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In a cheese factory, 4.5-kg blocks of cheese are cut manually. For a large number of blocks, the standard deviation of the cutting process is measured and found to be 0.10 kg. The measurement was done with a scale with an accuracy of 1.5% of the full scale of 12 kg. Calculate the total uncertainty of the weight of the blocks of cheese at a 95 % confidence level.

Sagot :

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Answer:

0.2690 kg

Step-by-step explanation:

Given that:

For cutting a 4.5 kg block of cheese; the standard deviation:

[tex]S_x = 0.1 \ kg[/tex]

For a large no of blocks,  at 95% confidence level;

Using the t table; where the degree of freedom is infinity or > 30

t-distribution (t) = 2

∴

[tex]P_x = tS_x[/tex]

[tex]P_x = 2 \times 0.1[/tex]

[tex]P_x = 0.2 \ kg[/tex]

The system uncertainty as a result of the accuracy [tex]B_x[/tex] is:

[tex]B_x = 1.5\% \ of \ reading[/tex]

[tex]B_x = (\dfrac{1.5}{100})(12 kg)[/tex]

[tex]B_x = 0.18 \ kg[/tex]

The total uncertainty  for a single experiment with a 95% confidence interval will be:

[tex]w_s = ( B_x^2 + P_x^2)^{\frac{1}{2}[/tex]

[tex]w_s = (0.18^2 + 0.2^2)^{\frac{1}{2}[/tex]

[tex]\mathbf{w_s = 0.2690 \ kg}[/tex]

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