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4 votes
14.

Two jewelers were asked to measure the
mass of a gold nugget. The true mass of the
nugget is 0.856 grams (g). Each jeweler took
three measurements. The average of the
three measurements was reported as the
"official” measurement with the following
results:
Jeweler A: 0.863 g, 0.869 g, 0.859 g
Jeweler B: 0.875 g, 0.834 g, 0.858 g
Which jeweler's official measurement was
more accurate? Which jeweler's measure-
ments were more precise? In each case, what
was the error and percent error in the official
measurement?​

User Wnafee
by
3.7k points

1 Answer

2 votes

Answer:

Jeweler B = more accurate

Jeweler A = more precise

Error:

0.008, 0

% error :

0.934% ; 0

Step-by-step explanation:

Given that:

True mass of nugget = 0.856

Jeweler A: 0.863 g, 0.869 g, 0.859 g

Jeweler B: 0.875 g, 0.834 g, 0.858 g

Official measurement (A) = 0.863 + 0.869 + 0.859 = 2.591 / 3 = 0.864

Official measurement (B) = 0.875 + 0.834 + 0.858 = 2.567 / 3 = 0.8556

Accuracy = closeness of a measurement to the true value

Accuracy = true value - official measurement

Jeweler A's accuracy :

0.856 - 0.864 = - 0.008

Jeweler B's accuracy :

0.856 - 0.856 = 0.00

Therefore, Jeweler B's official measurement is more accurate as it is more close to the true value of the gold nugget.

However, Jeweler A's official measurement is more precise as each Jeweler A's measurement are closer to one another than Jeweler B's measurement which are more spread out.

Error:

Jeweler A's error :

0.864 - 0.856 = 0.008

% error =( error / true value) × 100

% error = (0.008/0.856) × 100% = 0.934%

Jeweler B's error :

0.856 - 0.856 = 0 ( since the official measurement as been rounded to match the decimal representation of the true value)

% error = 0%

User Dertoni
by
3.8k points