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nickersia
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A random variable X follows a normal distribution with mean 20. (standard deviation 5)
Find probability 14=<X=<26
 one year ago
 one year ago
nickersia Group Title
A random variable X follows a normal distribution with mean 20. (standard deviation 5) Find probability 14=<X=<26
 one year ago
 one year ago

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nickersia Group TitleBest ResponseYou've already chosen the best response.0
\[14 \le x \le 26\]
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
Do you know how to work out the zscores for 14 and 26? \[z=\frac{X\mu}{\sigma}\]
 one year ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
Yes, but I'm not sure should I put 6 and 6 for z? That way I got 50 for x, which makes me no sense
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
\[z _{1}=\frac{1420}{5}=?\] \[z _{2}=\frac{2620}{5}=?\]
 one year ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
I got 76.98% as an answer. z= 1.2 and 1.2 So, z score from beginning of the curve to the s.d. of 1.2 is 0.8849, which meas that on the right it's 0.1151 and same on the left, so it's 0.2302. 23.02% are lower than 14 and bigger that 26, so 76.98% is X. Am I right?
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
A standard normal distribution table gives the following values for cumulative probability: z = 1.2 : p = 0.8849 z = 1.2 : p = 0.1151 You can check these results on the table at the following link (choose 'normal.pdf' from the menu: http://www.math.bgu.ac.il/~ngur/Teaching/probability/
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
\[P(14<X <26)=0.88490.1151=?\]
 one year ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
0.7698 which is 76.98% So, I suppose I got it right :) That's basically the same thing I did, thank you very much! :)
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.1
You're welcome :)
 one year ago
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