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In pharmacologic research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study. Suppose typical blood-glucose levels are normally distributed, with mean \(=90 \mathrm{mg} / \mathrm{dL}\) and standard deviation \(=38 \mathrm{mg} / \mathrm{dL}\). If the normal range is \(65-120\) mg/dL, then what percentage of values will fall in the normal range?

Short Answer

Expert verified
53.06% of the blood-glucose levels are within the normal range.

Step by step solution

01

Understand the Normal Distribution

Given that blood-glucose levels are normally distributed, we know the mean ( ext{mg/dL} 0) and the standard deviation ( 1 ext{mg/dL} 8). We need to find the percentage of values that fall within the normal range of 65-120 mg/dL.
02

Calculate the Z-scores

To find the percentage of values within the normal range, calculate the Z-scores for the limits of this range. The Z-score is given by: \[ Z = \frac{X - 0}{ 8} \]Calculated Z-scores:For 65 mg/dL: \[ Z_{65} = \frac{65 - 90}{38} = -0.66 \]For 120 mg/dL: \[ Z_{120} = \frac{120 - 90}{38} = 0.79 \]
03

Use the Z-table to Find Probabilities

Using a standard normal distribution table (Z-table), find the probabilities corresponding to these Z-scores. For \( Z = -0.66 \), the probability \( P(Z < -0.66) \approx 0.2546 \).For \( Z = 0.79 \), the probability \( P(Z < 0.79) \approx 0.7852 \).
04

Calculate the Probability Between the Two Z-scores

To find the percentage of blood-glucose levels within the normal range, subtract the probability of the lower Z-score from the probability of the upper Z-score:\[P(65 < X < 120) = P(Z < 0.79) - P(Z < -0.66)\]\[P(65 < X < 120) = 0.7852 - 0.2546 = 0.5306\]
05

Convert Probability to Percentage

Multiply the resulting probability by 100 to convert it to a percentage:\[0.5306 imes 100 = 53.06\%\]Thus, 53.06% of the blood-glucose levels lie within the normal range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
Normal distribution is a key concept in statistics that serves as the foundation for many types of data analysis. It is often referred to as a bell curve due to its characteristic shape.
When a dataset follows a normal distribution, most of the data points cluster around a central mean, with values tapering off symmetrically on both sides. This implies that:
  • About 68% of the data falls within one standard deviation of the mean.
  • About 95% falls within two standard deviations.
  • About 99.7% falls within three standard deviations.
In clinical settings, such as monitoring blood-glucose levels, understanding normal distribution allows researchers to assess what constitutes typical values and identify outliers that may indicate potential issues.
A "normal range" for a lab measurement is usually set by this distribution, which helps in making informed decisions and comparisons.
Z-score Calculation
The Z-score is a statistical measure used for standardizing individual data points within a distribution. This allows you to understand how far and in what direction a data point deviates from the mean, measured in units of standard deviation.
To calculate a Z-score, use the formula:\[ Z = \frac{X - \mu}{\sigma} \]Where:
  • \( X \) is the data point.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation of the dataset.
A Z-score can be positive or negative:
  • A positive Z-score indicates the data point is above the mean.
  • A negative Z-score indicates it is below the mean.
  • A Z-score of 0 means the data point is exactly at the mean.
Using Z-scores makes it easier to compare different data points from the same or different datasets, regardless of their original units or scales.
Probability
In statistics, probability is the measure that quantifies the likelihood that an event will occur. It ranges between 0 (an impossible event) and 1 (a certain event), but can also be expressed as a percentage.
When dealing with normally distributed data and Z-scores, probability helps us find the proportion of data points that lie within a specified range. Here's how:
  • Using Z-scores, you determine the position of a data point relative to the distribution.
  • Then, you use a Z-table to find the probability of that Z-score or lower (cumulative probability).
  • To find the probability between two Z-scores, you subtract the lower cumulative probability from the higher.
For example, if you have Z-scores of -0.66 and 0.79, their cumulative probabilities can be found using the Z-table. The difference between these cumulative probabilities gives the probability that a data point falls within that range.
Clinical Chemistry
Clinical chemistry is a branch of clinical pathology dealing with bodily fluids to diagnose and manage diseases. In a typical clinical chemistry lab, tests are carried out on blood, urine, and other body fluids to gain insight into a patient’s health status and detect any abnormalities.
Blood-glucose level testing, as highlighted in the exercise, is a common procedure. It helps in assessing metabolic functions and diagnosing conditions like diabetes. Blood-glucose levels, when normally distributed, can provide a baseline against which patient results are compared.
Overall, clinical chemistry makes use of many statistical methods, including normal distribution and probability calculations, to interpret test results accurately and reliably.
It plays a vital role in the modern healthcare system by helping clinicians make evidence-based decisions.

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Most popular questions from this chapter

In pharmacologic research a variety of clinical chemistry measurements are routinely monitored closely for evidence of side effects of the medication under study. Suppose typical blood-glucose levels are normally distributed, with mean \(=90 \mathrm{mg} / \mathrm{dL}\) and standard deviation \(=38 \mathrm{mg} / \mathrm{dL}\). Frequently, tests that yield abnormal results are repeated for confirmation. What is the probability that for a normal person a test will be at least 1.5 times as high as the upper limit of normal on two separate occasions?

Well-known racial differences in blood pressure exist between Caucasian and African American adults. These differences generally do not exist between Caucasian and African American children. Because aldosterone levels have been related to blood-pressure levels in adults in previous research, an investigation was performed to look at aldosterone levels among African American children and Caucasian children [2]. If the mean plasma-aldosterone level in Caucasian children is 400 pmol/L with standard deviation of 218 pmol/L, then what percentage of Caucasian children have levels \(\leq\) \(300 \mathrm{pmol} / \mathrm{L}\) if normality is assumed?

In the control group of the Diabetes Prevention Trial, the mean change in BMI was 0 units with a standard deviation of \(6 \mathrm{kg} / \mathrm{m}^{2}\). What is the probability that a random control group participant would lose at least 1 BMI unit over 24 months?

A study was conducted assessing the effect of observer training on measurement of blood pressure based on NHANES data from \(1999-2000\) (Ostchega, et al., [1] ). A goal was that the difference in recorded blood pressure between the observer and trainer be \(\leq 2 \mathrm{mm}\) Hg in absolute value. It was reported that the mean difference in systolic blood pressure (SBP) between observers and trainers (i.e., mean observer SBP minus trainer SBP) = mean (\Delta) was \(0.189 \mathrm{mm} \mathrm{Hg}\) with \(\mathrm{sd}=2.428 \mathrm{mm} \mathrm{Hg}\). If we assume that the distribution of \(\Delta\) is normally distributed, then what is the 90 percentile (i.e., the upper decile) and 10 percentile (i.e., the lower decile) of the distribution?

Physicians recommend that children with type-I (insulindependent) diabetes keep up with their insulin shots to minimize the chance of long-term complications. In addition, some diabetes researchers have observed that growth rate of weight during adolescence among diabetic patients is affected by level of compliance with insulin therapy. Suppose 12 -year-old type-l diabetic boys who comply with their insulin shots have a weight gain over 1 year that is normally distributed, with mean \(=12\) lbs. and variance \(=12\) lbs. What is the probability that compliant type-l diabetic 12-year-old boys will gain at least 15 lbs. over 1 year?

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