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z = (X-Mean)/Standard deviationThe weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random.
(a) Find the probability that this tin contains more than 168 g
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z value corresponding to X=168 is
z = (168-160)/5 = + 1.6
The area under the standard normal curve right to z = + 1.6 indicates the required probability.
This area lies in the extreme right tail of the normal curve.
The area corresponding to z = 1.6 is 0.4452
The area right to z = 0 is 0.5000
The area beyond z = + 1.6 is or
P(X > 168) = 0.5000 - 0.4452 = 0.0548