- Identify parameters for the binomial distribution: \(n=162\), \(p_n=0.289\).
- Calculate the mean \(\mu = n p_n = 46.818\) and standard deviation \(\sigma = \sqrt{n p_n (1 - p_n)} \approx 5.7695\).
- Apply the continuity correction for \(P(X > 50)\) to get \(P(Y > 50.5)\) and find the z-score: \(z = \frac{50.5 - 46.818}{5.7695} \approx 0.64\).
- Determine the probability using the standard normal distribution: \(P(Z > 0.64) = 1 - \Phi(0.64) = \boxed{0.2611}\).
Explanation
- Understand the problem and define parameters
We are given that the population of State College, Pennsylvania, is \(71.1\%\) college students. This means the probability that a randomly chosen person is a college student is \(p_s = 0.711\). Consequently, the probability that a person is not a college student is \(p_n = 1 - 0.711 = 0.289\). We consider a random sample of size \(n = 162\). Let \(X\) be the random variable representing the number of people in the sample who are not college students. \(X\) follows a binomial distribution \(B(n, p_n)\) with \(n = 162\) and \(p_n = 0.289\). Our objective is to find the probability \(P(X > 50)\).
- Calculate the mean and standard deviation
Since \(n\) is large, we can approximate the binomial distribution \(B(n, p_n)\) using a normal distribution \(N(\mu, \sigma^2)\).
The mean is calculated as:
\[\mu = n \cdot p_n = 162 \cdot 0.289 = 46.818\]
The standard deviation is calculated as:
\[\sigma = \sqrt{n \cdot p_n \cdot (1 - p_n)} = \sqrt{162 \cdot 0.289 \cdot (1 - 0.289)} = \sqrt{162 \cdot 0.289 \cdot 0.711} \approx 5.7695\]
- Apply continuity correction and find the z-score
To find \(P(X > 50)\) using the normal approximation, we apply a continuity correction. Since \(X\) is a discrete random variable, \(P(X > 50)\) is equivalent to \(P(X \ge 51)\). For the continuous approximation, we use \(P(Y > 50.5)\), where \(Y \sim N(46.818, 5.7695^2)\). We convert this to a z-score as follows:
\[z = \frac{x - \mu}{\sigma} = \frac{50.5 - 46.818}{5.7695} \approx \frac{3.682}{5.7695} \approx 0.638\]
Rounding to two decimal places as requested, we get \(z = 0.64\).
- Determine the final probability
Now, we find the probability \(P(Z > 0.64) = 1 - \Phi(0.64)\), where \(\Phi\) is the cumulative distribution function for the standard normal distribution. Using standard normal distribution tables or calculations, \(\Phi(0.64) \approx 0.7389\). Therefore:
\[P(Z > 0.64) = 1 - 0.7389 = 0.2611\]
Thus, the probability that more than \(50\) people in the sample are not college students is approximately \(0.2611\).
Examples
This method of using the normal approximation to the binomial distribution is frequently used in public health or market research to estimate characteristics of a population based on a sample. For instance, if you want to know the likelihood of exceeding a certain number of customers who prefer a specific service in a sample size of 162, you would apply the same statistical approach: \(\mu = np\), \(\sigma = \sqrt{np(1-p)}\), and \(z = \frac{x - \mu}{\sigma}\) to draw inferences about the population.