HW 3.6 (Normal distribution) The fill volume of an automated filling machine use

Question

HW 3.6 (Normal distribution) The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed with a mean of 12.4 fluid ounces and a standard deviation of 0.1 fluid ounce. (a) What is the probability that a fill volume is less than 12 fluid ounces? (b) If all cans less than 12.1 or more than 12.6 ounces are scrapped, what proportion of cans is scrapped. (c) Suppose the variance is unknown, and we know the probability that a fill volume exceeds 13 is 0.05, find the variance.

Explanation / Answer

Mean = 12.4

Stdev = .1

a.

P(X<12)

= P(Z<12-12.4/.1)

= P(Z<-4)

= ~0

b.

P(X<12.1 or X>12.6)

= P(Z<-3 or Z>2)

= .025+.0015 = .0265

c.

P(X>13) = .05
Z=1.645
P(Z> 13-12.4/s) = .05
13-12.4 / s = 1.645
s = .6/1.645 = .365
variance = .365^2 = .133

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