1.32 Factory defective rate. A factory quality control manager decides to investigate the percentage of defective items produced each day. Within a given work week (Monday through Friday) the percentage of defective items produced was 2%, 1.4%, 4%, 3%, 2.2%. (a) Calculate the mean for these data. (b) Calculate the standard deviation for these data, showing each step in detail.

a) the mean percentage of defective item produced is 2.52 %

b) the standard deviation of percentage of defective item produced is 1.01%

Step-by-step explanation:

Given that;

the percentage of defective items produced was 2%, 1.4%, 4%, 3%, 2.2%.

sample size n = 5

a) Calculate the mean for these data

mean percentage of defective item produced will be;

= ∑x / n

= ∑x / n = ( 2% + 1.4% + 4% + 3% + 2.2% ) / 5

 = 12.6 / 5

 = 2.52 %

Therefore, the mean percentage of defective item produced is 2.52 %

b) Calculate the standard deviation for these data

Formula for standard deviation is;

S = √( (∑(x- )²) / (n-1) )

so we make a table;

x                   ( x - )%                ( x - )²%

2%                 -0.52                         0.2704

1.4%               -1.12                           1.2544

4%                  1.48                          2.1904  

3%                  0.48                         0.2304

2.2%.             -0.32                         0.1024

summation                                     4.048

so (∑(x- )² = 4.048%  

so we substitute the value into our equation;

S = √( (∑(x- )²) / (n-1) )

S = √( (4.048%) / (5-1) )

S = √( 4.048% / 4 )

S = √( 1.0121

S =  1.00598 % ≈ 1.01%

Therefore, the standard deviation of percentage of defective item produced is 1.01%

the correct answer is a.) bc< mn 5 stars if this you : )

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v