Understanding the Probability Density Function (PDF) and Deciles

The question provides the Probability Density Function (PDF), \(f(x)\), for the age of babies, defined as \(f(x)=\dfrac{3}{4}x(2-x)\) for the range \(0<x<2\). A PDF describes the likelihood of a continuous random variable falling within a particular range of values.

We are asked to find the 5th decile point of X. Deciles divide a probability distribution into 10 equal parts. The 5th decile is the value below which 50% of the distribution lies. This is equivalent to the median of the distribution.

To find the \(k\)-th decile (where \(k\) is between 1 and 9), we need to find the value \(d_k\) such that the cumulative probability up to \(d_k\) is equal to \(k/10\). The cumulative probability is given by the Cumulative Distribution Function (CDF), \(F(x)\), which is the integral of the PDF from the lower limit of the distribution up to \(x\).

For the 5th decile, we need to find the value \(d_5\) such that \(P(X \le d_5) = 0.5\). This means we need to solve the equation:

\( \int_{0}^{d_5} f(x) dx = 0.5 \)

Substituting the given PDF:

\( \int_{0}^{d_5} \dfrac{3}{4}x(2-x) dx = 0.5 \)

Calculating the 5th Decile Value

Now, let's perform the integration step by step to find the value of \(d_5\).

The integral is:

\( \int_{0}^{d_5} \dfrac{3}{4}(2x-x^2) dx = 0.5 \)

We can pull the constant \(\dfrac{3}{4}\) outside the integral:

\( \dfrac{3}{4} \int_{0}^{d_5} (2x-x^2) dx = 0.5 \)

Now, integrate the polynomial term by term:

• The integral of \(2x\) is \(2 \times \dfrac{x^{1+1}}{1+1} = 2 \times \dfrac{x^2}{2} = x^2\).
• The integral of \(x^2\) is \(\dfrac{x^{2+1}}{2+1} = \dfrac{x^3}{3}\).

So, the antiderivative of \(2x-x^2\) is \(x^2 - \dfrac{x^3}{3}\). Now, we evaluate the definite integral from 0 to \(d_5\):

\( \dfrac{3}{4} \left[ x^2 - \dfrac{x^3}{3} \right]_{0}^{d_5} = 0.5 \)

Substitute the upper and lower limits:

\( \dfrac{3}{4} \left( (d_5^2 - \dfrac{d_5^3}{3}) - (0^2 - \dfrac{0^3}{3}) \right) = 0.5 \)

This simplifies to:

\( \dfrac{3}{4} \left( d_5^2 - \dfrac{d_5^3}{3} \right) = 0.5 \)

Multiply both sides by \(\dfrac{4}{3}\):

\( d_5^2 - \dfrac{d_5^3}{3} = 0.5 \times \dfrac{4}{3} \)

\( d_5^2 - \dfrac{d_5^3}{3} = \dfrac{1}{2} \times \dfrac{4}{3} = \dfrac{4}{6} = \dfrac{2}{3} \)

Now, rearrange the equation to solve for \(d_5\):

\( d_5^2 - \dfrac{d_5^3}{3} - \dfrac{2}{3} = 0 \)

Multiply the entire equation by 3 to remove fractions:

\( 3d_5^2 - d_5^3 - 2 = 0 \)

Rearrange into standard cubic form:

\( d_5^3 - 3d_5^2 + 2 = 0 \)

We need to find the roots of this cubic equation that are within the range \(0 < d_5 < 2\). We can test simple values, starting with integers that might be factors of the constant term (2), such as \(\pm 1, \pm 2\).

• Let's test \(d_5 = 1\): \( (1)^3 - 3(1)^2 + 2 = 1 - 3(1) + 2 = 1 - 3 + 2 = 0 \)

Since substituting \(d_5 = 1\) results in 0, \(d_5 = 1\) is a root of the equation. The range of X is \(0 < x < 2\). Since \(1\) lies within this range, it is a valid value for the 5th decile. Other roots of the cubic equation can be found (e.g., \(d_5 = 1\) is a repeated root, and there is another root at \(d_5 = 1 + \sqrt{3}\) and \(d_5 = 1 - \sqrt{3}\)), but only the root(s) within the domain \(0 < x < 2\) are relevant. The value \(d_5 = 1\) is the only one within the specified range.

Thus, the 5th decile point of X is 1.

Revision Table: Key Concepts for Deciles and PDF

Additional Information on Finding Percentiles and Quartiles

Finding deciles is a specific case of finding percentiles or quartiles.

• Percentiles: Percentiles divide a distribution into 100 equal parts. The \(p\)-th percentile (\(P_p\)) is the value such that \(P(X \le P_p) = p/100\). The 5th decile is the 50th percentile.
• Quartiles: Quartiles divide a distribution into 4 equal parts. The 1st quartile (\(Q_1\)) is the 25th percentile, the 2nd quartile (\(Q_2\)) is the 50th percentile (which is also the median and the 5th decile), and the 3rd quartile (\(Q_3\)) is the 75th percentile.

To find any percentile or quartile for a continuous distribution with PDF \(f(x)\), you would set up the equation \( \int_{-\infty}^{value} f(t) dt = probability \) and solve for 'value', where 'probability' is the cumulative probability corresponding to the desired percentile or quartile (e.g., 0.25 for Q1, 0.50 for Q2/5th decile/median, 0.75 for Q3, 0.10 for 1st decile, 0.90 for 9th decile, etc.). The process involves integrating the PDF to get the CDF and then solving the CDF equation for the specific probability.