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The level of various substances in the blood influences our health. Here are measurements of the level of phosphate in the blood of a patient, in milligrams of phosphate per deciliter of blood, made on 6 consecutive visits to a clinic: 5.6,5.2,4.6 , \(4.9,5.7,6.4 .\) A graph of only 6 observations gives little information, so we proceed to compute the mean and standard deviation.

Short Answer

Expert verified
Mean = 5.4 mg/dL; Standard Deviation ≈ 0.586 mg/dL.

Step by step solution

01

Understanding the Exercise

The exercise requires us to calculate the mean and standard deviation of phosphate levels given a set of six observations: 5.6, 5.2, 4.6, 4.9, 5.7, and 6.4 milligrams per deciliter.
02

Calculate the Mean

To find the mean, sum all the observations and divide by the number of observations. Mean = \( \frac{5.6+5.2+4.6+4.9+5.7+6.4}{6} \)Mean = \( \frac{32.4}{6} \)Mean = 5.4 milligrams per deciliter.
03

Calculate the Deviations

For each observation, find the deviation from the mean by subtracting the mean from each value:- Deviation for 5.6: \( 5.6 - 5.4 = 0.2 \)- Deviation for 5.2: \( 5.2 - 5.4 = -0.2 \)- Deviation for 4.6: \( 4.6 - 5.4 = -0.8 \)- Deviation for 4.9: \( 4.9 - 5.4 = -0.5 \)- Deviation for 5.7: \( 5.7 - 5.4 = 0.3 \)- Deviation for 6.4: \( 6.4 - 5.4 = 1.0 \)
04

Square the Deviations

Square each deviation found in the previous step:- \((0.2)^2 = 0.04\)- \((-0.2)^2 = 0.04\)- \((-0.8)^2 = 0.64\)- \((-0.5)^2 = 0.25\)- \((0.3)^2 = 0.09\)- \((1.0)^2 = 1.0\)
05

Calculate the Variance

Calculate the variance by finding the average of the squared deviations. Sum the squared deviations and divide by the number of observations:Variance = \( \frac{0.04 + 0.04 + 0.64 + 0.25 + 0.09 + 1.0}{6} \)Variance = \( \frac{2.06}{6} \)Variance = 0.3433 (approximated)
06

Calculate the Standard Deviation

The standard deviation is the square root of the variance:Standard Deviation = \( \sqrt{0.3433} \)Standard Deviation ≈ 0.586 milligrams per deciliter.

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Key Concepts

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

Mean Calculation
The mean, often called the average, is a fundamental statistical concept used to summarize a data set into a single value. In our example, we have blood phosphate levels measured over six visits: 5.6, 5.2, 4.6, 4.9, 5.7, and 6.4 mg/dL. To find the mean, we sum up these values and divide by the number of observations. This process is straightforward:
\[ \text{Mean} = \frac{5.6 + 5.2 + 4.6 + 4.9 + 5.7 + 6.4}{6} \]
Calculating this gives us a mean of 5.4 mg/dL.
  • The mean provides a central value for the data.
  • It helps us understand the typical phosphate level in the patient's blood.
Remember, the mean can be affected by very high or low values, or outliers. However, in a small data set like this, those extremes can skew the mean significantly.
Variance
Variance is another core concept of statistical analysis. It measures how far each number in the data set is from the mean, thus indicating the spread out nature of the data. To find the variance, we need to:
  • Calculate each observation's deviation from the mean (difference between the observation and the mean).
  • Square these deviations to ensure all values are positive.
  • Find the average of these squared deviations.
In our example:
The squared deviations are calculated as follows:\((0.2)^2 = 0.04, (-0.2)^2 = 0.04, (-0.8)^2 = 0.64, (-0.5)^2 = 0.25, (0.3)^2 = 0.09, (1.0)^2 = 1.0\).
Adding up these squared deviations, we get 2.06.
Thus, the variance is:
\[ \text{Variance} = \frac{2.06}{6} = 0.3433 \]
Variance tells us the degree to which the data points differ from the mean, helping to understand the variability in blood phosphate levels.
Statistical Analysis
Statistical analysis is a method of exploring datasets to derive insights. It employs techniques like mean and variance calculations to summarize and interpret the data. For the phosphate levels, statistical analysis involves:
  • Summarizing data through central measures like mean.
  • Assessing the spread or variability with variance and standard deviation.
  • Offering a numeric insight into variability, such as through standard deviation which is the square root of variance.
In this case, the standard deviation is:
\[ \text{Standard Deviation} = \sqrt{0.3433} \approx 0.586 \text{ mg/dL} \]
This tells us how spread out the phosphate levels are around the mean of 5.4 mg/dL.
Statistical analysis enables health professionals to assess whether the phosphate levels are within a normal range or if there is significant variation that requires further investigation.
Blood Phosphate Levels
Blood phosphate levels are crucial indicators of a person's health. Phosphate is key for vital functions, such as bone formation and energy storage. Measurements of these levels, like the six data points provided, assist medical professionals in diagnosing potential health issues.
- Consistent levels indicate stability and normal bodily function. - High or low levels may suggest metabolic or kidney disorders. Understanding phosphate measurement across several visits can reveal trends.
If levels are consistently high or low, or if they vary significantly during the visits, it could indicate an underlying health concern.
Being able to calculate mean and variance helps professionals to track changes over time, ensuring timely identification of potential issues.
This ensures patients receive appropriate and timely care, contributing to better health outcomes.

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Most popular questions from this chapter

The following partially complete two-way table shows the marginal distributions of gender and handedness for a sample of 100 high school students. $$ \begin{array}{lccc} \hline & \text { Male } & \text { Female } & \text { Total } \\ \text { Right } & x & & 90 \\ \text { Left } & & & 10 \\ \text { Total } & 40 & 60 & 100 \\ \hline \end{array} $$ If there is no association between gender and handedness for the members of the sample, which of the following is the correct value of \(x ?\) (a) 20 . (b) 30 . (c) 36 (d) 45 (e) Impossible to determine without more information.

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