- Identify the binomial parameters: \(n = 200\), \(p = 0.4872\), \(\mu = 97.44\), \(\sigma \approx 7.0688\).
- Apply continuity correction for \(P(X < 94)\) as \(P(X \le 93.5)\).
- Calculate the \(z\)-score: \(z = \frac{93.5 - 97.44}{7.06875} \approx -0.56\).
- Use the standard normal table to find \(P(Z < -0.56) \approx 0.2877\).
- The final probability is \(\boxed{0.2877}\).
Explanation
- Analyze the problem and data conditions
We are interested in the number of people in a sample of \(n=200\) who do not use the top web browser. Given the top browser's market share is \(51.28\%\), the probability of a person not using this browser is \(p = 1 - 0.5128 = 0.4872\). The random variable \(X\), representing the count of people who do not use the top browser, follows a binomial distribution \(X \sim \text{Binomial}(n=200, p=0.4872)\). To approximate this distribution using the normal distribution, we first check the conditions \(np = 200 \times 0.4872 = 97.44\) and \(n(1-p) = 200 \times 0.5128 = 102.56\). Since both values are greater than \(5\), the normal approximation is appropriate.
- Calculate the mean and standard deviation
The mean of the distribution is given by \(\mu = np = 97.44\). The standard deviation is \(\sigma = \sqrt{np(1-p)} = \sqrt{97.44 \times 0.5128} \approx \sqrt{49.9672} \approx 7.0688\).
- Apply continuity correction and find the z-score
We want to find the probability that \(X < 94\). Because we are approximating a discrete distribution (binomial) with a continuous one (normal), we use the continuity correction. The event \(X < 94\) in the binomial case is equivalent to \(X \le 93.5\) in the normal case. Now, we calculate the \(z\)-score for \(x = 93.5\):
\[z = \frac{x - \mu}{\sigma} = \frac{93.5 - 97.44}{7.06875} = \frac{-3.94}{7.06875} \approx -0.5574\]
Following the instruction to round to two decimal places, we use \(z = -0.56\).
- Find the final probability
Using the standard normal distribution table or a calculator, we look up the cumulative probability \(P(Z < -0.56)\). The value is approximately \(0.2877\). Thus, the probability that fewer than 94 people do not use the top web browser is approximately \(0.2877\).
Examples
This approach is frequently used in market research to predict consumer behavior or preferences within a population. For example, if a company wants to estimate how many users prefer a competitor's product, they can use this statistical method to determine the likelihood of certain outcomes in their user feedback surveys, helping them make informed business decisions based on sample data.