If we have 100 carbonated beverage cans, we assume that filling of cans is norma

Question

If we have 100 carbonated beverage cans, we assume that filling of cans is normally distributed with a mean of 12.2 fluid ounces and a standard deviation of 0.15 fluid ounce. Use standard normal random variables to resolve these probabilities. (a) What is the probability that a fill volume is less than 12.1 fluid ounces? (b) If all cans less than 12.0 or greater than 12.5 ounces are scrapped, what proportion of cans is scrapped? (c) Determine specifications that arc symmetric about the mean that include 99% of all cans.

Explanation / Answer

a. P(X<12.1)
= P(Z< (12.1 - 12.2) / (.15)
= P(Z<-.667)
=

b.
P( X<12 or X>12.5)
= P( Z< (12-12.2) / (.15)
=P(Z<-1.33 or Z> 2)
=.05+.0918 = .1418

c. The symmetric about mean that include 99% of all cans is :

= 12.2 +/- .15*2.575
= 11.814 to 12.586

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