Question

Normal Approximation to the Binomial A recent study reported that 85% of students prefer texting to talking on the phone. If 400 students are questioned, what is the probability that more than 300 of these students prefer texting?

Answer 1

The normal approximation to the binomial distribution was applied to find the probability that more than 300 out of 400 students prefer texting over talking on the phone. The calculated z-score was -4.71, and using standard normal distribution, the probability of more than 300 students preferring texting is very small.

Step-by-step explanation:

The question concerns the use of the normal approximation to the binomial distribution to find the probability that more than 300 out of 400 students prefer texting over talking on the phone, given that 85% of students prefer texting. To solve this problem, we calculate the mean (μ) and the standard deviation (σ) for the binomial distribution and then convert the problem to a z-score which compares our result to the standard normal distribution. Since the question requires a normal approximation, we apply the continuity correction by looking for the probability of more than 300.5 students (instead of 300) favoring texting.

First, we find the mean μ = n * p = 400 * 0.85 = 340 and the standard deviation σ = √(n * p * (1-p)) = √(400 * 0.85 * 0.15) ≈ 8.37. Then we calculate the z-score using Z = (x - μ) / σ = (300.5 - 340) / 8.37 = -4.71. Using standard normal distribution tables or software, we find that the probability of a z-score greater than -4.71 is extremely high, practically 1, meaning the probability that more than 300 students prefer texting is accordingly very small (since we're looking at the upper tail of the distribution).