Let P, Q, R represent mean, median and mode. If for some distribution \(5 P=4 Q=\frac{R}{2}\) then what is \(\frac{P+Q}{2 P+0.7 R}\) equal to ?
Understanding Mean, Median, and Mode Relationship
The question provides a relationship between the mean (P), median (Q), and mode (R) of a distribution: \(5 P = 4 Q = \frac{R}{2}\). We are asked to evaluate the expression \(\frac{P+Q}{2 P+0.7 R}\).
To solve this, we can express P, Q, and R in terms of a single variable using the given relationship.
Expressing P, Q, and R in Terms of a Constant
Let the common value of the given relationship be a constant, say \(k\). So, we have:
- \(5P = k \implies P = \frac{k}{5}\)
- \(4Q = k \implies Q = \frac{k}{4}\)
- \(\frac{R}{2} = k \implies R = 2k\)
Now we have P, Q, and R all expressed in terms of \(k\).
Evaluating the Expression
The expression we need to evaluate is \(\frac{P+Q}{2 P+0.7 R}\).
Substitute the values of P, Q, and R in terms of \(k\) into the expression:
Numerator: \(P + Q = \frac{k}{5} + \frac{k}{4}\)
Denominator: \(2P + 0.7R = 2 \left(\frac{k}{5}\right) + 0.7 (2k) = \frac{2k}{5} + 1.4k\)
Simplifying the Numerator and Denominator
Let's simplify the numerator:
\(P + Q = \frac{k}{5} + \frac{k}{4}\)
To add these fractions, we find a common denominator, which is 20.
\(P + Q = \frac{4k}{20} + \frac{5k}{20} = \frac{4k + 5k}{20} = \frac{9k}{20}\)
Now, let's simplify the denominator:
\(2P + 0.7R = \frac{2k}{5} + 1.4k\)
We can write \(1.4k\) as \(\frac{14}{10}k\) or \(\frac{7}{5}k\).
\(2P + 0.7R = \frac{2k}{5} + \frac{7k}{5} = \frac{2k + 7k}{5} = \frac{9k}{5}\)
Alternatively, for the denominator:
\(2P + 0.7R = \frac{2k}{5} + 1.4k\)
Convert \(\frac{2k}{5}\) to a decimal: \(\frac{2}{5}k = 0.4k\).
\(2P + 0.7R = 0.4k + 1.4k = 1.8k\)
Writing \(1.8k\) as a fraction: \(1.8k = \frac{18}{10}k = \frac{9}{5}k\). Both methods give the same result for the denominator.
Calculating the Final Value
Now, substitute the simplified numerator and denominator back into the expression:
\(\frac{P+Q}{2 P+0.7 R} = \frac{\frac{9k}{20}}{\frac{9k}{5}}\)
To divide by a fraction, we multiply by its reciprocal:
\(\frac{9k}{20} \times \frac{5}{9k}\)
Assuming \(k \neq 0\) (which must be true if P, Q, and R represent non-zero statistical measures related by this equation), the \(k\) terms cancel out, and the \(9\) terms cancel out:
\(\frac{\cancel{9k}}{20} \times \frac{5}{\cancel{9k}} = \frac{5}{20}\)
Simplify the fraction \(\frac{5}{20}\) by dividing both the numerator and denominator by 5:
\(\frac{5 \div 5}{20 \div 5} = \frac{1}{4}\)
So, the value of the expression \(\frac{P+Q}{2 P+0.7 R}\) is \(\frac{1}{4}\).
Summary of Calculation Steps
| Step | Description | Expression/Value |
|---|---|---|
| 1 | Given relationship | \(5P = 4Q = \frac{R}{2} = k\) |
| 2 | Express P, Q, R in terms of k | \(P = \frac{k}{5}, Q = \frac{k}{4}, R = 2k\) |
| 3 | Numerator \(P+Q\) | \(\frac{k}{5} + \frac{k}{4} = \frac{9k}{20}\) |
| 4 | Denominator \(2P+0.7R\) | \(2(\frac{k}{5}) + 0.7(2k) = \frac{2k}{5} + 1.4k = \frac{9k}{5}\) |
| 5 | Expression Value | \(\frac{\frac{9k}{20}}{\frac{9k}{5}} = \frac{9k}{20} \times \frac{5}{9k}\) |
| 6 | Final Simplification | \(\frac{5}{20} = \frac{1}{4}\) |
Conclusion on Mean, Median, Mode Relationship
Given the specific relationship \(5 P=4 Q=\frac{R}{2}\) between the mean (P), median (Q), and mode (R), we found that the value of the expression \(\frac{P+Q}{2 P+0.7 R}\) is \(\frac{1}{4}\). This calculation relied purely on the given algebraic relationship, not on the general properties of mean, median, and mode for specific distributions.
Revision Table: Key Concepts Reviewed
| Term | Definition | Symbol in Problem |
|---|---|---|
| Mean | The average of a dataset. | P |
| Median | The middle value in a dataset sorted numerically. | Q |
| Mode | The value that appears most frequently in a dataset. | R |
Additional Information: Mean, Median, and Mode
Mean, median, and mode are measures of central tendency. They provide a single value that attempts to describe the center of a set of data.
- The mean is calculated by summing all values and dividing by the number of values. It is sensitive to extreme values (outliers).
- The median is the middle value when the data is ordered. It is not affected by outliers. For an even number of data points, it's the average of the two middle values.
- The mode is the value that occurs most often. A dataset can have one mode (unimodal), multiple modes (multimodal), or no mode.
For symmetrical distributions (like the normal distribution), the mean, median, and mode are all equal. For skewed distributions, these measures differ, and their relative positions (e.g., mean > median > mode for positively skewed) can give insight into the shape of the distribution. The empirical relationship between mean, median, and mode, which states that for moderately skewed distributions, Mode \(\approx\) 3 Median - 2 Mean, is a general approximation and not directly used in this specific problem which provides a distinct, explicit relationship.
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