\begin{tabular}{|c|c|} \hline \begin{tabular}{c} Accidents per day \\ $\left(x_{i}\right)$ \end{tabular} & \begin{tabular}{c} Frequency \\ $\left(f_{i}\right)$ \end{tabular} \\ \hline 0 & 5 \\ 1 & 8 \\ 2 & 6 \\ 3 & 3 \\ 4 & 2 \\ 5 & 0 \\ 6 & 1 \\ \hline \end{tabular} \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Accidents per day \\ $\left(x_{i}\right)$ \end{tabular} & \begin{tabular}{c} Frequency \\ $\left(f_{i}\right)$ \end{tabular} \\ \hline 0 & 5 \\ 1 & 8 \\ 2 & 6 \\ 3 & 3 \\ 4 & 2 \\ 5 & 0 \\ 6 & 1 \\ \hline \end{tabular} determine the mean, median, mode, and standard deviation. With full solution numerical and solutions only and also answer also that table in tabulated form as well. Numerical answer only And Find the variance, sd, and cv. With a solution the subject is epidiomology bio statistics. In the bachelor of medical laboratory. With a full solution and no words just solutions that how it get and answers. Solutions of variance and answer, second the SD with full solutions answer, last cv with solutions and answer. Full solutions and answer 

Let's solve the problem step by step with full numerical solutions.

### Step 1: Calculate the mean ($\bar{x}$).

\[
\text{Mean}, \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}
\]

Given dataLet's solve the problem step by step with full numerical solutions.

### Step 1: Calculate the mean ($\bar{x}$).

\[
\text{Mean}, \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}
\]

Given 

\[
x_i = \{0, 1, 2, 3, 4, 5, 6\}, \quad f_i = \{5, 8, 6, 3, 2, 0, 1\}
\]

Now, compute the weighted sum:

\[
\sum (x_i \cdot f_i) = (0 \times 5) + (1 \times 8) + (2 \times 6) + (3 \times 3) + (4 \times 2) + (5 \times 0) + (6 \times 1) = 0 + 8 + 12 + 9 + 8 + 0 + 6 = 43
\]

Total frequency:

\[
\sum f_i = 5 + 8 + 6 + 3 + 2 + 0 + 1 = 25
\]

Thus, the mean:

\[
\bar{x} = \frac{43}{25} = 1.72
\]

---

### Step 2: Calculate the median.

To find the median, we need to find the cumulative frequency:

\[
\text{Cumulative Frequency:} \quad \{5, 13, 19, 22, 24, 24, 25\}
\]

Since the total frequency is 25, the median lies at the $\left(\frac{25+1}{2}\right)$-th value, which is the 13th value.

Looking at the cumulative frequencies, the 13th value falls in the class where $x_i = 1$.

Thus, the median is:

\[
\text{Median} = 1
\]

---

### Step 3: Calculate the mode.

The mode is the value with the highest frequency. Here, $f_1 = 8$ is the highest frequency, which corresponds to $x_1 = 1$.

Thus, the mode is:

\[
\text{Mode} = 1
\]

---

### Step 4: Calculate the variance.

The variance ($\sigma^2$) is given by:

\[
\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}
\]

We need to calculate $(x_i - \bar{x})^2$ for each value of $x_i$:

\[
(0 - 1.72)^2 = 2.9584, \quad (1 - 1.72)^2 = 0.5184, \quad (2 - 1.72)^2 = 0.0784
\]
\[
(3 - 1.72)^2 = 1.6384, \quad (4 - 1.72)^2 = 5.1984, \quad (5 - 1.72)^2 = 10.7584, \quad (6 - 1.72)^2 = 17.3184
\]

Now, compute the weighted sum:

\[
\sum f_i (x_i - \bar{x})^2 = (5 \times 2.9584) + (8 \times 0.5184) + (6 \times 0.0784) + (3 \times 1.6384) + (2 \times 5.1984) + (0 \times 10.7584) + (1 \times 17.3184)
\]
\[
= 14.792 + 4.1472 + 0.4704 + 4.9152 + 10.3968 + 0 + 17.3184 = 52.0392
\]

Thus, the variance:

\[
\sigma^2 = \frac{52.0392}{25} = 2.081568
\]

---

### Step 5: Calculate the standard deviation (SD).

The standard deviation ($\sigma$) is the square root of the variance:

\[
\sigma = \sqrt{2.081568} = 1.4436
\]

---

### Step 6: Calculate the coefficient of variation (CV).

The coefficient of variation is given by:

\[
CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.4436}{1.72} \times 100 = 83.84\%
\]

---

### Final Numerical Results:

1. Mean: $ \bar{x} = 1.72 $
2. Median: 1
3. Mode: 1
4. Variance: $ \sigma^2 = 2.081568 $
5. Standard Deviation: $ \sigma = 1.4436 $
6. Coefficient of Variation: $ CV = 83.84\% $

If you have any questions or want more details, let me know! Here are some related questions:

1. How does changing the frequency distribution affect the standard deviation?
2. Why is the coefficient of variation used in epidemiology?
3. How is the median useful in summarizing accident data?
4. What happens to variance if the frequencies are more spread out?
5. Can there be more than one mode in this dataset?

**Tip:** The coefficient of variation is useful when comparing variability across datasets with different means.

Determine the mean, median, mode, and standard deviation for the following data set: Accidents per day (x_i): {0, 1, 2, 3, 4, 5, 6}, Frequency (f_i): {5, 8, 6, 3, 2, 0, 1}. Additionally, find the variance, standard deviation, and coefficient of variation (CV) with full solutions provided. The subject is epidemiology and biostatistics at the bachelor level in medical laboratory studies.

Learn how to calculate the mean, median, mode, variance, standard deviation, and coefficient of variation for accident frequency data. This step-by-step guide provides full solutions, applying key statistical concepts in epidemiology and biostatistics, ideal for undergraduate medical laboratory students.

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