Probability: Given the Mean and Standard deviation

AI Thread Summary
The discussion revolves around calculating the probability of an addict with an autoimmune disorder reporting the use of 16 or more drugs, given a normal distribution with a mean of 6.9 and a standard deviation of 4.1. The initial calculation yields a z-score of 2.21, leading to a probability of 0.014 or 1.4%. A clarification is provided that the correct formulation should be P[x≥16] = 1 - P[x<16], which confirms the initial answer. The final consensus is that the original calculation is indeed correct. Understanding the z-score and its application in probability is essential for accurate results.
Nikki10
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Homework Statement



The distribution of a drugs among addicts with autoimmune disorder is N(6.9, 16.81). What is the probability that one of these addicts enters your office and reports taking 16 drugs or more?

The Attempt at a Solution



z = (16-6.9)/4.1 = 2.21
P(z >16) = 0.014 from the table
so P(z > 16) = 0.014 or 1.4%

Is this correct. It seems to simple and maybe I don't understand the concept
 
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your answer is correct but you should have written
P[x\geq 16]=1-P[x &lt; 16]

P[x\geq 16]=1-P[z &lt; 2.21]= 1-0.986

so your answer is correct...
 
thank you
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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