Final answer:
The probability that at least 30% of customers in a sample of 50 will make a purchase from the website is 0.0015 or 0.15%, determined by using the normal approximation to the binomial distribution and calculating the corresponding z-score.
Step-by-step explanation:
To determine the probability that a random sample of 50 customers will have at least 30% making a purchase, given that only 15% of customers visiting the website make a purchase, we will use the normal approximation to the binomial distribution. Since the sample size is large, we'll assume that the sample proportion, p', is approximately normally distributed.
The probability of a customer making a purchase is p = 0.15, and so the probability of not making a purchase is q = 1 - p = 0.85. For a sample of n = 50, the expected number of customers making a purchase is np = 50 * 0.15 = 7.5, and the variance is npq = 50 * 0.15 * 0.85 = 6.375. The standard deviation is therefore the square root of the variance, which is about 2.525.
To find the z-score for a proportion of 30%, we first convert this proportion to the number of customers in our sample, which is 0.3 * 50 = 15. The z-score is then (X - np) / (sqrt(npq)), which is (15 - 7.5) / 2.525 = 7.5 / 2.525 = 2.97 approximately.
Using the provided probability values, P(z < 2.97) = 0.9985, which means nearly all the area under the curve is to the left of our z-score. However, we want the area to the right since we are looking for the probability of at least this many customers making a purchase.
The probability we want is 1 - P(z < 2.97) = 1 - 0.9985 = 0.0015. Therefore, the probability that at least 30% of the customers in a sample of 50 will make a purchase after visiting the website is 0.0015 or 0.15%.
Note that this is a simplified approach and assumes a normal distribution, which should be appropriate given the large enough sample size and under specific conditions for the central limit theorem to hold.