A company purchases a small metal bracket in containers of 5,000 each. Ten containers have arrived at the unloading facility, and 250 brackets are selected at random from each container. The fraction nonconforming in each sample are 0, 0, 0, 0.004, 0.008, 0.020, 0.004, 0, 0, and 0.008. Do the data from this shipment indicate statistical control

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.

pls refer ro the top for answer

d 600 because 15*10*2=300

300+300=600

step-by-step explanation: