the amount of sodium in a certain brand of low-salt dinner is 650mg, the standard deviation is 32 mg. assume the variable is normally distributed.
is a sample of 10 dinners is selected, find the probability that the mean of the sample will be larger than 665 mg
Probability question URGENT
2011-11-04 2:38 am
回答 (1)
2011-11-04 6:39 am
✔ 最佳答案
Let X be the amount of the sample,i.e. X~N(650,(32^2)/10)
P(X>665)
= P(Z>(665-650)/(32/√10))
= P(Z>1.4823)
= 0.5 - 0.4309
= 0.0691
參考: Hope the solution can help you^^”
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