Question

Let \(X_1,X_2,X_3\) be a random sample of size \(3\) from \(U(1,2)\) distribution. Then \(P(X_{(3)}>1.5)\) equals (rounded off to two decimal places).

Hint:

For the maximum order statistic \(X_{(n)}\), use \[ P(X_{(n)}\leq x)=\left(F(x)\right)^n \] when the observations are independent and identically distributed.

Answer 1

Correct Answer: 0.88

Step 1: Understand the notation.
\(X_{(3)}\) denotes the largest order statistic among
\[ X_1,X_2,X_3 \] Thus,
\[ P(X_{(3)}>1.5) \] means the probability that at least one observation exceeds \(1.5\).

Step 2: Use the complement rule.
\[ P(X_{(3)}>1.5) = 1-P(X_{(3)}\leq 1.5) \] Now,
\[ X_{(3)}\leq 1.5 \] means all three observations are at most \(1.5\).
Therefore,
\[ P(X_{(3)}\leq 1.5) = P(X_1\leq 1.5,\;X_2\leq 1.5,\;X_3\leq 1.5) \] Since the sample is independent,
\[ = \left(P(X_1\leq 1.5)\right)^3 \]

Step 3: Compute \(P(X_1\leq 1.5)\).
Since
\[ X_1\sim U(1,2), \] the distribution is uniform on an interval of length \(1\).
Hence,
\[ P(X_1\leq 1.5) = \frac{1.5-1}{2-1} = 0.5 \] Therefore,
\[ P(X_{(3)}\leq 1.5) = (0.5)^3 = 0.125 \]

Step 4: Find the required probability.
\[ P(X_{(3)}>1.5) = 1-0.125 \] \[ = 0.875 \] Rounded off to two decimal places,
\[ 0.88 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{0.88} \]