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[SOLVED] Normal distribution questions

karush

Well-known member
Jan 31, 2012
2,776
Let \(\displaystyle X\) be normally distributed with \(\displaystyle \mu =100cm\) and \(\displaystyle \sigma =5 cm\)

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

View attachment 1010

(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 \(\displaystyle X>105\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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.
 

karush

Well-known member
Jan 31, 2012
2,776
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 \(\displaystyle P(X>105)\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
yes area not volume

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

karush

Well-known member
Jan 31, 2012
2,776
(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??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
(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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
(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?
 

karush

Well-known member
Jan 31, 2012
2,776
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I used:

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

in my statement, as :

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