The duration of a flash from a camera is timed at 0.004 seconds accurate to 1 / 1000 of a second calculate the upper and lower bounds for the duration of the flash using in quality

0.0035 ≤ x ≤ 0.0045

Step-by-step explanation:

Given that:

The error = 1/1000 = 0.001 / 2 = 0.0005 s

Hence, 0.004 ± Error

0.004 ± 0.0005

Lower boundary = (0.004 - 0.001) = 0.0035

Upper boundary = (0.004 + 0.001) = 0.0045

0.0035 ≤ x ≤ 0.0045

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.