We are given that 35% of customers shop online, so the probability of a customer shopping online is 0.35. For a sample of 100 customers, we can model the number of customers shopping online as a binomial distribution with parameters \(n = 100\) and \(p = 0.35\).
To approximate this binomial distribution, we use a normal distribution with the same mean and variance. The mean \(\mu\) is given by:
\(\mu = np = 100 \times 0.35 = 35\)
The variance \(\sigma^2\) is given by:
\(\sigma^2 = np(1-p) = 100 \times 0.35 \times 0.65 = 22.75\)
Thus, the standard deviation \(\sigma\) is \(\sqrt{22.75}\).
We need to find \(P(X > 39)\). Using the normal approximation, we apply a continuity correction and find \(P(X > 39.5)\).
We standardize this to find the corresponding \(z\)-score:
\(z = \frac{39.5 - 35}{\sqrt{22.75}} = 0.943\)
We then find \(P(Z > 0.943)\) using the standard normal distribution table:
\(P(Z > 0.943) = 1 - P(Z < 0.943) = 1 - 0.8272 = 0.173\)
Therefore, the probability that more than 39 customers shop online is 0.173.
