[/tex].
### 2. Standardization Using Z-Scores
The next step is to convert the given cholesterol levels to z-scores using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
For the limits of the borderline-high range:
- Lower limit: [tex]\(200 \text{ mg/dL}\)[/tex]
[tex]\[ z_{lower} = \frac{200 - 200}{30} = 0 \][/tex]
- Upper limit: [tex]\(239 \text{ mg/dL}\)[/tex]
[tex]\[ z_{upper} = \frac{239 - 200}{30} = 1.3 \][/tex]
Thus, the range [tex]\(200 \text{ mg/dL} \leq X \leq 239 \text{ mg/dL}\)[/tex] corresponds to the z-score range [tex]\(0 \leq z \leq 1.3\)[/tex].
### 3. Identifying the Statements
We are given four statements with different z-score ranges:
1. [tex]\(P(-1.3 \leq z \leq 0)\)[/tex]
- This range does not correspond to the borderline-high range.
2. [tex]\(P(-0.77 \leq z \leq 1)\)[/tex]
- This range does not correspond to the borderline-high range.
3. [tex]\(P(0 \leq z \leq 1.3)\)[/tex]
- This matches the z-score range for the borderline-high cholesterol levels.
4. [tex]\(P(0 \leq z \leq 0.77)\)[/tex]
- This range does not encompass the full range of borderline-high cholesterol levels.
### 4. Conclusion
By examining the z-score ranges, it is clear that the statement which represents the percent of American adults with total cholesterol scores in the borderline-high range ([tex]$200-239 \text{ mg/dL}$[/tex]) is:
[tex]\[P(0 \leq z \leq 1.3)\][/tex]
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