A quality control manager at a manufacturing facility has taken four samples wit

ID: 418628 • Letter: A

Question

A quality control manager at a manufacturing facility has taken four samples with four observations each of the diameter of a part.

Samples of Part Diameter in Inches

(a) Compute the mean of each sample. (Round answers to 3 decimal places, e.g. 15.250.)

(b) Compute an estimate of the mean and standard deviation of the sampling distribution. (Round answers to 4 decimal places, e.g. 15.2500.)

(c) Develop control limits for 3 standard deviations of the product diameter. (Round answers to 2 decimal places, e.g. 15.25.)

CAN SOMEONE PLEASE HELP ME ANSWER THIS QUESTION CORRECTLY/NEATLY WITH ALL THE WORK AS SOON AS POSSIBLE

1 2 3 4 6.3 6.0 6.2 5.8 5.9 5.8 6.2 6.3 6.3 5.8 5.9 6.0 5.7 5.9 5.8 6.2

a: Mean of samples

Sum of samples / No: of samples

Thus for 1: Mean = (6.3+ 5.9+6.3+5.7)/ 4 = 6.050

for 2: Mean = 5.875, for 3:- 6.025, for 4:- 6.075

b:)-Mean of all sample distribution

= Mean of ( mean of sample 1+ mean of sample 2+ mean of sample 3+ mean of sample 4)= (6.050+5.875+6.025+6.075) / 4 = 24.025/4 = 6.0063

Standard deviation :- Square root of ( (x-(x bar^2)/n) Where x are obeservation and x bar is mean of the samples.

By applying formula the standard deviation

Sample 1: Std deviation sample 1:- 9.502

Std deviation sample 2:- 5.0523;

Std deviation sample 3:- 5.1701

Std deviation sample 4:- 5.3433

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