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S1 Normal Distribution Question

Hi

I'm finishing off the S1 book today, and I am stuck on this one question. Now I have tried all the methods I know, but seem to keep getting the wrong answer.

The random variable X~N(20,12)

Find the value of a and the value of b such that:
a) P(X<a)=0.40
b) P(X>b)=0.6915
c) Write down P(b<X<a)

Now in the book, the answers for they got for a and b where 19.1 and 18.3, but at this point I was slightly :s-smilie: . Thanks
Reply 1
Avatar for 237829
237829
Have you normalised your variable X, and found the corresponding P(Z < f(a)) etc?
Reply 2
Avatar for Slowbro93
Slowbro93
OP
20
I normalised it and then made it so I could work out the vairables, but when I did the working out, the book said it was wrong. Do you think it could be the actual answer in the book thats wrong.
Reply 3
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237829
What answers did you get?
Reply 4
Avatar for Slowbro93
Slowbro93
OP
20
I got 20.9 for a (3sf) and 21.7 for b (3sf), but I know b is wrong, but I thought a was right.
Reply 5
Avatar for 237829
237829
Hmm, I got the textbook answers. Now, you understand that as P(X<a)=0.4, and therefore P(Z<f(a))=0.4 implies that z is to the left of the mean, i.e. is a negative value. When you use standard normal distribution tables, you don't actually get values less than 0.5000. By considering the symmetry of the curve, what is P(Z>-f(a))= to?
Reply 6
Avatar for 237829
237829
X<a⇒P(Z<a−2012)=0.4 X<a \Rightarrow P(Z <\dfrac{a-20}{\sqrt{12}}) = 0.4 . But this implies z < 0. Therefore P(Z<a−2012)=0.4⇒P(Z<20−a12)=0.6 P(Z <\dfrac{a-20}{\sqrt{12}}) = 0.4 \Rightarrow P(Z<\dfrac{20-a}{\sqrt 12})=0.6 .
P(Z<0.251)=0.6⇒a=20−0.251×12=19.131 P(Z< 0.251) = 0.6 \Rightarrow a=20 - 0.251 \times \sqrt{12} = 19.131

P(X>b)⇒P(Z>b−2012)=0.6915⇒P(Z<20−b12)=0.6915 P(X>b) \Rightarrow P(Z>\dfrac{b-20}{\sqrt{12}})=0.6915 \Rightarrow P(Z< \dfrac{20-b}{\sqrt{12}})=0.6915
P(Z<0.5)=0.6915⇒b=20−0.5×12=18.26P(Z<0.5)=0.6915 \Rightarrow b = 20 - 0.5 \times \sqrt{12} = 18.26

P(b<X<a)=0.4−(1−0.6915)=0.0195 P( b < X < a) = 0.4 - (1-0.6915) = 0.0195
Reply 7
Avatar for Slowbro93
Slowbro93
OP
20
jbeacom600
X<a⇒P(Z<a−2012)=0.4 X<a \Rightarrow P(Z <\dfrac{a-20}{\sqrt{12}}) = 0.4 . But this implies z < 0. Therefore P(Z<a−2012)=0.4⇒P(Z<20−a12)=0.6 P(Z <\dfrac{a-20}{\sqrt{12}}) = 0.4 \Rightarrow P(Z<\dfrac{20-a}{\sqrt 12})=0.6 .
P(Z<0.251)=0.6⇒a=20−0.251×12=19.131 P(Z< 0.251) = 0.6 \Rightarrow a=20 - 0.251 \times \sqrt{12} = 19.131

P(X>b)⇒P(Z>b−2012)=0.6915⇒P(Z<20−b12)=0.6915 P(X>b) \Rightarrow P(Z>\dfrac{b-20}{\sqrt{12}})=0.6915 \Rightarrow P(Z< \dfrac{20-b}{\sqrt{12}})=0.6915
P(Z<0.5)=0.6915⇒b=20−0.5×12=18.26P(Z<0.5)=0.6915 \Rightarrow b = 20 - 0.5 \times \sqrt{12} = 18.26

P(b<X<a)=0.4−(1−0.6915)=0.0195 P( b < X < a) = 0.4 - (1-0.6915) = 0.0195


Thanks, I see where I have gone wrong now. Rep all the way!
Reply 8
Avatar for 237829
237829
Glad I could help, and thanks very much :smile:

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