measures of dispersion
2011-01-25 3:14 am
For some boxes of chocolates, the mean and standard deviation of their weights are 248 g and 3.2 g respectively. Any box of chocolates with its weight between 244.8 g and 251.2 g meets the required standard. If the weights of these boxes of chocolates are normally distributed, what percentage of these boxes of chocolates do not meet the required standard?
回答 (2)
2011-01-25 4:05 am
✔ 最佳答案
The standard score for 244.8= (244.8-248)/3.2=-1The standard score for 251.2=(251.2-248)/3.2=1Probability that the weighs do not meet standard=Pr(Z>1)+Pr(Z<-1)=0.3173 
2011-01-25 4:11 am
μ = 248g, σ = 32g
P(244.8 < X < 251.2)
= P[(244.8 - 248)/3.2 < X < (251.2-248)/3.2]
= P(-1 < Z < 1)
= 0.6827
So, the percentage of these boxes of chocolates which do not meet the requirement standard
= 1 - 0.6827
= 31.73%
P(244.8 < X < 251.2)
= P[(244.8 - 248)/3.2 < X < (251.2-248)/3.2]
= P(-1 < Z < 1)
= 0.6827
So, the percentage of these boxes of chocolates which do not meet the requirement standard
= 1 - 0.6827
= 31.73%
收錄日期: 2021-04-20 00:00:41
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