Statistics Exercise with Solution

Given the following data:
- Mean lifespan of light bulbs, \(\mu = 1200\) hours
- Standard deviation, \(\sigma = 100\) hours
- Sample size, \(n = 40\)
- Sample mean to compare against, \(\bar{x} = 1210\) hours

We aim to find the probability that the average lifespan of bulbs in the sample will be more than 1210 hours.

First, we calculate the z-score using the formula:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Plugging in the given values:
\[ z = \frac{1210 - 1200}{\frac{100}{\sqrt{40}}} \approx 0.63 \]

Next, to find the probability that the average lifespan is more than 1210 hours, we look for the area to the right of this z-score in the standard normal distribution. This probability, \(P\), is given by:
\[ P = 1 - \Phi(z) \]

Where \(\Phi(z)\) is the cumulative distribution function (CDF) of the standard normal distribution evaluated at z.

Given our calculated z-score:
\[ P = 1 - \Phi(0.63) \approx 0.264 \]

Therefore, the probability that the average lifespan of the bulbs in the sample will be more than 1210 hours is approximately 0.264.

The correct answer is B) 0.2635.