A company manufactures light bulbs which have a mean life of 5000 hours and a standard deviation of 100 hours. The life length of these light bulbs follows a normal distribution. What is the probability that a light bulb will have a life length of less than 4800 hours?
Answer
Now, we look up this z-score in the z-table or use a calculator to find the area to the left of -2. The area to the left of -2 in the z-distribution is 0.02275 or 2.275%.
Step 1 ：First, we calculate the z-score of 4800, which is given by \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Step 2 ：Substituting the known values, we get \(Z = \frac{4800 - 5000}{100} = -2\).
Step 3 ：Now, we look up this z-score in the z-table or use a calculator to find the area to the left of -2. The area to the left of -2 in the z-distribution is 0.02275 or 2.275%.
