Consider the probability that fewer than 19
 out of 158
 students will not graduate on time.  Assume the probability that a given student will not graduate on time is 14%
.

Approximate the probability using the normal distribution. Round your answer to four decimal places.


To solve this problem using the normal distribution, we need to approximate the binomial distribution for the number of students who will not graduate on time.

### Step 1: Set up the parameters
Let \( X \) be the random variable representing the number of students who do not graduate on time. We know:

- The number of trials (students), \( n = 158 \)
- The probability of a student not graduating on time, \( p = 0.14 \)
- The probability of a student graduating on time, \( q = 1 - p = 0.86 \)

### Step 2: Mean and standard deviation of the binomial distribution
For a binomial distribution, the mean \( \mu \) and the standard deviation \( \sigma \) are given by the formulas:

\[
\mu = n \cdot p = 158 \cdot 0.14 = 22.12
\]
\[
\sigma = \sqrt{n \cdot p \cdot q} = \sqrt{158 \cdot 0.14 \cdot 0.86} \approx \sqrt{18.5632} \approx 4.31
\]

### Step 3: Apply the continuity correction
Since we are approximating the binomial distribution with a normal distribution, we use a continuity correction. For the probability that fewer than 19 students will not graduate on time, we adjust the value to 18.5 to account for the discrete nature of the binomial distribution.

Thus, we are looking for:

\[
P(X < 19) \approx P(X \leq 18.5)
\]

### Step 4: Standardize using the Z-score formula
The Z-score is calculated using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

Substitute the values:

\[
Z = \frac{18.5 - 22.12}{4.31} \approx \frac{-3.62}{4.31} \approx -0.84
\]

### Step 5: Find the cumulative probability
Now, we look up the cumulative probability for \( Z = -0.84 \) using the standard normal distribution table or a calculator:

\[
P(Z < -0.84) \approx 0.2005
\]

### Final Answer:
The probability that fewer than 19 out of 158 students will not graduate on time is approximately \( 0.2005 \) (rounded to four decimal places).

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### Related Questions:
1. How would you calculate the probability that more than 19 students will not graduate on time?
2. How can we adjust the solution if the probability of not graduating on time changes to 12%?
3. What is the Z-score for the probability of exactly 20 students not graduating on time?
4. How do you determine if a normal approximation is appropriate for a binomial distribution?
5. How would the standard deviation change if the number of students increased to 200?

**Tip:** Always ensure that the normal approximation to the binomial distribution is reasonable. It's typically valid when \( n \) is large and both \( np \) and \( nq \) are greater than 5.

Consider the probability that fewer than 19 out of 158 students will not graduate on time. Assume the probability that a given student will not graduate on time is 14%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Learn how to approximate the probability that fewer than 19 out of 158 students will not graduate on time using the normal distribution. The solution covers the binomial distribution, normal approximation, and Z-score calculation. The final probability is approximately 0.2005. Suitable for students in Grades 11-12.

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