Q 97. - The Practice of Statistics for AP

Question

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: $5.6, 5.2, 4.6, 4.9, 5.7 \mathrm{and} 6.4$. A graph of only $6$ observations gives little information, so we proceed to compute the mean and standard deviation.

(a) Find the standard deviation from its definition. That is, find the deviations of each observation from the mean, square the deviations, and then obtain the variance and the standard deviation.

(b) Interpret the value of six you obtained in (a)

Step-by-Step Solution

Answer

Part (a) The standards deviation is $0.6419 mg/dl$ and the value of the mean is  $5.4mg/dl$

Part (b) The level of phosphate from the mean is$0.6419 mg/dI$

Part (a) Step 1: Given information

Given an array of data :

$5.6, 5.2, 4.6, 4.9, 5.7 \mathrm{and} 6.4$

Part (a) Step 2: Concept

The steps for calculating the standard deviation are as follows:

Calculate the average first.

Step 2: Determine each score's standard deviation.

Step 3: Calculate the square root of each standard deviation.

Step 4: Add up the squares to get the total.

Step 5: Work out how much of a difference there is.

Step 6: Determine the square root of the variance.

Part (a) Step 3: Calculation

${s }_x= \sqrt{0.412} \approx 0.6419mg/dl$The number of values in the sample size is:$n=6$

The mean is calculated by dividing the total number of values by the sample size: $\overline{x} =\frac{5.6+5.2+4.6+4.9+5.7+6.4}{6} = 5.4mg/dl$

If $xi$ is the data values, then $(xi - \overline{x})$ is the deviation ${(xi -\overline{ x})}^2$  the squared deviation:

The variance is calculated by dividing the sum of the squared deviations by $n-1$

${s_x}^2 =\frac{0.04+0.04+0.64+0.25+0.09+1}{6-1} = \frac{2.06}{5}= 0.412$

The square root of the variance is the standard deviation: $s_x =\sqrt{ 0.412} \approx 0.6419mg/dl$

Hence, the standard deviation is $0.6419mg/dl$

Part (b) Step 1: Explanation

The phosphate level typically varies from the mean by $0.6419 mg/dl$