SOLUTION: a machine is packaging nominal weight is 800 grams packets and it has been found that over a long period the actual weights put in the packets has been normally distributed with a

Question 1162169: a machine is packaging nominal weight is 800 grams packets and it has been found that over a long period the actual weights put in the packets has been normally distributed with a standard deviation of 5 grams .a company wishes to ensure that no more than 1% of the packets has the weight of less than 800
grams ;at what mean should a machine be set?
the weekly outputs of packets average 2,400,000 and the cost in pence,of the producing a packet of weight w gram is given by relationship c=6+0.5w
if installation anew machinery , the standard deviation could be reduced to 2 grams,and the mean allowed to fall just far enough to give the same percentage bellow 800 grams as before , determine the average saving in dollars pr week
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the critical z-score is -2.326347877.
the area to the left of this z-score will be .01, which is 1%.

you want less than or equal to 1% of all the packets made to be equal to 800.

the z-score formula is z = (x - m) / x
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation.

in this formula, z = -2.326347877, s = 5, and x = 800.
you want to solve for m.

formula becomes -2.326347877 = (800 - m) / 5.
multiply both sides of the equation by 5 to get:
-2.326347877 * 5 = 800 - m
add m to both sides of the equation and subtract -2.326347877 * 5 from both sides of the equation to get:
m = 800 - (-2.326347877 * 5) = 811.6317394.

that's what the mean needs to be so that less than or equal to 1% of the packet weighs less than 800 when the standard deviation is 5.

if the standard deviation is equal to 2, then the formula to find m becomes:
-2.326347877 = (800 - m) / 2.
solve for m as before to get:
m = 800 - (-2.326347877 * 2) = 804.6526958

that's what the mean needs to be so that less than or equal to 1% of the packet weighs less than 800 when the standard deviation is 2.

an average of 2,400,000 packets are produced each week.

when the mean is 811.6317394, the cost is 2,400,000 * (6 + .5 * 811.6317394) = 988,358,087.3 dollars.
when the mean is 804.6526958, the cost is 2,400,000 * (6 + .5 * 804.6526958) = 979,983,234.9 dollars.

the savings are equal to 988,358,087.3 dollars minus 979,983,234.9 dollars = 8,374,852.358 dollars.

that's 8.3754852358 million dollars, which can also be shown as 8.3754852358 * 10^6 dollars.

i used statistical calculators to confirm the manual calculations.

here's what they showed.

they confirmed the manual calculations were correct.

in both cases, the number of packets weighing less than 800 grams was less than or equal to 1%.

the calculators can be found at:

https://www.omnicalculator.com/statistics/normal-distribution

http://davidmlane.com/hyperstat/z_table.html
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