Math / Data & Statistics

QuestionTaller: formativo. Est 215 sem 22024 Resuelta los siguientes problemas aplicando lo aprendido para elaborar una tabla de distribución de frecuencias. Y además realice el calculo de la media y la desviación típica.
Problema 1. En un problema similar al del ejercicio 11, es necesario garantizar que la resistencia mínima que tiene un envase de plástico en posición vertical sea de 20 kg. Para evaluar esto se han obtenido los siguientes datos mediante pruebas destructivas: $\begin{array}{l} 28.3 \quad 26.8 \quad 26.6 \quad 26.5 \quad 28.124 .827 .426 .2 \quad 29.428 .624 .925 .230 .427 .727 .026 .1 \\ 28.126 .928 .027 .625 .629 .527 .627 .326 .227 .727 .225 .926 .528 .326 .529 .1 \\ 23.729 .726 .829 .528 .426 .328 .128 .727 .025 .526 .927 .227 .625 .528 .327 .4 \\ 28.825 .025 .3 \quad 27.725 .228 .627 .928 .7 \end{array}$

Studdy Solution

STEP 1

1. The resistance data is given and needs to be organized into a frequency distribution table.
2. Frequency distribution will help in calculating the mean and standard deviation.
3. The mean is the average resistance value.
4. The standard deviation measures the spread of the resistance values around the mean.

STEP 2

1. Organize the data into a frequency distribution table.
2. Calculate the mean of the resistance values.
3. Calculate the variance and standard deviation of the resistance values.

STEP 3

First, we need to sort the given resistance data for easier analysis:
$\begin{array}{l} 28.3, \ 26.8, \ 26.6, \ 26.5, \ 28.124, \ 27.827, \ 27.426, \ 27.2, \ 29.428, \ 28.624, \ 28.925, \ 28.230, \ 28.427, \ 28.727, \ 28.026, \ 28.1, \\ 28.126, \ 27.928, \ 27.027, \ 27.625, \ 27.629, \ 27.527, \ 27.627, \ 27.326, \ 27.227, \ 27.727, \ 27.225, \ 27.926, \ 27.528, \ 27.326, \ 27.529, \ 27.1, \\ 23.729, \ 27.726, \ 27.829, \ 27.528, \ 27.426, \ 27.328, \ 27.128, \ 27.727, \ 27.025, \ 27.526, \ 27.927, \ 27.227, \ 27.625, \ 27.528, \ 27.327, \ 27.4, \\ 28.825, \ 27.025, \ 28.3, \ 27.725, \ 27.228, \ 27.627, \ 27.928, \ 27.7 \end{array}$

STEP 4

Create intervals for the frequency distribution table. Assume we choose a class width of 1 kg for simplicity.
$\begin{array}{l} 23-24 \\ 24-25 \\ 25-26 \\ 26-27 \\ 27-28 \\ 28-29 \\ 29-30 \\ \end{array}$

STEP 5

Count the frequency of data points within each interval:
$\begin{array}{c|c} \text{Interval} & \text{Frequency} \\ \hline 23-24 & 1 \\ 24-25 & 0 \\ 25-26 & 0 \\ 26-27 & 4 \\ 27-28 & 33 \\ 28-29 & 16 \\ 29-30 & 1 \\ \end{array}$

STEP 6

To calculate the mean, sum all resistance values and divide by the total number of values. Let $n$ be the total number of values.
$n = 55$

STEP 7

Calculate the sum of all resistance values:
$\text{Sum} = 28.3 + 26.8 + 26.6 + 26.5 + 28.124 + 27.827 + 27.426 + 27.2 + 29.428 + 28.624 + 28.925 + 28.230 + 28.427 + 28.727 + 28.026 + 28.1 + \ldots + 27.7$
$\text{Sum} = 1524.25$

STEP 8

Calculate the mean ( $\mu$ ):
$\mu = \frac{\text{Sum}}{n} = \frac{1524.25}{55} \approx 27.7136$

STEP 9

Next, we calculate the variance. First, find the squared deviation of each value from the mean, then sum these squared deviations:
$\sum (x_i - \mu)^2$
Where $x_i$ are the individual resistance values.

STEP 10

Calculate the squared deviations for each value and sum them up:
$\sum (x_i - 27.7136)^2 = (28.3 - 27.7136)^2 + (26.8 - 27.7136)^2 + \ldots + (27.7 - 27.7136)^2$
$\text{Sum of squared deviations} \approx 132.874$

STEP 11

Calculate the variance ( $\sigma^2$ ):
$\sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = \frac{132.874}{55} \approx 2.416$

STEP 12

Calculate the standard deviation ( $\sigma$ ):
$\sigma = \sqrt{\sigma^2} = \sqrt{2.416} \approx 1.555$
The mean resistance value is approximately $27.7136$ kg, and the standard deviation is approximately $1.555$ kg.

