Math / Data & Statistics

Question1. En la siguiente tabla de se muestran los años de servicio de una muestra de 100 empleados de un banco. Completa la tabla como en los ejemplos de la guía. Luego, calcula la desviación estándar y la varianza. \begin{tabular}{|c|c|} \hline Años & $N^{\circ}$ Empleados \\ \hline $0-2$ & 40 \\ \hline $3-5$ & 25 \\ \hline $6-8$ & 20 \\ \hline $9-11$ & 10 \\ \hline $12-14$ & 5 \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? We need to fill in some missing info in a table about how long bank employees have worked there, then calculate the standard deviation and variance of the years of service. Watch out! Remember that we're dealing with ranges of years, not single values, so we need to use the midpoint of each range in our calculations.
Also, don't mix up variance and standard deviation – standard deviation is the square root of the variance!

STEP 2

1. Complete the table
2. Calculate the mean
3. Calculate the variance
4. Calculate the standard deviation

STEP 3

Let's add columns for the midpoint ( $x_i$ ), $f_i \cdot x_i$ , and $f_i \cdot x_i^2$ to our table.
The midpoint is the middle of each range of years. $f_i$ is the number of employees in each range. $f_i \cdot x_i$ is the product of the number of employees and the midpoint, and $f_i \cdot x_i^2$ is the product of the number of employees and the square of the midpoint.
We'll need these for our calculations later!

STEP 4

$\begin{array}{c c c c c} \text{Años} & f_i & x_i & f_i \cdot x_i & f_i \cdot x_i^2 \\ 0-2 & 40 & 1 & 40 & 40 \\ 3-5 & 25 & 4 & 100 & 400 \\ 6-8 & 20 & 7 & 140 & 980 \\ 9-11 & 10 & 10 & 100 & 1000 \\ 12-14 & 5 & 13 & 65 & 845 \\ \text{Total} & 100 & & 445 & 3265 \\ \end{array}$

STEP 5

Now, let's calculate the mean, $\bar{x}$ , which is the average years of service.
To do this, we'll divide the sum of $f_i \cdot x_i$ by the total number of employees.

STEP 6

$\bar{x} = \frac{\sum f_i \cdot x_i}{\sum f_i} = \frac{445}{100} = 4.45$ So, the mean years of service is 4.45 years!

STEP 7

Next, we'll calculate the variance, $s^2$ , which measures how spread out the data is.
The formula for variance is: $s^2 = \frac{\sum f_i \cdot x_i^2}{N} - \bar{x}^2$ Where $N$ is the total number of employees.

STEP 8

Plugging in our values, we get: $s^2 = \frac{3265}{100} - (4.45)^2 = 32.65 - 19.8025 = 12.8475$ So, the variance is 12.8475!

STEP 9

Finally, let's calculate the standard deviation, $s$ , which is simply the square root of the variance.

STEP 10

$s = \sqrt{s^2} = \sqrt{12.8475} \approx 3.58$ The standard deviation is approximately 3.58!

STEP 11

The completed table is shown above in step 2.1.2.
The standard deviation is approximately 3.58 years, and the variance is 12.8475.

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Studdy Solution

STEP 1

STEP 2

1. Complete the table
2. Calculate the mean
3. Calculate the variance
4. Calculate the standard deviation

STEP 3

STEP 4

$\begin{array}{c c c c c} \text{Años} & f_i & x_i & f_i \cdot x_i & f_i \cdot x_i^2 \\ 0-2 & 40 & 1 & 40 & 40 \\ 3-5 & 25 & 4 & 100 & 400 \\ 6-8 & 20 & 7 & 140 & 980 \\ 9-11 & 10 & 10 & 100 & 1000 \\ 12-14 & 5 & 13 & 65 & 845 \\ \text{Total} & 100 & & 445 & 3265 \\ \end{array}$

STEP 5

Now, let's calculate the mean, $\bar{x}$ , which is the average years of service.
To do this, we'll divide the sum of $f_i \cdot x_i$ by the total number of employees.

