Let the random variable \(X\) represent the number of people who choose a mobile phone made by Company B. \(X\) follows a binomial distribution with parameters \(n = 130\) and \(p = 0.35\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 130 \times 0.35 = 45.5\)

\(\sigma^2 = np(1-p) = 130 \times 0.35 \times 0.65 = 29.575\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{29.575} \approx 5.438\).

We need to find \(P(X \geq 50)\). Using the normal approximation, we standardize:

\(P(X \geq 50) = P\left( Z \geq \frac{49.5 - 45.5}{5.438} \right) = P(Z \geq 0.7355)\)

Using the standard normal distribution table, \(P(Z \geq 0.7355) = 1 - \Phi(0.7355) \approx 1 - 0.7691 = 0.231\).