text: # [SOLVED]-aux.19 Normal distribution
  
  #### karush
  
  ##### Well-known member
  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 $$\displaystyle X>105$$
  
  Last edited:
  
  #### Prove It
  
  ##### Well-known member
  MHB Math Helper
  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
  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
  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
  (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
  (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
  
  Staff member
  (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
  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
  
  $$\displaystyle P(X>105)=0.16$$
  $$\displaystyle 0.5-P(X>105)=0.5-0.16=0.34$$
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