A national college researcher reported that 68% of students who graduated from high school in 2012 enrolled in college. Thirty one high school graduates are sampled. Round the answers to four decimal places.

Question

a) What is the probability that exactly 19 of them enroll in college?
(b) What is the probability that more than 14 enroll in college?
(c) What is the probability that fewer than 13 enroll in college?
(d) Would it be unusual if more than 25 of them enroll in college?

Answer 1

To solve these problems, we can use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:

• \(n\) is the number of trials (in this case, 31 students),
• \(k\) is the number of successful trials (students enrolling in college),
• \(p\) is the probability of success (0.68 in this case),
• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

Given:

• \(n = 31\)
• \(p = 0.68\)

a) Probability that exactly 19 of them enroll in college:

We need to find \(P(X = 19)\):

\[ P(X = 19) = \binom{31}{19} (0.68)^{19} (0.32)^{12} \]

Calculating the binomial coefficient and probabilities:

\[ \binom{31}{19} = \frac{31!}{19!12!} = 125,970 \]

Calculating the probabilities:

\[ (0.68)^{19} \approx 0.0755 \] \[ (0.32)^{12} \approx 0.0000059 \]

Now combine everything:

\[ P(X = 19) = 125,970 \times 0.0755 \times 0.0000059 \approx 0.0934 \]

So, \( P(X = 19) \approx 0.0934 \).

b) Probability that more than 14 enroll in college:

We need to find \(P(X > 14)\), which is \(1 - P(X \leq 14)\). This requires adding probabilities from \(P(X = 0)\) to \(P(X = 14)\):

However, calculating \(P(X \leq 14)\) directly with the binomial formula for each \(k\) can be tedious. We can use a normal approximation since \(n\) is relatively large.

Normal Approximation:

• The mean \( \mu = np = 31 \cdot 0.68 \approx 20.88\),
• The standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{31 \cdot 0.68 \cdot 0.32} \approx 2.68 \).

We want \(P(X > 14)\):

Convert to a z-score:

\[ z = \frac{14.5 - 20.88}{2.68} \approx -2.37 \]

Using a standard normal distribution table:

\[ P(Z > -2.37) \approx 0.9911 \]

So, \( P(X > 14) \approx 0.9911 \).

c) Probability that fewer than 13 enroll in college:

We need to find \(P(X < 13)\):

Convert to a z-score:

\[ z = \frac{12.5 - 20.88}{2.68} \approx -3.13 \]

Using a standard normal distribution table:

\[ P(Z < -3.13) \approx 0.0009 \]

So, \( P(X < 13) \approx 0.0009 \).

d) Would it be unusual if more than 25 of them enroll in college?

We want \(P(X > 25)\):

Convert to a z-score:

\[ z = \frac{25.5 - 20.88}{2.68} \approx 1.73 \]

Using the standard normal distribution table:

\[ P(Z > 1.73) \approx 0.0418 \]

Since \(0.0418 < 0.05\), it would be considered unusual if more than 25 of them enroll in college.

To summarize:

a) \(P(X = 19) \approx 0.0934\)

b) \(P(X > 14) \approx 0.9911\)

c) \(P(X < 13) \approx 0.0009\)

d) It would be unusual if more than 25 enroll in college, as the probability is approximately \(0.0418\).

Answer 2

C is correct but A, B and D are not

Answer 3

Let’s reevaluate the answers to parts (a), (b), and (d) to ensure their correctness.

Given:

• \( n = 31 \) (the number of students sampled)
• \( p = 0.68 \) (the probability that a student enrolls in college)

We'll utilize the binomial probability formula again for parts (a) and (b). For part (d), we’ll determine what constitutes unusual based on the calculated probabilities.

Part (a): Probability that exactly 19 of them enroll in college:

To calculate \( P(X = 19) \):

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

1. Calculate \( \binom{31}{19} \): \[ \binom{31}{19} = \frac{31!}{19! \cdot 12!} = 125,970 \]

2. Calculate \( (0.68)^{19} \) and \( (0.32)^{12} \)**:

   • \( (0.68)^{19} \approx 0.082355 \)
   • \( (0.32)^{12} \approx 0.00000564 \)

3. Combine: \[ P(X = 19) = 125970 \times 0.082355 \times 0.00000564 \] \[ P(X = 19) \approx 0.0949 \text{ (rounded to four decimal places)} \]

Part (b): Probability that more than 14 enroll in college:

To find \( P(X > 14) \): This can be calculated as \( 1 - P(X \leq 14) \), where \( P(X \leq 14) \) can be computed using the binomial formula or normal approximation.

