Esercizi sulla deviazione standard

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Esercizi sulla deviazione standard

Teoria della Deviazione Standard

La deviazione standard è una misura della dispersione o variabilità di un insieme di dati. Indica quanto i valori di un insieme si discostano dalla media. Una deviazione standard bassa indica che i dati sono vicini alla media, mentre una deviazione standard alta indica che i dati sono più sparsi.

Formula della Deviazione Standard

Per un insieme di dati campionari, la deviazione standard è calcolata come segue:

  1. Calcola la media (μ):

    \mu = \frac{1}{n} \sum_{i=1}^{n} x_i
    μ=1n∑i=1nxi\mu = \frac{1}{n} \sum_{i=1}^{n} x_i

    dove nnn è il numero totale di osservazioni e x_ixix_i sono i valori.

  2. Calcola la varianza (σ²):

    \sigma^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \mu)^2
    σ2=1n−1∑i=1n(xi−μ)2\sigma^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \mu)^2

  3. Calcola la deviazione standard (σ):

    \sigma = \sqrt{\sigma^2}
    σ=σ2\sigma = \sqrt{\sigma^2}

Esercizi sulla Deviazione Standard

Esercizio 1: Calcolo della Deviazione Standard

Considera il seguente insieme di dati:

\{4, 8, 6, 5, 3\}
{4,8,6,5,3}\{4, 8, 6, 5, 3\}

Svolgimento:

  1. Calcola la media:

    \mu = \frac{4 + 8 + 6 + 5 + 3}{5} = \frac{26}{5} = 5.2
    μ=4+8+6+5+35=265=5.2\mu = \frac{4 + 8 + 6 + 5 + 3}{5} = \frac{26}{5} = 5.2

  2. Calcola la varianza:

    \begin{align*}
\sigma^2 & = \frac{1}{5-1} [(4 - 5.2)^2 + (8 - 5.2)^2 + (6 - 5.2)^2 + (5 - 5.2)^2 + (3 - 5.2)^2] \\
& = \frac{1}{4} [(-1.2)^2 + (2.8)^2 + (0.8)^2 + (-0.2)^2 + (-2.2)^2] \\
& = \frac{1}{4} [1.44 + 7.84 + 0.64 + 0.04 + 4.84] \\
& = \frac{1}{4} [14.8] = 3.7
\end{align*}
    σ2=15−1[(4−5.2)2+(8−5.2)2+(6−5.2)2+(5−5.2)2+(3−5.2)2]=14[(−1.2)2+(2.8)2+(0.8)2+(−0.2)2+(−2.2)2]=14[1.44+7.84+0.64+0.04+4.84]=14[14.8]=3.7\begin{align*} \sigma^2 & = \frac{1}{5-1} [(4 - 5.2)^2 + (8 - 5.2)^2 + (6 - 5.2)^2 + (5 - 5.2)^2 + (3 - 5.2)^2] \\ & = \frac{1}{4} [(-1.2)^2 + (2.8)^2 + (0.8)^2 + (-0.2)^2 + (-2.2)^2] \\ & = \frac{1}{4} [1.44 + 7.84 + 0.64 + 0.04 + 4.84] \\ & = \frac{1}{4} [14.8] = 3.7 \end{align*}

  3. Calcola la deviazione standard:

    \sigma = \sqrt{3.7} \approx 1.923
    σ=3.7≈1.923\sigma = \sqrt{3.7} \approx 1.923

Risultato:

La deviazione standard dell’insieme di dati è circa σ ≈ 1.923σ≈1.923σ ≈ 1.923.

