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%.
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%.
