For the cumulative distribution function \(F(x) = \left\{ {\begin{array}{*{20}{c}} {0;x < - 1}\\ {\frac{1}{2}{{(x + 1)}^2}; - 1 \le x < 0}\\ {1 - \frac{{{{(1 - x)}^2}}}{2};0 \le x < 1}\\ {1.1 < x < \infty } \end{array}} \right.\) the upper quartile point is
1 - \(\sqrt{0.5}\)
Finding the Upper Quartile from a Cumulative Distribution Function (CDF)
The upper quartile point (often denoted as \(Q_3\) or \(q_3\)) of a probability distribution is the value \(x\) such that the cumulative probability up to \(x\) is 0.75. In other words, \(F(x) = 0.75\), where \(F(x)\) is the cumulative distribution function (CDF).
The given cumulative distribution function is defined as:
\[ F(x) = \left\{ {\begin{array}{*{20}{c}} {0;} & {x < - 1}\\ {\frac{1}{2}{{(x + 1)}^2};} & {- 1 \le x < 0}\\ {1 - \frac{{{{(1 - x)}^2}}}{2};} & {0 \le x < 1}\\ {1;} & {x \ge 1} \end{array}} \right. \]
Note: The provided CDF had a segment \(1.1 < x < \infty\). A valid CDF for a continuous distribution must reach 1 as \(x \to \infty\) and be non-decreasing. Assuming the CDF reaches 1 at \(x=1\) and stays 1 thereafter, we've used \(x \ge 1\) with \(F(x)=1\) for the last segment, which is consistent with \(F(1) = 1 - \frac{1}{2}(1-1)^2 = 1\).
To find the upper quartile, we need to find the value of \(x\) for which \(F(x) = 0.75\). We need to check each segment of the CDF to see where 0.75 falls within the range of probabilities for that segment.
- For \(x < -1\), \(F(x) = 0\). The value 0.75 is not in this range.
- For \(-1 \le x < 0\), \(F(x) = \frac{1}{2}(x + 1)^2\). At \(x = -1\), \(F(-1) = 0\). As \(x\) increases towards 0, \(F(x)\) increases. At \(x = 0\), \(F(0) = \frac{1}{2}(0 + 1)^2 = \frac{1}{2} = 0.5\). So, for \(-1 \le x < 0\), the range of \(F(x)\) is \(0 \le F(x) < 0.5\). The value 0.75 is not in this range.
- For \(0 \le x < 1\), \(F(x) = 1 - \frac{1}{2}(1 - x)^2\). At \(x = 0\), \(F(0) = 1 - \frac{1}{2}(1 - 0)^2 = 1 - \frac{1}{2} = 0.5\). As \(x\) increases towards 1, \(F(x)\) increases. At \(x = 1\), \(F(1) = 1 - \frac{1}{2}(1 - 1)^2 = 1 - 0 = 1\). So, for \(0 \le x < 1\), the range of \(F(x)\) is \(0.5 \le F(x) < 1\). The value 0.75 falls within this range.
- For \(x \ge 1\), \(F(x) = 1\). The value 0.75 is not in this range.
Since 0.75 falls in the range of the third segment, the upper quartile must be found by setting \(F(x) = 0.75\) for \(0 \le x < 1\):
\[ 1 - \frac{1}{2}(1 - x)^2 = 0.75 \]
Now, we solve this equation for \(x\):
Subtract 1 from both sides:
\[ - \frac{1}{2}(1 - x)^2 = 0.75 - 1 \]
\[ - \frac{1}{2}(1 - x)^2 = -0.25 \]
Multiply both sides by -2:
\[ (1 - x)^2 = -0.25 \times -2 \]
\[ (1 - x)^2 = 0.5 \]
Take the square root of both sides:
\[ \sqrt{(1 - x)^2} = \sqrt{0.5} \]
\[ |1 - x| = \sqrt{0.5} \]
Since we are looking for \(x\) in the range \(0 \le x < 1\), the term \((1 - x)\) will be positive (if \(x=0\), \(1-x=1\); if \(x\) is close to 1, \(1-x\) is positive and close to 0). Thus, \(|1 - x| = 1 - x\).