Esercizio 2: Calcolo della Deviazione Standard con Valori Negativi

Considera il seguente insieme di dati:

\{-3, -1, -4, -2, -5\}
{−3,−1,−4,−2,−5}\{-3, -1, -4, -2, -5\}

Svolgimento:

  1. Calcola la media:

    \mu = \frac{-3 + (-1) + (-4) + (-2) + (-5)}{5} = \frac{-15}{5} = -3
    μ=−3+(−1)+(−4)+(−2)+(−5)5=−155=−3\mu = \frac{-3 + (-1) + (-4) + (-2) + (-5)}{5} = \frac{-15}{5} = -3

  2. Calcola la varianza:

    \begin{align*}
\sigma^2 & = \frac{1}{5-1} [(-3 - (-3))^2 + (-1 - (-3))^2 + (-4 - (-3))^2 + (-2 - (-3))^2 + (-5 - (-3))^2] \\
& = \frac{1}{4} [(0)^2 + (2)^2 + (-1)^2 + (1)^2 + (-2)^2] \\
& = \frac{1}{4} [0 + 4 + 1 + 1 + 4] = \frac{10}{4} = 2.5
\end{align*}
    σ2=15−1[(−3−(−3))2+(−1−(−3))2+(−4−(−3))2+(−2−(−3))2+(−5−(−3))2]=14[(0)2+(2)2+(−1)2+(1)2+(−2)2]=14[0+4+1+1+4]=104=2.5\begin{align*} \sigma^2 & = \frac{1}{5-1} [(-3 - (-3))^2 + (-1 - (-3))^2 + (-4 - (-3))^2 + (-2 - (-3))^2 + (-5 - (-3))^2] \\ & = \frac{1}{4} [(0)^2 + (2)^2 + (-1)^2 + (1)^2 + (-2)^2] \\ & = \frac{1}{4} [0 + 4 + 1 + 1 + 4] = \frac{10}{4} = 2.5 \end{align*}

  3. Calcola la deviazione standard:

    σ = \sqrt{2.5} ≈ 1.581
    σ=2.5≈1.581σ = \sqrt{2.5} ≈ 1.581

Risultato:

La deviazione standard dell’insieme di dati è circa σ ≈ 1.581σ≈1.581σ ≈ 1.581.

Esercizio 3: Calcolo della Deviazione Standard con Valori Decimali

Considera il seguente insieme di dati:

\{10.5, 11.0, 9.8, 10.0, 10.7\}
{10.5,11.0,9.8,10.0,10.7}\{10.5, 11.0, 9.8, 10.0, 10.7\}

Svolgimento:

  1. Calcola la media:

    μ = \frac{10.5 + 11.0 + 9.8 + 10.0 + 10.7}{5} = \frac{52}{5} = 10.4
    μ=10.5+11.0+9.8+10.0+10.75=525=10.4μ = \frac{10.5 + 11.0 + 9.8 + 10.0 + 10.7}{5} = \frac{52}{5} = 10.4

  2. Calcola la varianza:

    \begin{align*}
σ^2 & = \frac{1}{5-1} [(10.5 - 10.4)^2 + (11.0 - 10.4)^2 + (9.8 - 10.4)^2 + (10.0 - 10.4)^2 + (10.7 - 10.4)^2] \\ 
    & = \frac{1}{4} [(0.1)^2 + (0.6)^2 + (-0.6)^2 + (-0.4)^2 + (0.3)^2] \\ 
    & = \frac{1}{4} [0.01 + 0.36 + 0.36 + 0.16 + 0.09] \\ 
    & = \frac{0.98}{4} ≈ 0.245
\end{align*}
    σ2=15−1[(10.5−10.4)2+(11.0−10.4)2+(9.8−10.4)2+(10.0−10.4)2+(10.7−10.4)2]=14[(0.1)2+(0.6)2+(−0.6)2+(−0.4)2+(0.3)2]=14[0.01+0.36+0.36+0.16+0.09]=0.984≈0.245\begin{align*} σ^2 & = \frac{1}{5-1} [(10.5 - 10.4)^2 + (11.0 - 10.4)^2 + (9.8 - 10.4)^2 + (10.0 - 10.4)^2 + (10.7 - 10.4)^2] \\ & = \frac{1}{4} [(0.1)^2 + (0.6)^2 + (-0.6)^2 + (-0.4)^2 + (0.3)^2] \\ & = \frac{1}{4} [0.01 + 0.36 + 0.36 + 0.16 + 0.09] \\ & = \frac{0.98}{4} ≈ 0.245 \end{align*}

  3. Calcola la deviazione standard:

    σ ≈ √0.245 ≈ 0,495.
    σ≈√0.245≈0,495.σ ≈ √0.245 ≈ 0,495.

