Using this approximation, find the probability of 45 or 46 students receiving a [tex]$B$[/tex] or above on the final exam in an accounting class. You may use the portion of the Standard Normal Table below.

\begin{tabular}{c|cccccccccc}
[tex]$z$[/tex] & [tex]$0 . 0 0$[/tex] & [tex]$0 . 0 1$[/tex] & [tex]$0 . 0 2$[/tex] & [tex]$0 . 0 3$[/tex] & [tex]$0 . 0 4$[/tex] & [tex]$0 . 0 5$[/tex] & [tex]$0 . 0 6$[/tex] & [tex]$0 . 0 7$[/tex] & [tex]$0 . 0 8$[/tex] & [tex]$0 . 0 9$[/tex] \\
\hline [tex]$0 . 5$[/tex] & 0.6915 & 0.6950 & 0.6985 & 0.7019 & 0.7054 & 0.7088 & 0.7123 & 0.7157 & 0.7190 & 0.7224 \\
[tex]$0 . 6$[/tex] & 0.7257 & 0.7291 & 0.7324 & 0.7357 & 0.7389 & 0.7422 & 0.7454 & 0.7486 & 0.7517 & 0.7549 \\
[tex]$0 . 7$[/tex] & 0.7580 & 0.7611 & 0.7642 & 0.7673 & 0.7704 & 0.7734 & 0.7764 & 0.7794 & 0.7823 & 0.7852 \\
[tex]$0 . 8$[/tex] & 0.7881 & 0.7910 & 0.7939 & 0.7967 & 0.7995 & 0.8023 & 0.8051 & 0.8078 & 0.8106 & 0.8133 \\
[tex]$0 . 9$[/tex] & 0.8159 & 0.8186 & 0.8212 & 0.8238 & 0.8264 & 0.8289 & 0.8315 & 0.8340 & 0.8365 & 0.8389 \\
1.0 & 0.8413 & 0.8438 & 0.8461 & 0.8485 & 0.8508 & 0.8531 & 0.8554 & 0.8577 & 0.8599 & 0.8621 \\
1.1 & 0.8643 & 0.8665 & 0.8686 & 0.8708 & 0.8729 & 0.8749 & 0.8770 & 0.8790 & 0.8810 & 0.8830
\end{tabular}

- Round the final answer to two decimal places.



Answer :

To solve this problem, we need to determine the probability that the number of students receiving a [tex]$B$[/tex] or above on the final exam is between 45 and 46 in an accounting class. Let's break it down step by step.

1. Understand the parameters given:
- Sample size (\( n \)) = 85
- Population mean (\( \mu \)) = 22
- Population standard deviation (\( \sigma \)) = 13
- We're interested in the interval [19, 23] for the lower and upper bounds.

2. Convert the raw scores (lower bound and upper bound) to z-scores:
- The formula for converting a raw score \( X \) to a z-score is:
[tex]\[ z = \frac{X - \mu}{\sigma / \sqrt{n}} \][/tex]

3. Calculate the z-scores:
- For the lower bound (19):
[tex]\[ z_{\text{lower}} = \frac{19 - 22}{13 / \sqrt{85}} = -2.13 \][/tex]
- For the upper bound (23):
[tex]\[ z_{\text{upper}} = \frac{23 - 22}{13 / \sqrt{85}} = 0.71 \][/tex]

4. Use the Standard Normal Table to find the cumulative probabilities (Φ) for the z-scores:
- For \( z_{\text{lower}} = -2.13 \):
[tex]\[ \Phi(-2.13) \approx 1 - \Phi(2.13) = 1 - 0.9834 = 0.0166 \][/tex]
- For \( z_{\text{upper}} = 0.71 \):
[tex]\[ \Phi(0.71) = 0.7611 \][/tex]

5. Calculate the probability within the z-score range:
[tex]\[ P(19 \leq X \leq 23) = \Phi(0.71) - \Phi(-2.13) = 0.7611 - 0.0166 = 0.7445 \][/tex]

6. Round the final answer to two decimal places:
[tex]\[ \boxed{0.74} \][/tex]

In conclusion, the probability that between 45 or 46 students receive a [tex]$B$[/tex] or above is [tex]\( 0.74 \)[/tex] or 74%.