We want to find the probability that either 45 or 46 students out of 84 will receive a B or above on the final exam. Given that the number of students, \(X\), follows a normal distribution \(N(42, 4.6)\) with a mean \(\mu = 42\) and standard deviation \(\sigma = 4.6\), we need to calculate the probabilities for \(X = 45\) and \(X = 46\).
First, let's standardize the values using the z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
1. For \(X = 45\):
[tex]\[ Z = \frac{45 - 42}{4.6} \approx \frac{3}{4.6} \approx 0.652 \][/tex]
2. For \(X = 46\):
[tex]\[ Z = \frac{46 - 42}{4.6} \approx \frac{4}{4.6} \approx 0.870 \][/tex]
Now we need to determine the probabilities associated with these z-scores using the standard normal distribution table.
3. Finding the cumulative probability for \(Z = 0.652\):
Using the standard normal table, we need to find the value closest to \(Z = 0.652\). Looking at the z-table, we can see \(Z = 0.65\) is approximately 0.7422 (since 0.652 is very close to 0.65).
4. Finding the cumulative probability for \(Z = 0.870\):
Similarly, for \(Z = 0.870\), the corresponding probability from the table is 0.8078.
Next, we find the probability that \(X\) is between 45 and 46:
[tex]\[ P(45 \le X \le 46) = P(X \le 46) - P(X \le 45) \][/tex]
Using the cumulative probabilities:
[tex]\[ P(45 \le X \le 46) = 0.8078 - 0.7422 = 0.0656 \][/tex]
Thus, the probability that either 45 or 46 students will receive a B or above on the final exam is approximately [tex]\(0.0656\)[/tex] or [tex]\(6.56\%\)[/tex].