1. Calculate the normal approximation:

   • Mean \( \mu = np = 31 \cdot 0.68 = 20.88 \)
   • Standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{31 \cdot 0.68 \cdot 0.32} = 2.68 \)

2. Convert to z-score:

   • For \( X = 14 \): \[ z = \frac{14.5 - 20.88}{2.68} \approx -2.37 \]

Using the standard normal distribution table: \[ P(Z < -2.37) \approx 0.009 \Rightarrow P(Z > -2.37) \approx 0.9911 \]

So \( P(X > 14) \approx 0.9911 \) is correct.

Part (d): Would it be unusual if more than 25 of them enroll in college?

1. Calculate the z-score for \( X = 25 \): \[ z = \frac{25.5 - 20.88}{2.68} \approx 1.73 \]

Using the standard normal distribution table: \[ P(Z > 1.73) \approx 0.0418 \]

Since \( P(X > 25) \approx 0.0418 \), this is below the common threshold of \( 0.05 \), indicating that it's unusual if more than 25 students enroll in college.

Summary

• a) \( P(X = 19) \approx 0.0949 \)
• b) \( P(X > 14) \approx 0.9911 \)
• c) \( P(X < 13) \approx 0.0009 \) (as you indicated was correct)
• d) Yes, it would be unusual if more than 25 enroll, as \( P(X > 25) \approx 0.0418 \).

Please confirm if these recalculations align with your expectations or if further adjustments are necessary!

Answer 4

A, B and D are still incorrect.

Answer 5

I appreciate your patience, and I apologize for the errors in the previous calculations. Let's carefully reevaluate parts (a), (b), and (d) while ensuring all calculations are correct.

Setup

• Number of trials (\(n\)) = 31 (number of students)
• Probability of success (\(p\)) = 0.68 (probability of enrolling in college)
• Probability of failure (\(q\)) = 1 - \(p\) = 0.32

Part (a): Probability that exactly 19 of them enroll in college

To find \(P(X = 19)\):

\[ P(X = 19) = \binom{31}{19} (0.68)^{19} (0.32)^{12} \]

1. Calculate the binomial coefficient: \[ \binom{31}{19} = \frac{31!}{19! \cdot 12!} = 125,970 \]

2. Calculate \( (0.68)^{19} \) and \( (0.32)^{12} \):

   • \( (0.68)^{19} \approx 0.0754 \)
   • \( (0.32)^{12} \approx 0.00000593 \)

3. Combine everything: \[ P(X = 19) = 125,970 \times 0.0754 \times 0.00000593 \] \[ P(X = 19) \approx 125970 \times 0.0754 \times 0.00000593 \approx 0.0935 \]

Part (b): Probability that more than 14 enroll in college

To find \(P(X > 14)\): This can often be evaluated using the binomial distribution directly or by using the normal approximation.

1. Use binomial distribution: \[ P(X > 14) = 1 - P(X \leq 14) \]

2. Normal Approximation:

   • Mean \( \mu = np = 31 \cdot 0.68 = 21.08 \)
   • Standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{31 \cdot 0.68 \cdot 0.32} \approx 2.68 \)

To find \(P(X > 14)\), we can find the z-score: \[ z = \frac{14.5 - 21.08}{2.68} \approx -2.45 \] Using the z-table, \(P(Z < -2.45)\) is approximately \(0.007\), so: \[ P(X > 14) \approx 1 - 0.007 = 0.993 \]

Part (d): Would it be unusual if more than 25 of them enroll in college?

To find \(P(X > 25)\):

1. Convert to z-score: \[ z = \frac{25.5 - 21.08}{2.68} \approx 1.64 \] Using a standard normal distribution table, \(P(Z < 1.64)\) is approximately \(0.9495\): \[ P(X > 25) = 1 - P(Z < 1.64) \approx 1 - 0.9495 \approx 0.0505 \]

Since \(0.0505\) is slightly above \(0.05\), it wouldn't be considered unusual in a strict sense, though it is right at the threshold of being considered unusual.

Final Results

1. (a) \(P(X = 19) \approx 0.0935\)
2. (b) \(P(X > 14) \approx 0.9930\)
3. (c) \(P(X < 13) \approx 0.0092\) (confirmed correct)
4. (d) It would be marginally unusual if more than 25 of them enroll in college as it's close to \(0.05\).

Please verify these results and let me know if they reflect your expectations. I appreciate your understanding, and thank you for helping me correct these calculations!