Experience suggests that 4 percent of all college students have had a tonsillectomy.In a sample of 300 college students,we need to find the probability that at least 10 have had a tonsillectomy.It is acceptable to use the normal distribution to estimate this probability.

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Since the sample size is reasonably large (n=300>30), we can use the normal distribution to estimate the probability. We can use the formula for the sample proportion: p̂ = x/n where x is the number of students who had a tonsillectomy in the sample (we don't know this value yet), and n is the sample size (n=300). The expected value of p̂ is equal to the population proportion, so E(p̂) = 0.04. The standard deviation of p̂ is calculated as: σp̂ = sqrt[p(1-p)/n] where p is the population proportion (p = 0.04). So σp̂ = sqrt[(0.04)(1-0.04)/300] = 0.016 Now we can standardize the variable p̂ using the z-score formula: z = (p̂ - E(p̂))/σp̂ We want to find the probability that at least 10 students out of 300 have had a tonsillectomy, which means p̂ >= 0.0333 (since 0.0333*300 = 10). So we can calculate the z-score: z = (0.0333 - 0.04)/0.016 = -0.4375 The area to the left of this z-score represents the probability that fewer than 10 students in the sample have had a tonsillectomy. So we need to subtract this value from 1 to get the probability that at least 10 students have had a tonsillectomy: P(p̂ >= 0.0333) = 1 - P(p̂ < 0.0333) = 1 - P(z < -0.4375) Using a standard normal table or calculator, we can find that P(z < -0.4375) = 0.3308. Therefore, P(p̂ >= 0.0333) = 1 - 0.3308 = 0.6692. So the likelihood that at least 10 students in a sample of 300 have had a tonsillectomy is about 0.6692 or 66.92%.

Experience Suggests That 4 Percent of All College Students Have