Question: Assume that the random variable $X$ is normally distributed, with mean $\mu = 51$ and standard deviation $\sigma = 10$. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded.
Upper $P(X \leq 42)$
1. **State the problem:** We want to find the probability that the random variable $X$ is less than or equal to $42$, given that $X$ is normally distributed with mean $\mu = 51$ and standard deviation $\sigma = 10$.
2. **Formula used:** For a normal distribution, the probability $P(X \leq x)$ can be found by converting $X$ to the standard normal variable $Z$ using the formula:
$$Z = \frac{X - \mu}{\sigma}$$
Then, $P(X \leq x) = P\left(Z \leq \frac{x - \mu}{\sigma}\right)$.
3. **Calculate the $Z$-score:**
$$Z = \frac{42 - 51}{10} = \frac{-9}{10} = -0.9$$
4. **Find the probability:**
We want $P(Z \leq -0.9)$. Using standard normal distribution tables or a calculator, this probability is approximately $0.1841$.
5. **Interpretation:**
This means there is about an $18.41\%$ chance that the random variable $X$ takes a value less than or equal to $42$.
6. **Summary:**
$$P(X \leq 42) = P\left(Z \leq -0.9\right) \approx 0.1841$$
This is the area under the normal curve to the left of $42$, which corresponds to the shaded region in the normal distribution curve.