Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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 :

ajeigbeibraheem

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]

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.