Question

Soft drink cans are filled by an automated filling machine and the standard deviation is 0.5 fluid ounce. Assume that the fill volumes of the cans are independent, normal random variables.

Round your answers to four decimal places (e.g. 98.7654).

What is the standard deviation of the average fill volume of 100 cans?

If the mean fill volume 12.1 ounces, what is the probability that the average fill volume of the 100 cans is less than 12 fluid ounces?

What should the mean fill volume equal so that the probability that the average of 100 cans is less than 12 ounces is 0.005?

If the mean fill volume 12.1 ounces, what should the standard deviation of fill volume equal so that the average of 100 cans is less than 12 fluid ounces is 0.005?

Answer 1

Giren, Standard deviation : 50.5. Stan for 100 cars, Standard deviation as come out of -0.05. Jo 5:0:05 we know, 25 RH WN 10,). Given u = 12-2, n=roo, S=0.05. o plX $12) • PIXES 12.65) :Plzs..0.05 ) :P(24-2) - 12) = 1-512): 1-0.97.72 -0.0228 pis 12) = 0: 0228. opi x < 12): 0.005 *Ples Dieren ) : 1 - 312.58) > Plzsbetona = $(-2.58) » Chevette : -2.58) comparing we get, 0-05 -2.58 t: 12.129. DP (8412) = 0.005 oy presies 12-12-1) = (-2.58) :-) B (- ) = (-2.58) comparing we get S = 0.0388

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