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3 votes
Suppose the scores of students on a Statistics course are

Normally distributed with a mean of 588 and a standard deviation of
46.
What percentage of the students scored between 588 and 680 on the
exam

2 Answers

3 votes

Answer: 68%

Explanation:

To find the percentage of students who scored between 588 and 680 on the exam, we can use the properties of the normal distribution and z-scores.

1. **Calculate Z-scores:**

The formula for calculating the z-score is: \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For 588:

\(z_{588} = \frac{588 - 588}{46} = 0\)

For 680:

\(z_{680} = \frac{680 - 588}{46} \approx 2.00\)

2. **Find the Percentage:**

To find the percentage between these two z-scores, we can use a standard normal distribution table (also known as a Z-table).

The Z-table provides the cumulative probability up to a certain z-score. Looking up the z-score of 2.00 in the Z-table, we find that the cumulative probability is approximately 0.9772.

The cumulative probability up to the z-score of 0 (which corresponds to the mean) is 0.5000.

3. **Calculate the Percentage:**

The percentage of students scoring between 588 and 680 is the difference between these two cumulative probabilities:

Percentage = \(0.9772 - 0.5000 \approx 0.4772\)

To express this as a percentage, we multiply by 100:

Percentage ≈ \(0.4772 \times 100 \approx 47.72\)%.

Therefore, approximately 47.72% of the students scored between 588 and 680 on the exam.

User Gruangly
by
8.9k points
1 vote

Answer:

Step-by-step explanation: Im not sure you should search it up.

User Luca Angeletti
by
8.1k points

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