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This problem deals with a normal distribution for final exam scores, where the mean \(\mu = 61\%\) and the standard deviation \(\sigma = 11\%\). Let's go through each part:

### Part (a)
We are asked to find the probability that a randomly chosen exam mark is at least 70%. This requires finding the **z-score** for 70% and then using a standard normal distribution table.

The z-score formula is:
\[
z = \frac{x - \mu}{\sigma}
\]
Where:
- \( x = 70\% \)
- \( \mu = 61\% \)
- \( \sigma = 11\% \)

\[
z = \frac{70 - 61}{11} = \frac{9}{11} \approx 0.8182
\]

Now, using the standard normal distribution table or a calculator, we can find the area to the left of \(z = 0.8182\), which is approximately 0.7939. Therefore, the area to the right, representing the probability of scoring at least 70%, is:

\[
P(X \geq 70) = 1 - 0.7939 = 0.2061
\]

Thus, the probability is approximately **0.2061**.

### Part (b)
We are asked to find the proportion of students who will not pass the final exam if the passing mark is 50%.

We again use the z-score formula, with \( x = 50\% \):

\[
z = \frac{50 - 61}{11} = \frac{-11}{11} = -1
\]

From the z-table, the area to the left of \( z = -1 \) is approximately 0.1587. Therefore, the proportion of students who will not pass is **0.1587**.

### Part (c)
This part asks for the minimum final exam mark that corresponds to the top 3% of students. We are looking for the z-score that corresponds to the 97th percentile (since the top 3% corresponds to a cumulative area of 0.97).

From the z-table, the z-score corresponding to 0.97 is approximately \(z = 1.88\).

Using the z-score formula to solve for \(x\):
\[
1.88 = \frac{x - 61}{11}
\]
\[
x - 61 = 1.88 \times 11 \approx 20.68
\]
\[
x = 61 + 20.68 = 81.68
\]

Thus, the minimum mark needed for an A is approximately **81.68%**.

### Part (d)
We are asked to find the probability that 4 out of 28 exams will have a grade of less than 50%. The probability of getting less than 50% is \(P(X < 50)\), which we found in part (b) to be **0.1587**.

This is a **binomial probability problem**, where:
- \(n = 28\),
- \(k = 4\),
- \(p = 0.1587\).

The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
Substituting in the values:
\[
P(X = 4) = \binom{28}{4} (0.1587)^4 (1 - 0.1587)^{24}
\]

Let me calculate that for you:

\[
P(X = 4) \approx 0.1923
\]

So, the probability that 4 of the 28 exams will have a grade less than 50% is approximately **0.1923**.

---

Would you like further clarifications or more details on any of these steps? Here are some related questions you might consider:

1. How do z-scores relate to normal distributions?
2. Can you explain how to interpret the z-table?
3. What is the difference between cumulative and point probabilities in a normal distribution?
4. How would the answers change if the mean exam score was 65% instead of 61%?
5. How is the binomial distribution related to the normal distribution?

**Tip**: When solving normal distribution problems, always check if you need to find the area to the left or right of the z-score to avoid mistakes.

The professor of an introductory calculus class has stated that, historically, the distribution of final exam grades in the course resembles a normal distribution with a mean final exam mark of μ = 61% and a standard deviation of σ = 11%. 
(a) What is the probability that a randomly chosen final exam mark in this course will be at least 70%? Answer to four decimals.
(b) In order to pass this course, a student must have a final exam mark of at least 50%. What proportion of students will not pass the final exam? Use four decimals in your answer.
(c) The top 3% of students writing the final exam will receive a letter grade of at least A in the course. To two decimal places, find the minimum final exam mark needed to earn a letter grade of at least A in the course.
(d) Suppose this professor randomly picked 28 final exams, observing the earned mark on each. What is the probability that 4 of these exams will have a grade of less than 50%? Use four decimals in your answer.

Explore probability questions on final exam scores using normal and binomial distributions with detailed solutions. Learn to calculate probabilities for passing, scoring above 70%, achieving top grades, and specific grade outcomes in samples.

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