Risultato:

La deviazione standard dell’insieme di dati è circa σ ≈ 0,495σ≈0,495σ ≈ 0,495.

English version

Standard Deviation Exercises

Standard Deviation Theory

Standard deviation is a measure of the dispersion or variability of a data set. It indicates how much the values ​​in a set differ from the mean. A low standard deviation indicates that the data are close to the mean, while a high standard deviation indicates that the data are more spread out.

Standard Deviation Formula

For a sample data set, the standard deviation is calculated as follows:

  1. Calculate the mean (μ):
\mu = \frac{1}{n} \sum_{i=1}^{n} x_i
μ=1n∑i=1nxi\mu = \frac{1}{n} \sum_{i=1}^{n} x_i

where nnn is the total number of observations and x_ixix_i are the values.

  1. Calculate the variance (σ²):
\sigma^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \mu)^2
σ2=1n−1∑i=1n(xi−μ)2\sigma^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \mu)^2
  1. Calculate the standard deviation (σ):
\sigma = \sqrt{\sigma^2}
σ=σ2\sigma = \sqrt{\sigma^2}

Standard Deviation Exercises

Exercise 1: Calculating the Standard Deviation

Consider the following set of data:

\{4, 8, 6, 5, 3\}
{4,8,6,5,3}\{4, 8, 6, 5, 3\}

Procedure:

  1. Calculate the mean:
\mu = \frac{4 + 8 + 6 + 5 + 3}{5} = \frac{26}{5} = 5.2
μ=4+8+6+5+35=265=5.2\mu = \frac{4 + 8 + 6 + 5 + 3}{5} = \frac{26}{5} = 5.2
  1. Calculate the variance:
\begin{align*}
\sigma^2 & = \frac{1}{5-1} [(4 - 5.2)^2 + (8 - 5.2)^2 + (6 - 5.2)^2 + (5 - 5.2)^2 + (3 - 5.2)^2] \\
& = \frac{1}{4} [(-1.2)^2 + (2.8)^2 + (0.8)^2 + (-0.2)^2 + (-2.2)^2] \\
& = \frac{1}{4} [1.44 + 7.84 + 0.64 + 0.04 + 4.84] \\
& = \frac{1}{4} [14.8] = 3.7
\end{align*}
σ2=15−1[(4−5.2)2+(8−5.2)2+(6−5.2)2+(5−5.2)2+(3−5.2)2]=14[(−1.2)2+(2.8)2+(0.8)2+(−0.2)2+(−2.2)2]=14[1.44+7.84+0.64+0.04+4.84]=14[14.8]=3.7\begin{align*} \sigma^2 & = \frac{1}{5-1} [(4 - 5.2)^2 + (8 - 5.2)^2 + (6 - 5.2)^2 + (5 - 5.2)^2 + (3 - 5.2)^2] \\ & = \frac{1}{4} [(-1.2)^2 + (2.8)^2 + (0.8)^2 + (-0.2)^2 + (-2.2)^2] \\ & = \frac{1}{4} [1.44 + 7.84 + 0.64 + 0.04 + 4.84] \\ & = \frac{1}{4} [14.8] = 3.7 \end{align*}
  1. Calculate the standard deviation:
\sigma = \sqrt{3.7} \approx 1.923
σ=3.7≈1.923\sigma = \sqrt{3.7} \approx 1.923

Result:

The standard deviation of the data set is approximately σ ≈ 1.923σ≈1.923σ ≈ 1.923.

