Question

Withdrawal symptoms may occur when a person using a painkiller suddenly stops using it. For a special type of painkiller, withdrawal symptoms occur in 5% of the cases. A random sample of 1900 people who have stopped using the painkiller is going to be taken. Let p be the proportion of people in the sample who experience withdrawal symptoms.

(a) Find the mean of p (µp).

(b) Find the standard deviation of p (σp).

(c) Compute an approximation for P(p > 0.06), which is the probability that more than 6% of the people in the sample experience withdrawal symptoms. Round your answer to four decimal places.

Answer 1

The mean of the sample proportion p is 0.05, its standard deviation is approximately 0.004887, and the probability that more than 6% of the people in the sample experience withdrawal symptoms is approximately 0.0202.

Step-by-step explanation:

Solving the Problem Step-by-Step

(a) Mean of p (µp): The mean of the sample proportion p is equal to the population proportion. Since 5% of the cases show withdrawal symptoms, the mean of p is 0.05.

(b) Standard Deviation of p (σp): The standard deviation of the sample proportion is given by √(pq/n), where p is the population proportion, q is the complement of the population proportion (1-p), and n is the sample size. In this case, σp = √(0.05*0.95/1900) ≈ 0.004887.

(c) Probability P(p > 0.06): To compute an approximation for the probability that more than 6% of the people in the sample experience withdrawal symptoms, we use the normal distribution approximation. First, find the z-score which is (p - µp) / σp. The z-score for p = 0.06 is (0.06 - 0.05) / 0.004887 ≈ 2.048. Using the standard normal distribution table or a calculator, we can find the probability that corresponds to this z-score. The area to the right of this z-score gives us P(p > 0.06), which is approximately 0.0202 after rounding to four decimal places.