(i) We know that 10% of the bulbs last longer than 5130 hours. This corresponds to a z-score of 1.282 in a standard normal distribution. Using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 5130\), \(\sigma = 40.6\), and \(z = 1.282\), we solve for \(\mu\):

\(1.282 = \frac{5130 - \mu}{40.6}\)

\(\mu = 5130 - 1.282 \times 40.6\)

\(\mu = 5080\) (rounded to the nearest hour)

(ii) To find the probability that a light bulb fails to last for 5000 hours, we calculate the z-score:

\(z = \frac{5000 - 5078}{40.6} = -1.921\)

The probability \(P(X < 5000) = \Phi(-1.921) = 0.0273\) or 2.73%.

(iii) For 600 light bulbs, the mean number lasting longer than 5130 hours is \(\mu = 60\) and variance \(\sigma^2 = 54\). We approximate using a normal distribution:

\(P(\text{fewer than 65}) = \Phi\left(\frac{64.5 - 60}{\sqrt{54}}\right)\)

\(= \Phi(0.6123)\)

\(= 0.730\) or 0.73