Car seat belts are tested for strength, and it is found the results are normally distributed with a mean of 2400 lb and a standard deviation of 400 lb. Using the 68-95-99.7 rule, find the percentage of scores between 1600 lb and 3200 lb.
95.45 %
see:
http://davidmlane.com/hyperstat/z_table.html
I have no idea what the rule is.
To find the percentage of scores between 1600 lb and 3200 lb, we will use the 68-95-99.7 rule for a normal distribution.
According to the 68-95-99.7 rule:
- Approximately 68% of the data falls within one standard deviation of the mean
- Approximately 95% of the data falls within two standard deviations of the mean
- Approximately 99.7% of the data falls within three standard deviations of the mean
Given a mean of 2400 lb and a standard deviation of 400 lb, we can calculate the z-scores for the given values of 1600 lb and 3200 lb.
To calculate the z-score, use the formula:
z = (x - μ) / σ
where x is the given score, μ is the mean, and σ is the standard deviation.
For the lower limit (1600 lb):
z_lower = (1600 - 2400) / 400
= -800 / 400
= -2
For the upper limit (3200 lb):
z_upper = (3200 - 2400) / 400
= 800 / 400
= 2
Now, we can find the percentage of scores between these z-scores.
Using a standard normal distribution table or a calculator, we can find that the percentage of scores between -2 and 2 is approximately 95.4%.
Therefore, approximately 95.4% of the scores fall between 1600 lb and 3200 lb.