Question

Determine the area of a flat piece of metal according to the data in the table below. The answer will need to be correctly stated as: (mean value ± σ_m) units. For example: Area = (3.2 ± 0.2 cm²).

Length, cm	Width, cm
6.2	8.2
6.5	8.0
6.3	8.6
6.7	8.4
6.4	8.1
6.8	8.5

(a)

Determine the mean, standard deviation, and the standard deviation of the mean for the measurements. (Hint: Use Logger Pro to help you make the calculations. Enter your mean values to at least four decimal places, and enter your standard deviations to at least five decimal places.)

Answer 1

Mean value for length,

= (6.2 + 6.5 + 6.3 + 6.7 + 6.4 + 6.8) / 6 = 6.483 cm

standard deviation of the length,

= sqrt[(6.2 - 6.483)² + (6.5 - 6.483)² + (6.3 - 6.483)² .......+ (6.8 - 6.483)² / (6 - 1)]

= 0.2316 cm

standard deviation of the mean of the length,

= 0.2316 / sqrt (6) = 0.09457 cm

So,  the length and its error ,

L = 6.483 0.09457 cm

Mean value of width, = 8.3 cm

standard deviation of the width, = 0.2366 cm

standard deviation of the mean of width, = 0.096609 cm

So width and its error, W = 8.3 0.096609 cm

Area of the plate,

A = L*W = 6.483 * 8.3

A = 53.8114 cm^2

Independent error of area,

A = 53.8114 * sqrt [(0.09457 / 6.483)² + (0.096609 / 8.3)²] = 1.0042 cm^2

Area and its error,

A = 53.8114 1.0042 cm^2