Math Problem Statement

\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

Solution

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

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

Mean,xˉ=(xifi)fi\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 (xˉ\bar{x}).

Mean,xˉ=(xifi)fi\text{Mean}, \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}

Given

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

Now, compute the weighted sum:

(xifi)=(0×5)+(1×8)+(2×6)+(3×3)+(4×2)+(5×0)+(6×1)=0+8+12+9+8+0+6=43\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:

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

Thus, the mean:

xˉ=4325=1.72\bar{x} = \frac{43}{25} = 1.72


Step 2: Calculate the median.

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

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

Since the total frequency is 25, the median lies at the (25+12)\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 xi=1x_i = 1.

Thus, the median is:

Median=1\text{Median} = 1


Step 3: Calculate the mode.

The mode is the value with the highest frequency. Here, f1=8f_1 = 8 is the highest frequency, which corresponds to x1=1x_1 = 1.

Thus, the mode is:

Mode=1\text{Mode} = 1


Step 4: Calculate the variance.

The variance (σ2\sigma^2) is given by:

σ2=fi(xixˉ)2fi\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}

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

(01.72)2=2.9584,(11.72)2=0.5184,(21.72)2=0.0784(0 - 1.72)^2 = 2.9584, \quad (1 - 1.72)^2 = 0.5184, \quad (2 - 1.72)^2 = 0.0784 (31.72)2=1.6384,(41.72)2=5.1984,(51.72)2=10.7584,(61.72)2=17.3184(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:

fi(xixˉ)2=(5×2.9584)+(8×0.5184)+(6×0.0784)+(3×1.6384)+(2×5.1984)+(0×10.7584)+(1×17.3184)\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= 14.792 + 4.1472 + 0.4704 + 4.9152 + 10.3968 + 0 + 17.3184 = 52.0392

Thus, the variance:

σ2=52.039225=2.081568\sigma^2 = \frac{52.0392}{25} = 2.081568


Step 5: Calculate the standard deviation (SD).

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

σ=2.081568=1.4436\sigma = \sqrt{2.081568} = 1.4436


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

The coefficient of variation is given by:

CV=σxˉ×100=1.44361.72×100=83.84%CV = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.4436}{1.72} \times 100 = 83.84\%


Final Numerical Results:

  1. Mean: xˉ=1.72\bar{x} = 1.72
  2. Median: 1
  3. Mode: 1
  4. Variance: σ2=2.081568\sigma^2 = 2.081568
  5. Standard Deviation: σ=1.4436\sigma = 1.4436
  6. Coefficient of Variation: CV=83.84%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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Measures of Central Tendency
Variance
Standard Deviation
Coefficient of Variation

Formulas

Mean: \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}
Variance: \sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}
Standard Deviation: \sigma = \sqrt{\sigma^2}
Coefficient of Variation: CV = \frac{\sigma}{\bar{x}} \times 100

Theorems

Empirical Rule (for standard deviation)

Suitable Grade Level

Undergraduate (Bachelor level)