https://file.aiquickdraw.com/mathbot/file/1729041929169e40nb7rl.jpg

To calculate the **mean grade** in this course, we'll use the probability distribution provided in the table. The mean, denoted as \( \mu \), is the expected value of the grade, which can be calculated using the formula:

\[
\mu = \sum (x_i \times p_i)
\]

where \( x_i \) is the grade (0 for F, 1 for D, 2 for C, 3 for B, 4 for A), and \( p_i \) is the corresponding probability for each grade.

From the table:
- Grade \( 0 \) (F): Probability = 0.04
- Grade \( 1 \) (D): Probability = 0.04
- Grade \( 2 \) (C): Probability = 0.31
- Grade \( 3 \) (B): Probability = 0.43
- Grade \( 4 \) (A): Probability = 0.18

Now, calculating \( \mu \):

\[
\mu = (0 \times 0.04) + (1 \times 0.04) + (2 \times 0.31) + (3 \times 0.43) + (4 \times 0.18)
\]

\[
\mu = 0 + 0.04 + 0.62 + 1.29 + 0.72 = 2.67
\]

So, the mean grade in this course is **2.67**.

---

Next, for the **standard deviation** (denoted \( \sigma \)), we use the formula:

\[
\sigma = \sqrt{\sum (x_i - \mu)^2 \times p_i}
\]

We'll calculate each term \( (x_i - \mu)^2 \) and then find the weighted sum:

1. For Grade \( 0 \): \( (0 - 2.67)^2 = 7.1289 \)
2. For Grade \( 1 \): \( (1 - 2.67)^2 = 2.7889 \)
3. For Grade \( 2 \): \( (2 - 2.67)^2 = 0.4489 \)
4. For Grade \( 3 \): \( (3 - 2.67)^2 = 0.1089 \)
5. For Grade \( 4 \): \( (4 - 2.67)^2 = 1.7689 \)

Now multiply by the respective probabilities:

\[
\sigma^2 = (7.1289 \times 0.04) + (2.7889 \times 0.04) + (0.4489 \times 0.31) + (0.1089 \times 0.43) + (1.7689 \times 0.18)
\]

\[
\sigma^2 = 0.285156 + 0.111556 + 0.139159 + 0.046827 + 0.318402 = 0.9011
\]

\[
\sigma = \sqrt{0.9011} = 0.9492
\]

Given that the image shows \( \sigma = 0.9491 \), rounding differences explain the slight variation.

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Would you like more details or have further questions?

Here are 5 related questions to explore:
1. How would the mean change if the percentage of A grades increased?
2. What if the standard deviation were higher—what would that indicate about grade distribution?
3. Can you compute the variance directly from the standard deviation?
4. How can you interpret a mean of 2.67 in terms of the grading scale?
5. If the distribution were uniform, what would be the expected mean?

**Tip**: Remember, standard deviation tells you how spread out the grades are from the mean!

What is the mean and standard deviation of grades for a Statistics 101 course, given the probability distribution for grades?

Calculate the mean and standard deviation for a Statistics 101 course using the probability distribution of grades, including step-by-step calculations for both mean and standard deviation. This problem covers key statistical concepts such as expected value and spread of a distribution.

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