Consider a continuous random variable \(X\) which follows a normal distribution. A sample of size \(n = 25\) is drawn from the distribution of \(X\). Sample mean is 5. Sample standard deviation is 1.5. The probability of \(P[X = 5]\) is _____ (Rounded off to two decimal places).
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Remember that for any continuous random variable, the probability of it taking one exact value is always zero, regardless of the mean or standard deviation.
Step 1: Write the general normal density formula. A normal random variable with mean $\mu$ and standard deviation $\sigma$ has density $f(x) = \dfrac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, which is finite and bounded for every $x$. Step 2: Approximate the probability of a narrow interval around $x = 5$. For a small width $\Delta x$ centered at $x = 5$, $P(5 - \Delta x/2 \le X \le 5 + \Delta x/2) \approx f(5)\,\Delta x$, since $f(5)$ is a fixed finite number for any choice of $\mu$ and $\sigma$. Step 3: Let the interval width shrink to zero. Exact equality $X = 5$ corresponds to $\Delta x \to 0$. Taking this limit, \[ P[X = 5] = \lim_{\Delta x \to 0} f(5)\, \Delta x = f(5) \times 0 = 0 \] because $f(5)$ stays finite while $\Delta x$ shrinks to zero. Step 4: Conclude. This confirms, by a limiting argument instead of the direct integral argument, that the probability of any single exact value of a continuous random variable is zero. \[ \boxed{P[X = 5] = 0.00} \]
