Answer:
Do the data from this shipment indicate statistical control: No
Step-by-step explanation:
Calculating the mean of the sample, we have;
Mean (x-bar) = sum of individual sample/number of sample
= (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10
= 0.044/10
= 0.0044
Calculating the lower control limit (LCL) using the formula;
LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n
= 0.0044 - 3*√(0.0044(1-0.0044))
= 0.0044- (3*0.0042)
= 0.0044 - 0.01256
= -0.00816 ∠ 0
Calculating the upper control limit (UCL) using the formula;
UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n
= 0.0044 + 3*√(0.0044(1-0.0044))
= 0.0044+ (3*0.0042)
= 0.0044 + 0.01256
=0.01696∠ 0
Do the data from this shipment indicate statistical control: No
Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that the data from this shipment do not indicate statistical control.
Answer:
64
Step-by-step explanation:
p=6
(6+2)= 8
8x8= 64
First term is 13m with factors 13 and m
second term is -2n with factors -2 and n
Would it be C? im not too sure but thats my guess =)