This problem involves using the normal approximation to the binomial distribution since $n = 200$ is large. The random variable $X$, representing the number of cars equipped with studs, follows a binomial distribution:

$X \sim \text{Binomial}(n=200, p=0.65).$

We need to find $P(X \geq 130)$.

Step 1: Check conditions for normal approximation

The normal approximation to the binomial distribution is valid if both $np$ and $n(1-p)$ are greater than 5: $np = 200 \times 0.65 = 130, \quad n(1-p) = 200 \times 0.35 = 70.$ Both conditions are satisfied.

Step 2: Compute mean and standard deviation

For a binomial distribution: $\mu = np = 130, \quad \sigma = \sqrt{np(1-p)} = \sqrt{200 \times 0.65 \times 0.35} \approx 6.7912.$

Step 3: Apply the normal approximation

Under the normal approximation: $X \sim N(\mu = 130, \sigma^2 = 6.7912^2).$

To find $P(X \geq 130)$, apply the continuity correction: $P(X \geq 130) \approx P\left(Z \geq \frac{130 - 130 + 0.5}{6.7912}\right) = P\left(Z \geq \frac{0.5}{6.7912}\right).$

Step 4: Standardize and compute the z-score

$Z = \frac{0.5}{6.7912} \approx 0.0736.$

Using standard normal tables or a calculator: $P(Z \geq 0.0736) = 1 - P(Z \leq 0.0736).$

From the standard normal table: $P(Z \leq 0.0736) \approx 0.5294.$

Thus: $P(Z \geq 0.0736) = 1 - 0.5294 = 0.4706.$

Final Answer:

The probability that at least 130 of the 200 cars are equipped with studs is approximately: P(X \geq 130) \approx 0.4706 \, \text{(47.06%)}.

Let me know if you want more details or have any questions!

Related Questions:

1. What would the probability be if we increased the sample size to 300 cars?
2. How does the probability change if the proportion of cars with studs increases to 70%?
3. Can you calculate the probability that exactly 130 cars have studs using the binomial formula?
4. What is the z-score for $X = 125$ in this context?
5. How would you simulate this problem using random numbers?

Tip:

When using the normal approximation to the binomial, always remember to apply the continuity correction for more accurate results.