Questions: The life of light bulbs is distributed normally. The variance of the lifetime is 900 and the mean lifetime of a bulb is 580 hours. Find the probability of a bulb lasting for at most 610 hours. Round your answer to four decimal places.

The life of light bulbs is distributed normally. The variance of the lifetime is 900 and the mean lifetime of a bulb is 580 hours. Find the probability of a bulb lasting for at most 610 hours. Round your answer to four decimal places.

Transcript text: The life of light bulbs is distributed normally. The variance of the lifetime is 900 and the mean lifetime of a bulb is 580 hours. Find the probability of a bulb lasting for at most 610 hours. Round your answer to four decimal places.

Solution

Solution Steps

Step 1: Given Parameters

The lifetime of light bulbs is normally distributed with the following parameters:

Mean (\( \mu \)): 580 hours
Variance (\( \sigma^2 \)): 900 hours
Standard deviation (\( \sigma \)): \( \sqrt{900} = 30 \) hours

Step 2: Define the Range

We want to find the probability that a bulb lasts for at most 610 hours. Thus, the upper bound of our range is:

Upper bound: 610 hours

Step 3: Calculate the Z-Score

To find the probability, we first calculate the Z-score for the upper bound using the formula: \[ Z = \frac{X - \mu}{\sigma} \] Substituting the values: \[ Z_{end} = \frac{610 - 580}{30} = \frac{30}{30} = 1.0 \]

Step 4: Calculate the Probability

The probability that a bulb lasts at most 610 hours can be expressed as: \[ P(X \leq 610) = \Phi(Z_{end}) - \Phi(Z_{start}) \] Where \( Z_{start} \) approaches negative infinity (\(-\infty\)). Thus: \[ P(X \leq 610) = \Phi(1.0) - \Phi(-\infty) \] Using the standard normal distribution table, we find: \[ \Phi(1.0) \approx 0.8413 \quad \text{and} \quad \Phi(-\infty) = 0 \] Therefore: \[ P(X \leq 610) = 0.8413 - 0 = 0.8413 \]