Problem 27 - Algebra 2

Question

Jackson \(\qquad\) Washington \(\qquad\) King 170, 165, 140, 188, 195 \(\qquad\) 144, 177, 215, 225, 197 \(\qquad\) 166, 175, 196, 206, 219 Find the standard deviation of the weights for Jackson High.

Step-by-Step Solution

Answer

The standard deviation of the weights for Jackson High is approximately 21.58.

Step 1: Calculate the Mean

First, find the mean (average) of the weights for Jackson High. This involves summing all the weights and then dividing by the number of weights. The weights are 170, 165, 140, 188, 195.Calculate the sum: \(170 + 165 + 140 + 188 + 195 = 858\)The number of weights is 5, so the mean is: \[ \bar{x} = \frac{858}{5} = 171.6 \]

Step 2: Compute the Deviations from the Mean

For each weight, subtract the mean calculated in Step 1 to find the deviation. Deviations are:- \(170 - 171.6 = -1.6\)- \(165 - 171.6 = -6.6\)- \(140 - 171.6 = -31.6\)- \(188 - 171.6 = 16.4\)- \(195 - 171.6 = 23.4\)

Step 3: Square the Deviations

Square each of the deviations calculated in Step 2 to get rid of negative signs and emphasize larger deviations.Squared deviations are:- \((-1.6)^2 = 2.56\)- \((-6.6)^2 = 43.56\)- \((-31.6)^2 = 998.56\)- \((16.4)^2 = 268.96\)- \((23.4)^2 = 547.56\)

Step 4: Calculate the Variance

To find the variance, take the sum of the squared deviations from Step 3, and then divide by the number of weights minus one (n-1, for a sample standard deviation).Sum of squared deviations: \[2.56 + 43.56 + 998.56 + 268.96 + 547.56 = 1861.2\]Divide this sum by 4 (since there are 5 weights):\[s^2 = \frac{1861.2}{4} = 465.3\]

Step 5: Compute the Standard Deviation

The standard deviation is the square root of the variance obtained in Step 4.\[s = \sqrt{465.3} \approx 21.58\]

Key Concepts

Mean calculation

The first step in finding the standard deviation is to calculate the mean of the data set. The mean is essentially the average value and provides a central point around which we measure the spread of the data. To find the mean, you sum up all the data points and divide by the number of data points. For Jackson High, the weights provided are 170, 165, 140, 188, and 195.
- Firstly, add these numbers to find the total: \( 170 + 165 + 140 + 188 + 195 = 858 \) - Then, divide by the number of weights, which is 5: \( \bar{x} = \frac{858}{5} = 171.6 \) This result means that the average weight is 171.6. This value helps us understand the overall weight distribution of Jackson High.

Variance

Once we have the mean, the next step is to compute the variance, which describes how much the data points deviate from the mean on average. Unlike range, which simply takes the difference between the highest and lowest values, variance considers all data points. To calculate the variance, we need the squared deviations, which we will discuss next.
Variance for a sample is calculated by dividing the sum of squared deviations by the number of data points minus one. This adjustment prevents underestimation and provides a more accurate measure of spread. The formula used is: - \( s^2 = \frac{\text{sum of squared deviations}}{n-1} \) For Jackson High, after calculating the squared deviations, the sum was 1861.2, and there were 5 data points, so: - \( s^2 = \frac{1861.2}{4} = 465.3 \) This variance shows the average squared deviation from the mean, providing insight into the data's dispersion.

Squared deviations

Squared deviations play a crucial role in finding the variance and standard deviation. You calculate each deviation by subtracting the mean from each data point.
- If a weight is 170 and the mean is 171.6, the deviation is: \( 170 - 171.6 = -1.6 \) We square each deviation to make all values positive, enhancing the impact of larger deviations and allowing for more significant analysis.
- For instance, the squared deviation of -1.6 would be: \( (-1.6)^2 = 2.56 \) Perform this operation for each data point, and sum these squared deviations to get the total, which feeds into our variance calculation. This process is integral to understanding how each weight deviates from the mean over the entire dataset.

Sample standard deviation

The sample standard deviation provides a measure of variability within a dataset. After calculating the variance, the standard deviation offers a more intuitive number that expresses this dispersion since it uses the same units as the data itself.
- With the variance already determined as 465.3 for Jackson High, to find the sample standard deviation, take the square root of the variance: \( s = \sqrt{465.3} \approx 21.58 \) The result, approximately 21.58, tells us the average amount each weight differs from the mean weight. Unlike the variance, the standard deviation places this back into the original context of the data — in this case, pounds. This is why standard deviation is often more clear to interpret and discuss.