\[ 1 - x = \sqrt{0.5} \]
Now, solve for \(x\):
\[ x = 1 - \sqrt{0.5} \]
We should verify that this value of \(x\) is within the range \(0 \le x < 1\). Since \(0 < 0.5 < 1\), we have \(0 < \sqrt{0.5} < 1\). Therefore, \(0 < 1 - \sqrt{0.5} < 1\). The value \(x = 1 - \sqrt{0.5}\) is indeed in the correct interval.
The upper quartile point is \(1 - \sqrt{0.5}\).
| CDF Segment | Domain | Range of F(x) | Includes 0.75? |
|---|---|---|---|
| \(F(x) = 0\) | \(x < -1\) | \(0\) | No |
| \(F(x) = \frac{1}{2}(x + 1)^2\) | \(-1 \le x < 0\) | \([0, 0.5)\) | No |
| \(F(x) = 1 - \frac{1}{2}(1 - x)^2\) | \(0 \le x < 1\) | \([0.5, 1)\) | Yes |
| \(F(x) = 1\) | \(x \ge 1\) | \(1\) | No |
Revision Table: Cumulative Distribution Function (CDF) and Quartiles
| Concept | Definition | How to Find \(Q_3\) |
|---|---|---|
| CDF, \(F(x)\) | For a random variable \(X\), \(F(x) = P(X \le x)\). Gives the probability that \(X\) takes a value less than or equal to \(x\). | Find \(x\) such that \(F(x) = 0.75\). |
| First Quartile (\(Q_1\)) | Value \(x\) such that \(F(x) = 0.25\). | Solve \(F(x) = 0.25\). |
| Second Quartile (\(Q_2\), Median) | Value \(x\) such that \(F(x) = 0.50\). | Solve \(F(x) = 0.50\). |
| Upper Quartile (\(Q_3\)) | Value \(x\) such that \(F(x) = 0.75\). | Solve \(F(x) = 0.75\). |
Additional Information on CDFs and Quartiles
A cumulative distribution function \(F(x)\) for a continuous random variable has the following properties:
- \(0 \le F(x) \le 1\) for all \(x\).
- \(F(x)\) is non-decreasing.
- \(\lim_{x \to -\infty} F(x) = 0\) and \(\lim_{x \to \infty} F(x) = 1\).
- \(F(x)\) is right-continuous.
Quartiles divide the probability distribution into four equal parts. The first quartile \(Q_1\) is the value below which 25% of the data falls, the second quartile \(Q_2\) (median) is the value below which 50% of the data falls, and the third quartile \(Q_3\) is the value below which 75% of the data falls.
To find any quantile (like a quartile, median, or percentile) from a continuous CDF, you set \(F(x)\) equal to the desired probability (e.g., 0.25 for \(Q_1\), 0.50 for \(Q_2\), 0.75 for \(Q_3\)) and solve for \(x\) within the appropriate interval where the CDF value lies.
Important Questions from Probability Distribution
If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:
For the distribution with unknown θ
\(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)
We set the testing of hypothesis H 0 ∶ θ = 1 vs H 1 ∶ θ = 2. When the critical region X ≥ 0.4, the value of probability of type-II error is:
Let the joint probability density function of \( (X, Y) \) be
\[f(x, y) = \begin{cases} 6xy^2 & \text{if } 0 < x < 1, 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\]
Then \( P\left(\frac{1}{2} < X < \frac{3}{4}\right) \) is:
Let X and Y have the joint p.m.f. f(x, y) = x + y / 21, where x = 1, 2, 3 and y = 1, 2. The marginal p.m.f. of X is:
For a random variable X following Poisson distribution with parameter 5, the variance of X is:
