Given that $X$ has a normal distribution with a mean $\mu = 80$ and standard deviation $\sigma = 6$, we need to find the value of $a$ such that $P(X < a) = 0.90$.

Step 1: Translate the probability into a Z-score

Since $P(X < a) = 0.90$, we want the 90th percentile of the normal distribution.

From standard normal distribution tables or using a calculator, the Z-score corresponding to a cumulative probability of 0.90 is approximately: $Z \approx 1.28$

Step 2: Use the Z-score formula to solve for $a$

The Z-score formula is: $Z = \frac{a - \mu}{\sigma}$

Rearranging to solve for $a$: $a = Z \cdot \sigma + \mu$

Step 3: Substitute the values

Substitute $Z = 1.28$, $\sigma = 6$, and $\mu = 80$: $a = 1.28 \cdot 6 + 80$ $a = 7.68 + 80$ $a = 87.68$

Answer

The value of $a$ such that $P(X < a) = 0.90$ is:

• [d]. 87.68

Would you like further details on any part of the solution?

Here are five related questions you might find interesting:

1. How do we find Z-scores for different percentiles?
2. Why is the 90th percentile important in statistics?
3. How would the answer change if the mean were different?
4. What if the standard deviation were a different value?
5. How can we use normal distribution in real-life applications?

Tip: For normal distribution problems, always translate probabilities to Z-scores; this simplifies solving for unknown values.