Design specifications for filling a bottled soda claim that bottles should contain 350-360 milliliters of liquid. sample data indicate that the bottles contain an average of 355 milliliters of liquid, with a standard deviation of 2 milliliters. is the filling operation capable of meeting the design specifications?

It is high likely that the filling operation is capale of meeting design specifications.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Is the filling operation capable of meeting the design specifications?

It will be capable if it is highly likely that the specifications will be met. A probability is said to be high likely when it is of at least 95%.

In this case, the probability of containing between 350 and 360 ml of liquid is the pvalue of Z when X = 360 subtracted by the pvalue of Z when X = 350.

X = 360

has a pvalue of 0.9938.

X = 350

has a pvalue of 0.0062.

This means that there is a 0.9938 - 0.0062 = 0.9876 = 98.76% probability that the filling operation is capable of meeting the design specifications. It is high likely that the filling operation is capale of meeting design specifications.

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

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