A new extended-life light bulb has an average service life of 750 hours, with a standard deviation of 50 hours. If the service life of these light bulbs approximates a normal distribution, about what percent of the distribution will be between 600 and 900 hours?
a. 95 percent
` b. 68 percent
c. 34 percent
d. 99.7 percent
e. cannot be determined from the information given.
also please explain how you get the answer...
To find the percentage of the distribution between 600 and 900 hours, we need to calculate the area under the normal distribution curve within that range.
Step 1: Calculate the z-scores for both 600 and 900 hours.
The formula for the z-score is: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation.
For 600 hours:
z = (600 - 750) / 50 = -3
For 900 hours:
z = (900 - 750) / 50 = 3
Step 2: Look up the corresponding z-scores in the standard normal distribution table. The table will provide the percentage of the distribution between the mean and the given z-score.
According to the standard normal distribution table:
- For a z-score of -3, the percentage is approximately 0.0013 (0.13%).
- For a z-score of 3, the percentage is approximately 0.9987 (99.87%).
Step 3: Subtract the small percentage (0.13%) from the large percentage (99.87%) to calculate the percentage between 600 and 900 hours.
Percentage between 600 and 900 hours = 99.87% - 0.13% = 99.74%
Therefore, the correct answer is d. 99.7 percent.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z scores.