Question

An electrical firm manufactures a 100-watt light bulb, which, according to

specifications written on the package, has a mean life of 900 hours with a
standard deviation of 50 hours. At most, what percentage of the bulbs fail
to last even 700 hours? Assume that the distribution is symmetric about the
mean.

Answer 1

Using the z-score formula, the z-score for a light bulb failing at 700 hours is calculated to be -4. This z-score indicates a very small, negligible proportion of light bulbs would fail before 700 hours, substantially less than 1%.

Step-by-step explanation:

According to the problem, the electrical firm manufactures a 100-watt light bulb with a mean life of 900 hours and a standard deviation of 50 hours. To find the percentage of bulbs that fail before 700 hours, we can use the z-score formula:

Z = (X - μ) / σ

Where Z is the z-score, X is the value of interest (700 hours), μ is the mean (900 hours), and σ is the standard deviation (50 hours). Plugging in the values:

Z = (700 - 900) / 50

= -200 / 50

= -4
This z-score is extreme and would fall in the tail of the normal distribution curve.

Since the distribution is symmetric about the mean, we consult a standard normal distribution table or use a calculator to find that the proportion of the area that falls to the left of a z-score of -4 is very small, typically less than 0.01%.

Therefore, at most, substantially less than 1% of the light bulbs would fail to last even 700 hours.