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Should I use a certain table for this question or should I use a special formula. A random value has a normal distribution with the mean 102.9 and the standard deviation 4.7. What are the probabilities that this random variable will also take on a value

a. Less than 110.1;

b. Greater than 95.6;

c. Between 104.5 and 105.9;

d. Between 98.7 and 150?

I have worked so far by finding the z value and im going on to find it through the table.

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closed as off-topic by Did, user61527, colormegone, ml0105, Henry Swanson Apr 27 '14 at 23:51

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the formula for $P(X \leq x)$ is $\Phi\left( \frac{x-\mu}{\sigma} \right)$

And for $P(X \geq x)=1-\Phi\left( \frac{x-\mu}{\sigma} \right)$

And for $P(x_1 \leq X \leq x_2)=\Phi\left( \frac{x_2-\mu}{\sigma} \right)-\Phi\left( \frac{x_1-\mu}{\sigma} \right)$

$\Phi(.)$ is the function of the normal distribution.

greetings,

calculus

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  • $\begingroup$ can u please tell me the function because there are a few derived and underived ones so im not sure which one to use, also is the sigma standard deviation and the mu for mean? thanks :) $\endgroup$ – user146208 Apr 27 '14 at 21:57
  • $\begingroup$ Im not sure what you are meaning. You have to calculate the fraction in the brackets. $P(X < 110.1)= \Phi\left( \frac{110.1-102.9}{4.7} \right)=\Phi\left( 1.532 \right)$. Then look in the table of Standard Normal Distribution. Find the probability where $x=1.532$ $\endgroup$ – callculus Apr 27 '14 at 22:23

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