Let's solve this problem step-by-step.
### Step 1: Identify the given values
- Mean ([tex]\(\mu\)[/tex]): 82
- Standard deviation ([tex]\(\sigma\)[/tex]): 4
- Score ([tex]\(x\)[/tex]): 88
### Step 2: Calculate the z-score
The z-score formula is given by:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
Plug in the given values:
[tex]\[ z = \frac{88 - 82}{4} \][/tex]
[tex]\[ z = \frac{6}{4} \][/tex]
[tex]\[ z = 1.5 \][/tex]
So, the z-score corresponding to the score of 88 is 1.5.
### Step 3: Find the probability using the z-score
Refer to the standard normal distribution table (the chart on page 11 of the lesson). The z-score of 1.5 corresponds to a cumulative probability.
From the standard normal distribution table, a z-score of 1.5 gives us a cumulative probability of approximately 0.9332.
### Step 4: Conclusion
The probability that a randomly chosen student scores 88 or below is approximately 0.9332.
Therefore, the correct answer is:
a. .9332
