Question

Let \(X\) be \(N(5,10)\). Find \(P\left[0.04<(X-5)^{2}<38.4\right]\).

Short answer

The probability that the squared deviation from the mean falls within the range [0.04, 38.4] is approximately 0.42005.

Step-by-step solution

Identify Parameters

The problem tells us that \(X\) follows a normal distribution \(N(5,10)\). This means \(X\) has a mean (\(\mu\)) of 5 and a variance (\(\sigma^{2}\)) of 10, therefore a standard deviation (\(\sigma\)) of \(\sqrt{10}\), which is approximately \(3.16228\).

Standardize and apply the Z-score

Z-scores are used to identify how many standard deviations a value is away from the mean. We can standardize the given range using z-scores. First, rewrite the inequality \((X-\mu)^{2}\) as Z-scores by using the formula \(Z=\frac{(X-\mu)}{\sigma}\). So the left inequality becomes \(\left(\frac{(X-5)}{\sqrt{10}}\right)^{2}>0.04\) and the right inequality becomes \(\left(\frac{(X-5)}{\sqrt{10}}\right)^{2}<38.4\). Solving these gives z-values of \(\pm 0.2\) and \(\pm 3.2\) respectively.

Calculate the Probability

Using the Z-table or software to find the probabilities. The probability for \(\pm 0.2\) is \(0.57926\) and for \(\pm 3.2\) is \(0.99931\). So, \(P[0.04<(X-5)^{2}<38.4]\) will be obtained by finding the difference between these two probabilities \(0.99931-0.57926\).