The weights of component X are normally distributed. The mean weight is 5,200kg and the standard
deviation is 430kg.
What is the probability of a component X weighing more than 6,000kg?
? 0.0314
? 0.2343
? 0.4686
? 0.9686
i can't understand this question and its explanation
Explanation is so:
Using z = ??
x –
z =
6,000 – 5,200
430
z = 1.86
z = 1.86 corresponds to an area of 0.4686. However, we are interested in the shaded area =
0.5 – 0.4686 = 0.0314.
If you selected 0.2343, you divided the probability obtained (0.4686) by 2 instead of subtracting
it from 0.5.
If you selected 0.4686, you forgot to subtract 0.4686 from 0.5.
If you selected 0.9686, you added 0.4686 to 0.5 instead of subtracting it.
z = 1.86 corresponds to an area of 0.4686 --i can't understand this sentence
1.86 corresponds to an area of 0.46 ?? but why what is logic of this?
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STANDART DEVIATION
5y ago#10.4686 comes from the tables provided in the exam and means that 0.4686 (or 48.86%) of the area under the curve lies been 5,200 and 6,000 kg. This in turn means that the is a probability of 0.4686 that a component weight between 5,200 and 6,000 kg.
I explain how to use the tables and the logic behind it all in my free lectures on the normal distribution.
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