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Studdy Solution

STEP 1

STEP 2

1. Organize the data into a frequency distribution table.
2. Calculate the mean of the resistance values.
3. Calculate the variance and standard deviation of the resistance values.

STEP 3

STEP 4

Create intervals for the frequency distribution table. Assume we choose a class width of 1 kg for simplicity.
$\begin{array}{l} 23-24 \\ 24-25 \\ 25-26 \\ 26-27 \\ 27-28 \\ 28-29 \\ 29-30 \\ \end{array}$

STEP 5

Count the frequency of data points within each interval:
$\begin{array}{c|c} \text{Interval} & \text{Frequency} \\ \hline 23-24 & 1 \\ 24-25 & 0 \\ 25-26 & 0 \\ 26-27 & 4 \\ 27-28 & 33 \\ 28-29 & 16 \\ 29-30 & 1 \\ \end{array}$

STEP 6

To calculate the mean, sum all resistance values and divide by the total number of values. Let $n$ be the total number of values.
$n = 55$

STEP 7

Calculate the sum of all resistance values:
$\text{Sum} = 28.3 + 26.8 + 26.6 + 26.5 + 28.124 + 27.827 + 27.426 + 27.2 + 29.428 + 28.624 + 28.925 + 28.230 + 28.427 + 28.727 + 28.026 + 28.1 + \ldots + 27.7$
$\text{Sum} = 1524.25$

STEP 8

Calculate the mean ( $\mu$ ):
$\mu = \frac{\text{Sum}}{n} = \frac{1524.25}{55} \approx 27.7136$

STEP 9

Next, we calculate the variance. First, find the squared deviation of each value from the mean, then sum these squared deviations:
$\sum (x_i - \mu)^2$
Where $x_i$ are the individual resistance values.

STEP 10

Calculate the squared deviations for each value and sum them up:
$\sum (x_i - 27.7136)^2 = (28.3 - 27.7136)^2 + (26.8 - 27.7136)^2 + \ldots + (27.7 - 27.7136)^2$
$\text{Sum of squared deviations} \approx 132.874$

STEP 11

Calculate the variance ( $\sigma^2$ ):
$\sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = \frac{132.874}{55} \approx 2.416$

STEP 12

Calculate the standard deviation ( $\sigma$ ):
$\sigma = \sqrt{\sigma^2} = \sqrt{2.416} \approx 1.555$
The mean resistance value is approximately $27.7136$ kg, and the standard deviation is approximately $1.555$ kg.

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Studdy Solution

STEP 1

STEP 2

1. Organize the data into a frequency distribution table.2. Calculate the mean of the resistance values.3. Calculate the variance and standard deviation of the resistance values.

STEP 3

STEP 4

STEP 5

STEP 6

To calculate the mean, sum all resistance values and divide by the total number of values. Let n n n be the total number of values.n=55 n = 55 n=55

STEP 7

STEP 8

Calculate the mean (μ\muμ):μ=Sumn=1524.2555≈27.7136 \mu = \frac{\text{Sum}}{n} = \frac{1524.25}{55} \approx 27.7136 μ=nSum​=551524.25​≈27.7136

STEP 9

Next, we calculate the variance. First, find the squared deviation of each value from the mean, then sum these squared deviations:∑(xi−μ)2 \sum (x_i - \mu)^2 ∑(xi​−μ)2Where xi x_i xi​ are the individual resistance values.

STEP 10

STEP 11

Calculate the variance (σ2\sigma^2σ2):σ2=∑(xi−μ)2n=132.87455≈2.416 \sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = \frac{132.874}{55} \approx 2.416 σ2=n∑(xi​−μ)2​=55132.874​≈2.416

STEP 12

Calculate the standard deviation (σ\sigmaσ):σ=σ2=2.416≈1.555 \sigma = \sqrt{\sigma^2} = \sqrt{2.416} \approx 1.555 σ=σ2​=2.416​≈1.555The mean resistance value is approximately 27.7136 27.7136 27.7136 kg, and the standard deviation is approximately 1.555 1.555 1.555 kg.

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1. Organize the data into a frequency distribution table.
2. Calculate the mean of the resistance values.
3. Calculate the variance and standard deviation of the resistance values.

To calculate the mean, sum all resistance values and divide by the total number of values. Let $n$ be the total number of values.
$n = 55$

Calculate the mean ( $\mu$ ):
$\mu = \frac{\text{Sum}}{n} = \frac{1524.25}{55} \approx 27.7136$

Next, we calculate the variance. First, find the squared deviation of each value from the mean, then sum these squared deviations:
$\sum (x_i - \mu)^2$
Where $x_i$ are the individual resistance values.

Calculate the variance ( $\sigma^2$ ):
$\sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = \frac{132.874}{55} \approx 2.416$

Calculate the standard deviation ( $\sigma$ ):
$\sigma = \sqrt{\sigma^2} = \sqrt{2.416} \approx 1.555$
The mean resistance value is approximately $27.7136$ kg, and the standard deviation is approximately $1.555$ kg.