Normal distribution, finding a value question!

**youngb11** · October 13th 2011, 02:35 PM

For a normal distribution, with a mean of $10.5$ and a variance of $16.7$ , determine the value of $X_{0}$ so that $P(10.5 < x < X_{0}) = 0.4987$ .

I am very lost on this question, any and all help is greatly appreciated!

Answer is supposedly $X_{0}=22.76$

**pickslides** · October 13th 2011, 04:20 PM

Apply the fact that $P(10.5 < x < X_{0}) = P(x < X_{0}) - P(x<10.5 )$

**youngb11** · October 13th 2011, 04:35 PM

Originally Posted by pickslides

Apply the fact that $P(10.5 < x < X_{0}) = P(x < X_{0}) - P(x<10.5 )$

$0.4987 = P(x < X_{0}) - 0.5$

$0.9987 = P(x < X_{0})$

Is that correct so far? Not sure what I do next

**pickslides** · October 13th 2011, 04:41 PM

Looks great

$0.9987 = P(x < X_{0})$

$\displaystyle 0.9987 = P\left(z < \frac{X_{0}-10.5}{16.7}\right)$

Using the Normal Distribution Table

$\displaystyle 3.1 = \frac{X_{0}-10.5}{16.7}$

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