- Calculate the mean \(\mu = 30\) and standard deviation \(\sigma \approx 3.4641\).
- Apply the continuity correction for \(X=32\) to get the interval \([31.5, 32.5]\).
- Determine the standardized z-scores: \(z_1 = 0.43\) and \(z_2 = 0.72\).
- Find the area between the z-scores: \(\boxed{0.0978}\).
Explanation
- Calculate distribution parameters
To approximate the binomial distribution using the normal distribution, we first identify the parameters of the binomial distribution: \(n = 50\) trials and success probability \(p = 0.6\). The mean \(\mu\) and variance \(\sigma^2\) are calculated as follows:
- Mean: \(\mu = np = 50 \times 0.6 = 30\)
- Variance: \(\sigma^2 = np(1-p) = 50 \times 0.6 \times (1 - 0.6) = 30 \times 0.4 = 12\)
- Standard deviation: \(\sigma = \sqrt{12} \approx 3.4641\).
- Apply continuity correction
Since we are approximating a discrete distribution with a continuous normal distribution, we apply the continuity correction. To find the probability \(P(X = 32)\), we look for the area under the normal curve between the boundaries \(31.5\) and \(32.5\):
\[P(31.5 < X < 32.5)\]
- Calculate z-scores
Now, we calculate the z-scores for the boundaries \(31.5\) and \(32.5\) to standardize the values:
- Lower bound z-score: \(z_1 = \frac{31.5 - \mu}{\sigma} = \frac{31.5 - 30}{3.4641} \approx \frac{1.5}{3.4641} \approx 0.43\)
- Upper bound z-score: \(z_2 = \frac{32.5 - 30}{3.4641} \approx \frac{2.5}{3.4641} \approx 0.72\)
Rounding to two decimal places, we have \(z_1 = 0.43\) and \(z_2 = 0.72\).
- Determine the final probability
Using the standard normal distribution table or a statistical function, we find the cumulative probabilities for our z-scores:
- \(P(Z < 0.72) \approx 0.7642\)
- \(P(Z < 0.43) \approx 0.6664\)
The probability of the variable falling within the range is:
\[P(0.43 < Z < 0.72) = P(Z < 0.72) - P(Z < 0.43)\]
\[P(0.43 < Z < 0.72) = 0.7642 - 0.6664 = 0.0978\]
Thus, the approximated probability is \(\boxed{0.0978}\).
Examples
Normal approximation is widely used in quality control and statistics to quickly estimate the probability of specific outcomes in large samples without performing cumbersome calculations. For instance, a manufacturing company checking the number of defective items in a batch of 50 can use this method to understand the likelihood of seeing a specific number of defects and decide if their process is within acceptable limits.