Final answer:
The problem involves calculating the probability of a sample size of 199 households, with more than 74 but fewer than 111 having personal computers using the normal distribution.
Step-by-step explanation:
The question given is a probability question involving the use of the normal distribution to find the probability that a random sample of 199 households contains more than 74 but fewer than 111 households with personal computers, given that 60% of households have them. To solve this, we need to calculate the mean and standard deviation for the number of households with personal computers in the sample, which can be found as follows:
Mean (μ) = n * p = 199 * 60% = 119.4
Standard Deviation (σ) = sqrt(n * p * (1 - p)) = sqrt(199 * 60% * 40%) ≈ 6.85
Next, we calculate the z-scores for 74 and 111 households:
Z for 74 = (74 - 119.4) / 6.85 ≈ -6.63 (rounded to 2 decimal places)
Z for 111 = (111 - 119.4) / 6.85 ≈ -1.23 (rounded to 2 decimal places)
We then use the normal distribution to find the probabilities associated with these z-scores. We would typically use a statistical calculator or table to find:
P(74 < X < 111) = P(Z < -1.23) - P(Z < -6.63)
Since the probability associated with Z < -6.63 is virtually 0, the probability is mainly determined by Z < -1.23.
The final probability is then estimated using the normal distribution, usually with the help of statistical software or a standard normal distribution table.