1. Let $\displaystyle X$ be normally distributed with $\displaystyle \mu =100cm$ and $\displaystyle \sigma =5 cm$

$\displaystyle (a$) shade region $\displaystyle P(X>105)$

(b) Given that $\displaystyle P(X<d)=P(X>105)$, find the value of $\displaystyle d$.

wasn't sure if this meant that $\displaystyle d$ is the left of 105 which would be larger in volume than $\displaystyle X>105$

2. What volume? The probabilities are areas...

Anyway, it's asking you to set the two probabilities equal to each other. So if your area was measured from the left instead of from the right, what value on the x axis would you get to? Hint: The areas are equal and symmetrical about the mean.

Originally Posted by Prove It
What volume? The probabilities are areas...

Anyway, it's asking you to set the two probabilities equal to each other. So if your area was measured from the left instead of from the right, what value on the x axis would you get to? Hint: The areas are equal and symmetrical about the mean.
yes area not volume

so then $\displaystyle d=95$ if $\displaystyle p(X<d)$ for the same area as $\displaystyle P(X>105)$

4. Originally Posted by karush
yes area not volume

so then $\displaystyle d=95$ if $\displaystyle p(X<d)$ for the same area as $\displaystyle P(X>105)$
Correct

(c) Given that $\displaystyle P(X>105)=0.16$ (correct to $\displaystyle 2$ significant figures), find $\displaystyle P(d<X<105)$

so that is within $\displaystyle 68\%$ within
$\displaystyle 1$ standard deviation of the mean

or do just $\displaystyle (2)0.16 = 0.32$

not sure??

6. Originally Posted by karush
(c) Given that $\displaystyle P(X>105)=0.16$ (correct to $\displaystyle 2$ significant figures), find $\displaystyle P(d<X<105)$

so that is within $\displaystyle 68\%$ within
$\displaystyle 1$ standard deviation of the mean

or do just $\displaystyle (2)0.16 = 0.32$

not sure??
Yeah. It's 68% within 1 standard deviation.
But that means that P(d<X<105)=P(95<X<105)=0.68.

7. Originally Posted by karush
(c) Given that $\displaystyle P(X>105)=0.16$ (correct to $\displaystyle 2$ significant figures), find $\displaystyle P(d<X<105)$

so that is within $\displaystyle 68\%$ within
$\displaystyle 1$ standard deviation of the mean

or do just $\displaystyle (2)0.16 = 0.32$

not sure??
I would write (for clarity):

$\displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$

$\displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$

$\displaystyle 0.68=P(d<X<100)+0.34$

$\displaystyle P(d<X<100)=0.34$

What do you find?

Originally Posted by MarkFL
I would write (for clarity):

$\displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$

$\displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$

$\displaystyle 0.68=P(d<X<100)+0.34$

$\displaystyle P(d<X<100)=0.34$

What do you find?
i understand what you have here... but don't know how the 0.16 comes into this.

9. I used:

$\displaystyle P(X>105)=0.16$

in my statement, as :

$\displaystyle 0.5-P(X>105)=0.5-0.16=0.34$

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