STEP 6

$\bar{x} = \frac{\sum f_i \cdot x_i}{\sum f_i} = \frac{445}{100} = 4.45$ So, the mean years of service is 4.45 years!

STEP 7

Next, we'll calculate the variance, $s^2$ , which measures how spread out the data is.
The formula for variance is: $s^2 = \frac{\sum f_i \cdot x_i^2}{N} - \bar{x}^2$ Where $N$ is the total number of employees.

STEP 8

Plugging in our values, we get: $s^2 = \frac{3265}{100} - (4.45)^2 = 32.65 - 19.8025 = 12.8475$ So, the variance is 12.8475!

STEP 9

Finally, let's calculate the standard deviation, $s$ , which is simply the square root of the variance.

STEP 10

$s = \sqrt{s^2} = \sqrt{12.8475} \approx 3.58$ The standard deviation is approximately 3.58!

STEP 11

The completed table is shown above in step 2.1.2.
The standard deviation is approximately 3.58 years, and the variance is 12.8475.

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Studdy Solution

STEP 1

STEP 2

1. Complete the table2. Calculate the mean3. Calculate the variance4. Calculate the standard deviation

STEP 3

STEP 4

STEP 5

Now, let's **calculate the mean**, xˉ\bar{x}xˉ, which is the average years of service.To do this, we'll divide the sum of fi⋅xif_i \cdot x_ifi​⋅xi​ by the total number of employees.

STEP 6

xˉ=∑fi⋅xi∑fi=445100=4.45\bar{x} = \frac{\sum f_i \cdot x_i}{\sum f_i} = \frac{445}{100} = 4.45xˉ=∑fi​∑fi​⋅xi​​=100445​=4.45 So, the **mean** years of service is **4.45** years!

STEP 7

Next, we'll **calculate the variance**, s2s^2s2, which measures how spread out the data is.The formula for variance is: s2=∑fi⋅xi2N−xˉ2s^2 = \frac{\sum f_i \cdot x_i^2}{N} - \bar{x}^2s2=N∑fi​⋅xi2​​−xˉ2 Where NNN is the total number of employees.

STEP 8

Plugging in our values, we get: s2=3265100−(4.45)2=32.65−19.8025=12.8475s^2 = \frac{3265}{100} - (4.45)^2 = 32.65 - 19.8025 = 12.8475s2=1003265​−(4.45)2=32.65−19.8025=12.8475 So, the **variance** is **12.8475**!

STEP 9

Finally, let's **calculate the standard deviation**, sss, which is simply the square root of the variance.

STEP 10

s=s2=12.8475≈3.58s = \sqrt{s^2} = \sqrt{12.8475} \approx 3.58s=s2​=12.8475​≈3.58 The **standard deviation** is approximately **3.58**!

STEP 11

The completed table is shown above in step 2.1.2.The standard deviation is approximately **3.58** years, and the variance is **12.8475**.

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1. Complete the table
2. Calculate the mean
3. Calculate the variance
4. Calculate the standard deviation

Now, let's calculate the mean, $\bar{x}$ , which is the average years of service.
To do this, we'll divide the sum of $f_i \cdot x_i$ by the total number of employees.

$\bar{x} = \frac{\sum f_i \cdot x_i}{\sum f_i} = \frac{445}{100} = 4.45$ So, the mean years of service is 4.45 years!

Next, we'll calculate the variance, $s^2$ , which measures how spread out the data is.
The formula for variance is: $s^2 = \frac{\sum f_i \cdot x_i^2}{N} - \bar{x}^2$ Where $N$ is the total number of employees.

Plugging in our values, we get: $s^2 = \frac{3265}{100} - (4.45)^2 = 32.65 - 19.8025 = 12.8475$ So, the variance is 12.8475!

Finally, let's calculate the standard deviation, $s$ , which is simply the square root of the variance.

$s = \sqrt{s^2} = \sqrt{12.8475} \approx 3.58$ The standard deviation is approximately 3.58!

The completed table is shown above in step 2.1.2.
The standard deviation is approximately 3.58 years, and the variance is 12.8475.