Question 1169044: Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 144 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.
Probability that fewer than 37 voted
The probability that fewer than 37 of 144 eligible voters voted is ____.
(Round to four decimal places as needed.)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem using the normal approximation to the binomial distribution:
**1. Identify the parameters:**
* Number of trials (voters selected): $n = 144$
* Probability of success (a voter voted): $p = 0.22$
* Number of successes we're interested in (fewer than 37): $x < 37$
**2. Calculate the mean and standard deviation of the binomial distribution:**
* Mean: $\mu = np = 144 \times 0.22 = 31.68$
* Standard deviation: $\sigma = \sqrt{np(1-p)} = \sqrt{144 \times 0.22 \times (1-0.22)} = \sqrt{144 \times 0.22 \times 0.78} = \sqrt{24.6912} \approx 4.969$
**3. Apply the continuity correction:**
Since we are approximating a discrete distribution (binomial) with a continuous distribution (normal), we need to apply a continuity correction. "Fewer than 37" means we are interested in $x = 0, 1, 2, ..., 36$. To account for the continuous nature of the normal distribution, we extend this range slightly to the right, up to $36.5$. So, we want to find the probability that the normal random variable is less than $36.5$.
**4. Calculate the z-score:**
The z-score standardizes the value of interest with respect to the mean and standard deviation of the normal distribution:
$z = \frac{x - \mu}{\sigma} = \frac{36.5 - 31.68}{4.969} = \frac{4.82}{4.969} \approx 0.9700$
**5. Find the probability using the z-score:**
We want to find the probability $P(Z < 0.9700)$, where $Z$ is a standard normal random variable. We can look this value up in a standard normal distribution table or use a statistical calculator.
Looking up $z = 0.97$ in a standard normal table, we find the area to the left is approximately $0.8340$.
**Therefore, the probability that fewer than 37 of 144 eligible voters voted is approximately 0.8340.**
Final Answer: The final answer is $\boxed{0.8340}$
|
|
|
