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# Stat - Normal Distribution

有條normal distribution + probability既問題想問，

Workers's average(mean) income is $35.2,

Standard Deviation is $7.5

-> 現在抽9個workers.

Find the probability that at most 2個workers 既 incomes係less than $34, given that at least 6個workers既 incomes 係greater than or equal to $34

thanks ><

### 1 Answer

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- 翻雷滾天 風卷殘雲Lv 79 years agoFavorite Answer
First of all, the z-score of $34 is:

(34 - 35.2)/7.5 = -0.16

P(z < -0.16) = 0.4364

So when randomly picking up a worker, prob. that his incomes < $34 is 0.4364

Using Binomial distribution:

P(At most 2 workers income < $34) = 1 - 0.43649 - 9 x 0.43648 x (1 - 0.4364) - 9C2 x 0.43647 x (1 - 0.4364)2 = 0.999196

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