Final answer:
To determine the probability that the sample mean differs from the true mean by less than 5.4 months, calculate the Z-score and consult the Z-table. The Z-score is approximately 2.73, leading to a probability of about 0.9967, with the closest answer choice being 0.9772 (option d).
Step-by-step explanation:
The question pertains to the probability that the sample mean of 83 television sets will be within 5.4 months of the true mean life of television sets, given a population mean (μ) of 138 months and a variance (σ2) of 324 months2. First, we calculate the standard deviation (σ) which is the square root of the variance. So, σ = √324 = 18 months. Next, we find the standard error of the mean (SEM), which is given by σ/√n, where n is the sample size. Here, n = 83, making the SEM = 18/√83 ≈ 1.98 months. We now express our range of interest (within 5.4 months of the mean) as a Z-score by dividing by the SEM, yielding 5.4 / 1.98 ≈ 2.73. Using the Z-table, we find that the probability of being within ±2.73 standard deviations of the mean is approximately 0.9967 (98.34% on both sides minus 1.32% beyond both tails = 0.9967). This suggests that the closest answer choice would be 0.9772, corresponding to option d.