Question

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (Round your answer to two decimal places.) hours

Answer 1

To find the lifetime value such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages, we can use the inverse of the cumulative distribution function (CDF) of the normal distribution. The lifetime value is approximately 16.65 hours.

Step-by-step explanation:

To find the lifetime value such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages, we need to find the inverse of the cumulative distribution function (CDF) of the normal distribution. The mean value is 15 hours and the standard deviation is 1 hour. We can use a standard normal distribution table or a statistical calculator to find the z-score corresponding to a cumulative probability of 0.95 (1 - 0.05). Using the z-score, we can then find the corresponding lifetime value.

Step 1: Find the z-score using the cumulative probability of 0.95. The closest value we can find in the table is 1.645. This means that 95% of the area under the curve is to the left of this z-score.

Step 2: Use the z-score to find the lifetime value. The formula to calculate the value is: x = mean + (z-score * standard deviation). Substituting the values, we get: x = 15 + (1.645 * 1) = 16.645. Therefore, the lifetime value is approximately 16.65 hours.