Question

Approximately 10.3% of American high school students drop out of school before graduation. Choose 10 students entering high school at random. Find the probability that at least 6 graduate. Round your answer to the nearest thousandth.

Answer 1

To solve this problem, we can model the situation using a binomial distribution. We define success as a student graduating high school (which has a probability of \( p = 1 - 0.103 = 0.897 \)). The number of trials is \( n = 10 \), as we are choosing 10 students.

We want to find the probability that at least 6 students graduate. This can be expressed as:

\[
P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
\]

where \( X \) is the number of students graduating and follows a binomial distribution \( X \sim B(n = 10, p = 0.897) \).

The probability mass function of a binomial distribution is given by:

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

where \( \binom{n}{k} \) is the binomial coefficient.

Now, we can calculate each of these probabilities:

1. **For \( k = 6 \)**:
\[
P(X = 6) = \binom{10}{6} (0.897)^6 (0.103)^4
\]
\[
= 210 \cdot (0.897)^6 \cdot (0.103)^4 \approx 210 \cdot 0.536 \cdot 0.000115 \approx 0.012505
\]

2. **For \( k = 7 \)**:
\[
P(X = 7) = \binom{10}{7} (0.897)^7 (0.103)^3
\]
\[
= 120 \cdot (0.897)^7 \cdot (0.103)^3 \approx 120 \cdot 0.479 \cdot 0.001092 \approx 0.062887
\]

3. **For \( k = 8 \)**:
\[
P(X = 8) = \binom{10}{8} (0.897)^8 (0.103)^2
\]
\[
= 45 \cdot (0.897)^8 \cdot (0.103)^2 \approx 45 \cdot 0.428 \cdot 0.010609 \approx 0.182486
\]

4. **For \( k = 9 \)**:
\[
P(X = 9) = \binom{10}{9} (0.897)^9 (0.103)^1
\]
\[
= 10 \cdot (0.897)^9 \cdot (0.103)^1 \approx 10 \cdot 0.383 \cdot 0.103 \approx 0.394065
\]

5. **For \( k = 10 \)**:
\[
P(X = 10) = \binom{10}{10} (0.897)^{10} (0.103)^0
\]
\[
= 1 \cdot (0.897)^{10} \approx 0.343
\]

Now, we sum these probabilities:

\[
P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
\]

Calculating the total:

\[
P(X \geq 6) \approx 0.012505 + 0.062887 + 0.182486 + 0.394065 + 0.343963 \approx 0.995906
\]

Rounding to the nearest thousandth:

\[
P(X \geq 6) \approx 0.996
\]

Thus, the probability that at least 6 of the 10 randomly chosen students graduate is approximately **0.996**.