Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 10 hours. If the manufacturer of the battery is able to reduce the standard deviation of battery life from 10 to 9 hours, what would be the probability that 16 batteries randomly sampled from the population will have a sample mean life of between 70 and 80 hours?
A) 0.6127
B) 0.8124
C) 0.9736
D) 0.8812

Question

Quizplus · Accepted Answer

To find the probability that the sample mean life of 16 batteries is between 70 and 80 hours, we use the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size (n). With the new standard deviation of 9 hours, the standard error is 9 / sqrt(16) = 9 / 4 = 2.25 hours. We then use the Z-score formula for each boundary: Z = (X - μ) / σ, where X is the boundary value, μ is the mean, and σ is the standard error. For X = 70, Z = (70 - 75) / 2.25 = -2.22. For X = 80, Z = (80 - 75) / 2.25 = 2.22. Looking up these Z-scores in the standard normal distribution table or using a calculator, we find the probabilities corresponding to these Z-scores and subtract the lower from the upper to find the probability of the sample mean being between 70 and 80 hours. The result is approximately 0.9736, indicating a high probability.

Suppose the Life of a Particular Brand of Calculator Battery