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Math  /  Data & Statistics

Question2. Consider the following frequency distribution \begin{tabular}{|l|l|l|c|l|l|l|l|} \hline Class & 15−1915-19 & 20−2420-24 & 25−2925-29 & 30−3430-34 & 35−3935-39 & 40−4440-44 & 45−4945-49 \\ \hline Frequency & 10 & 22 & f1f_{1} & 40 & f2f_{2} & 18 & 12 \\ \hline \end{tabular}
The total frequency is 160 and the modal value is 31.0909 . Find; i) The value of f1f_{1} and f2f_{2} ii) Mode iii) Median iv) Coefficient of Quartile deviation v) Mean vi) Mean Absolute deviation vii) Standard deviation

Studdy Solution

STEP 1

1. The total frequency of the distribution is 160.
2. The modal class is given by the class interval 30−3430-34 with a modal value of 31.0909.
3. The class intervals are continuous and of equal width.
4. The frequencies f1f_1 and f2f_2 are unknown.
5. The formula for the mode in a grouped frequency distribution is used.
6. The median, mean, and measures of dispersion will be calculated using standard formulas for grouped data.

STEP 2

1. Determine the values of f1f_1 and f2f_2.
2. Calculate the mode.
3. Calculate the median.
4. Calculate the coefficient of quartile deviation.
5. Calculate the mean.
6. Calculate the mean absolute deviation.
7. Calculate the standard deviation.

STEP 3

Use the total frequency equation:
10+22+f1+40+f2+18+12=160 10 + 22 + f_1 + 40 + f_2 + 18 + 12 = 160
Simplify and solve for f1+f2f_1 + f_2:
102+f1+f2=160 102 + f_1 + f_2 = 160
f1+f2=58 f_1 + f_2 = 58

STEP 4

Use the modal class information. The mode formula for grouped data is:
Mode=L+(fm−fm−12fm−fm−1−fm+1)×h \text{Mode} = L + \left( \frac{f_m - f_{m-1}}{2f_m - f_{m-1} - f_{m+1}} \right) \times h
Where: - LL is the lower boundary of the modal class, - fmf_m is the frequency of the modal class, - fm−1f_{m-1} is the frequency of the class before the modal class, - fm+1f_{m+1} is the frequency of the class after the modal class, - hh is the class width.
Given, Mode = 31.0909, L=30L = 30, fm=40f_m = 40, fm−1=f1f_{m-1} = f_1, fm+1=f2f_{m+1} = f_2, and h=5h = 5.
Substitute the known values into the mode formula and solve for f1f_1 and f2f_2.
31.0909=30+(40−f12×40−f1−f2)×5 31.0909 = 30 + \left( \frac{40 - f_1}{2 \times 40 - f_1 - f_2} \right) \times 5
Solve this equation along with f1+f2=58f_1 + f_2 = 58 to find f1f_1 and f2f_2.

STEP 5

Calculate the mode using the values of f1f_1 and f2f_2 found in Step 1.

STEP 6

Calculate the median using the median formula for grouped data:
Median=L+(N2−CFfm)×h \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f_m} \right) \times h
Where: - LL is the lower boundary of the median class, - NN is the total frequency, - CFCF is the cumulative frequency before the median class, - fmf_m is the frequency of the median class, - hh is the class width.
Identify the median class and substitute the known values to find the median.

STEP 7

Calculate the coefficient of quartile deviation using:
Coefficient of Quartile Deviation=Q3−Q1Q3+Q1 \text{Coefficient of Quartile Deviation} = \frac{Q_3 - Q_1}{Q_3 + Q_1}
Find Q1Q_1 and Q3Q_3 using the quartile formulas for grouped data and substitute into the formula.

STEP 8

Calculate the mean using the formula for grouped data:
Mean=∑fi⋅xiN \text{Mean} = \frac{\sum f_i \cdot x_i}{N}
Where xix_i is the midpoint of each class interval.

STEP 9

Calculate the mean absolute deviation using:
Mean Absolute Deviation=∑∣xi−Mean∣⋅fiN \text{Mean Absolute Deviation} = \frac{\sum |x_i - \text{Mean}| \cdot f_i}{N}

STEP 10

Calculate the standard deviation using:
Standard Deviation=∑(xi−Mean)2⋅fiN \text{Standard Deviation} = \sqrt{\frac{\sum (x_i - \text{Mean})^2 \cdot f_i}{N}}

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