The number of cars sold at a dealership over several weeks is given below:
$14, 23, 31, 29, 33$
What is the standard deviation for this set of population data?
Standard deviation: $\sigma=\sqrt{\frac{\left(x_1-\mu\right)^2+\left(x_2-\mu\right)^2+\ldots+\left(x_N-\mu\right)^2}{N}}$
Answer (2)
Calculate the mean of the data set: μ = 5 14 + 23 + 31 + 29 + 33 = 26 .
Calculate the squared differences from the mean: ( 14 − 26 ) 2 = 144 , ( 23 − 26 ) 2 = 9 , ( 31 − 26 ) 2 = 25 , ( 29 − 26 ) 2 = 9 , ( 33 − 26 ) 2 = 49 .
Calculate the variance: 5 144 + 9 + 25 + 9 + 49 = 47.2 .
Calculate the standard deviation: σ = 47.2 ≈ 6.9 .
Explanation
Understand the problem and provided data We are given the number of cars sold at a dealership over several weeks: 14 , 23 , 31 , 29 , 33 . We need to find the standard deviation for this population data using the formula: σ = N ( x 1 − μ ) 2 + ( x 2 − μ ) 2 + … + ( x N − μ ) 2 where x i represents each data point, μ is the mean of the data, and N is the number of data points.
Calculate the mean First, we calculate the mean ( μ ) of the data set. The mean is the sum of all the data points divided by the number of data points: μ = 5 14 + 23 + 31 + 29 + 33 = 5 130 = 26
Calculate squared differences Next, we calculate the squared differences from the mean for each data point. This means we subtract the mean from each data point and then square the result:
( 14 − 26 ) 2 = ( − 12 ) 2 = 144
( 23 − 26 ) 2 = ( − 3 ) 2 = 9
( 31 − 26 ) 2 = ( 5 ) 2 = 25
( 29 − 26 ) 2 = ( 3 ) 2 = 9
( 33 − 26 ) 2 = ( 7 ) 2 = 49
Sum the squared differences Now, we sum the squared differences: i = 1 ∑ 5 ( x i − μ ) 2 = 144 + 9 + 25 + 9 + 49 = 236
Calculate the variance We divide the sum of squared differences by the number of data points ( N = 5 ) to find the variance: 5 ∑ i = 1 5 ( x i − μ ) 2 = 5 236 = 47.2
Calculate the standard deviation Finally, we take the square root of the variance to find the standard deviation: σ = 5 ∑ i = 1 5 ( x i − μ ) 2 = 47.2 ≈ 6.87
State the final answer The standard deviation for this set of population data is approximately 6.87 . Comparing this to the given options, the closest value is 6.9 .
Examples
Understanding standard deviation is very useful in finance. For example, if you are analyzing the stock prices of a company over a certain period, calculating the standard deviation can help you understand the volatility of the stock. A higher standard deviation indicates that the stock price is more volatile and thus riskier. This information can help investors make informed decisions about whether to invest in the stock.
The standard deviation of the number of cars sold at a dealership, given the data points 14 , 23 , 31 , 29 , 33 , is approximately 6.87 , which can be rounded to 6.9 . This was calculated by first finding the mean, then determining the squared differences from the mean, summing those, calculating the variance, and finally taking the square root to find the standard deviation.
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