Final answer:
The probability that the mean life of a sample of 84 monitors is greater than 96.3 months is 0.0087, or 0.87%, assuming the population mean is 95 months and the standard deviation is 5 months.
Step-by-step explanation:
To find the probability that the mean monitor life of a sample of 84 monitors would be greater than 96.3 months, given the claim that the monitors have a mean life of 95 months with a standard deviation of 5 months, we need to use the concept of the sampling distribution of the sample mean.
The standard error (SE) of the mean is calculated using the formula SE = σ/√n, where σ is the population standard deviation and n is the sample size. Plugging in the values, we get SE = 5 / √84 ≈ 0.5456.
Next, we calculate the z-score for 96.3 months using the formula z = (X - μ) / SE, where X is the sample mean, and μ is the population mean. So, z = (96.3 - 95) / 0.5456 ≈ 2.3839.
Finally, we look up the z-score in a standard normal distribution table or use a calculator with normal distribution functions to find the probability associated with the calculated z-score. The probability of a z-score greater than 2.3839 is about 0.0087. Therefore, the probability that a sample of 84 monitors has a mean life greater than 96.3 months is 0.0087, or 0.87%.