Exercise 2: Calculating the Standard Deviation with Negative Values

Consider the following data set:

\{-3, -1, -4, -2, -5\}
{−3,−1,−4,−2,−5}\{-3, -1, -4, -2, -5\}

Procedure:

  1. Calculate the mean:
\mu = \frac{-3 + (-1) + (-4) + (-2) + (-5)}{5} = \frac{-15}{5} = -3
μ=−3+(−1)+(−4)+(−2)+(−5)5=−155=−3\mu = \frac{-3 + (-1) + (-4) + (-2) + (-5)}{5} = \frac{-15}{5} = -3
  1. Calculate the variance:
\begin{align*}
\sigma^2 & = \frac{1}{5-1} [(-3 - (-3))^2 + (-1 - (-3))^2 + (-4 - (-3))^2 + (-2 - (-3))^2 + (-5 - (-3))^2] \\
& = \frac{1}{4} [(0)^2 + (2)^2 + (-1)^2 + (1)^2 + (-2)^2] \\
& = \frac{1}{4} [0 + 4 + 1 + 1 + 4] = \frac{10}{4} = 2.5
\end{align*}
σ2=15−1[(−3−(−3))2+(−1−(−3))2+(−4−(−3))2+(−2−(−3))2+(−5−(−3))2]=14[(0)2+(2)2+(−1)2+(1)2+(−2)2]=14[0+4+1+1+4]=104=2.5\begin{align*} \sigma^2 & = \frac{1}{5-1} [(-3 - (-3))^2 + (-1 - (-3))^2 + (-4 - (-3))^2 + (-2 - (-3))^2 + (-5 - (-3))^2] \\ & = \frac{1}{4} [(0)^2 + (2)^2 + (-1)^2 + (1)^2 + (-2)^2] \\ & = \frac{1}{4} [0 + 4 + 1 + 1 + 4] = \frac{10}{4} = 2.5 \end{align*}
  1. Calculate the standard deviation:
σ = \sqrt{2.5} ≈ 1.581
σ=2.5≈1.581σ = \sqrt{2.5} ≈ 1.581

Result:

The standard deviation of the data set is approximately σ ≈ 1.581σ≈1.581σ ≈ 1.581.

Exercise 3: Calculating the Standard Deviation with Decimal Values

Consider the following data set:

\{10.5, 11.0, 9.8, 10.0, 10.7\}
{10.5,11.0,9.8,10.0,10.7}\{10.5, 11.0, 9.8, 10.0, 10.7\}

Procedure:

  1. Calculate the mean:
μ = \frac{10.5 + 11.0 + 9.8 + 10.0 + 10.7}{5} = \frac{52}{5} = 10.4
μ=10.5+11.0+9.8+10.0+10.75=525=10.4μ = \frac{10.5 + 11.0 + 9.8 + 10.0 + 10.7}{5} = \frac{52}{5} = 10.4
  1. Calculate the variance:
\begin{align*}
σ^2 & = \frac{1}{5-1} [(10.5 - 10.4)^2 + (11.0 - 10.4)^2 + (9.8 - 10.4)^2 + (10.0 - 10.4)^2 + (10.7 - 10.4)^2] \\
& = \frac{1}{4} [(0.1)^2 + (0.6)^2 + (-0.6)^2 + (-0.4)^2 + (0.3)^2] \\
& = \frac{1}{4} [0.01 + 0.36 + 0.36 + 0.16 + 0.09] \\
& = \frac{0.98}{4} ≈ 0.245
\end{align*}
σ2=15−1[(10.5−10.4)2+(11.0−10.4)2+(9.8−10.4)2+(10.0−10.4)2+(10.7−10.4)2]=14[(0.1)2+(0.6)2+(−0.6)2+(−0.4)2+(0.3)2]=14[0.01+0.36+0.36+0.16+0.09]=0.984≈0.245\begin{align*} σ^2 & = \frac{1}{5-1} [(10.5 - 10.4)^2 + (11.0 - 10.4)^2 + (9.8 - 10.4)^2 + (10.0 - 10.4)^2 + (10.7 - 10.4)^2] \\ & = \frac{1}{4} [(0.1)^2 + (0.6)^2 + (-0.6)^2 + (-0.4)^2 + (0.3)^2] \\ & = \frac{1}{4} [0.01 + 0.36 + 0.36 + 0.16 + 0.09] \\ & = \frac{0.98}{4} ≈ 0.245 \end{align*}
  1. Calculate the standard deviation:
σ ≈ √0.245 ≈ 0.495.
σ≈√0.245≈0.495.σ ≈ √0.245 ≈ 0.495.

Result:

The standard deviation of the data set is approximately σ ≈ 0.495σ≈0.495σ ≈ 0.495.

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