Components used in a cellular telephone are manufactured with nominal dimension of 0.3 mm and lower and upper specification limits of 0.295 mm and 0.305 mm respectively. The and R control charts for this process are based on subgroups of size 3 and they exhibit statistical control, with the center line on the chart at 0.3015 mm and the center line on the R chart at 0.00154 mm. a) Estimate the mean and standard deviation of this process.
b) Suppose that parts below the LSL can be reworked, but parts above the USL must be scrapped. Estimate the proportion of scrap and rework produced by this process.
c) Suppose that the mean of this process can be reset by fairly simple adjustments. What value of the process mean would you recommend? Estimate the proportion of scrap and rework produced by the process at this new mean.

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).