Beer bottles are filled so that they contain an average of 330 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 4 ml.
Solution :
(a)
P(x < 325) = P[(x - ) / < (325 - 330) / 4]
= P(z < -1.25)
= 0.1056
(b)
_{} = / n = 4 / 6
P( < 328) = P(( - _{} ) / _{} < (328 - 330) / 4 / 6 )
= P(z < 1.225)
= 0.1103
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