MHB What is the value of d if P(X<d)=0.34?

AI Thread Summary
The discussion revolves around finding the value of d in a normal distribution where P(X<d)=0.34, given that X is normally distributed with a mean of 100 cm and a standard deviation of 5 cm. The participants clarify that to find d, one must equate the probabilities P(X<d) and P(X>105), leading to the conclusion that d equals 95 cm. They also explore the probability P(d<X<105) and confirm that it equals 0.68, which represents the area within one standard deviation of the mean. The calculations emphasize the symmetry of the normal distribution and the relationship between the areas under the curve. The final consensus is that the value of d is 95 cm, and P(d<X<105) is 0.68.
karush
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Let $$X$$ be normally distributed with $$\mu =100cm$$ and $$\sigma =5 cm$$

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

https://www.physicsforums.com/attachments/1010

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

wasn't sure if this meant that $$d$$ is the left of 105 which would be larger in volume than $$X>105$$
 
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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.
 
Prove It said:
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 $$d=95$$ if $$p(X<d)$$ for the same area as $$P(X>105)$$
 
karush said:
yes area not volume

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

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

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

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

not sure??
 
karush said:
(c) Given that $$P(X>105)=0.16$$ (correct to $$2$$ significant figures), find $$P(d<X<105)$$

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

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

not sure??

Yeah. It's 68% within 1 standard deviation.
But that means that P(d<X<105)=P(95<X<105)=0.68.
 
karush said:
(c) Given that $$P(X>105)=0.16$$ (correct to $$2$$ significant figures), find $$P(d<X<105)$$

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

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

not sure??

I would write (for clarity):

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

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

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

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

What do you find?
 
MarkFL said:
I would write (for clarity):

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

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

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

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

What do you find?

i understand what you have here... but don't know how the 0.16 comes into this.
 
I used:

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

in my statement, as :

$$0.5-P(X>105)=0.5-0.16=0.34$$
 
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