Hi I think I know what I need to do but I just need clarification. I'm doing my course distance learning so there isn't anyone to speak to straight away.

Here is the question.

c) Your analysis shows that the mean capacitance of a batch of 500 of the capacitors you have selected is 46 μF, with a standard deviation of 4 μF. Assuming the capacitors are normally distributed, determine the number of capacitors likely to have values between 42 μF and 50 μF.

Note: Use the z-table given in Appendix A when answering part c.

d) The Quality Assurance Department is anxious to improve the nominal value of the capacitors to ensure more of them fall within the rated tolerance band. The manufacturing process is adjusted in an attempt to bring about this improvement. Following the adjustment, a further sample of 100 capacitors is taken in order to determine if the adjustment has resulted in an improvement to the nominal value of the capacitors. Analysing this second sample, it is noted that the mean value is now 46.5 μF, with a standard deviation of 2 μF.

By comparing these new results of the capacitor values following the manufacturing adjustment, with the values obtained in part (c) recorded before the adjustment, show whether or not you agree with the hypothesis that the adjustment to the manufacturing process has had a beneficial effect by making the capacitor values closer to the desired manufactured nominal value of 47 μF.

Illustrate your answer with a hand drawn sketch of the normal distribution, with shaded regions representing the results of your analysis, which you can show to the Quality Assurance Manager.

Note: Use the z-table given in Appendix A when answering part d.

Here is the question.

c) Your analysis shows that the mean capacitance of a batch of 500 of the capacitors you have selected is 46 μF, with a standard deviation of 4 μF. Assuming the capacitors are normally distributed, determine the number of capacitors likely to have values between 42 μF and 50 μF.

Note: Use the z-table given in Appendix A when answering part c.

d) The Quality Assurance Department is anxious to improve the nominal value of the capacitors to ensure more of them fall within the rated tolerance band. The manufacturing process is adjusted in an attempt to bring about this improvement. Following the adjustment, a further sample of 100 capacitors is taken in order to determine if the adjustment has resulted in an improvement to the nominal value of the capacitors. Analysing this second sample, it is noted that the mean value is now 46.5 μF, with a standard deviation of 2 μF.

By comparing these new results of the capacitor values following the manufacturing adjustment, with the values obtained in part (c) recorded before the adjustment, show whether or not you agree with the hypothesis that the adjustment to the manufacturing process has had a beneficial effect by making the capacitor values closer to the desired manufactured nominal value of 47 μF.

Illustrate your answer with a hand drawn sketch of the normal distribution, with shaded regions representing the results of your analysis, which you can show to the Quality Assurance Manager.

Note: Use the z-table given in Appendix A when answering part d.

(edited 7 months ago)

Yeah we have . They are pretty vague to be fair. I'm 43 and not studied in a long time. I think I've got the first part but not sure of the second part. I just need a little guidance to be fair.

This is what I have done so far :

Z = 42-46/ 4 =-1 Standard deviation

Z = 50 -46 /4 = 1 Standard deviation

From Z-table : area = 0.34134

2 × 0.34134 = 0.68368

total area = 0.68268

0.68268 x 500 = 341.34

341 Capacitors are likely to between 42μF and 50μF

I'm just assuming this but my understanding is that for the second part they are asking to analyse the second sample with the following:

Capacitors = 100

Mean = 46.5 μF

Standard deviation = 2 μF

H0 = μ = 46 μF

H1 = μ < 46 μF

Is this how I start the second part. Im not sure how to draw the results either.

This is what I have done so far :

Z = 42-46/ 4 =-1 Standard deviation

Z = 50 -46 /4 = 1 Standard deviation

From Z-table : area = 0.34134

2 × 0.34134 = 0.68368

total area = 0.68268

0.68268 x 500 = 341.34

341 Capacitors are likely to between 42μF and 50μF

I'm just assuming this but my understanding is that for the second part they are asking to analyse the second sample with the following:

Capacitors = 100

Mean = 46.5 μF

Standard deviation = 2 μF

H0 = μ = 46 μF

H1 = μ < 46 μF

Is this how I start the second part. Im not sure how to draw the results either.

(edited 7 months ago)

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