# Thread: Normal Distribution Problem (Attempted Working Inside)

1. ## Normal Distribution Problem (Attempted Working Inside)

X~N(8,25)
P($\displaystyle {-a}\leq X \leq{a}$)=0.4

We need to find a.

I decided to use GC and standardize it to find a.
P($\displaystyle {(-a-8)/5}\leq {Z} \leq{(a-8)/5}$)=0.4
Since it is now standardized,

P($\displaystyle {(-a-8)/5}\leq {Z} \leq{(a-8)/5}$)
=1-2P($\displaystyle {Z}\geq{(a-8)/5})$=0.4
P($\displaystyle {Z}\geq{(a-8)/5})$=0.3
P($\displaystyle {Z}\leq{(a-8)/5})$=0.7

So I used Ti84 InvNorm Function to get $\displaystyle {(a-8)/5}$=0.5244
Which means $\displaystyle {a}$ is 10.622. However, the ans is 6.75.

Anyone can point out my mistake?

Thanks.

2. Originally Posted by qazxsw11111
X~N(8,25)
P($\displaystyle {-a}\leq X \leq{a}$)=0.4

We need to find a.

I decided to use GC and standardize it to find a.
P($\displaystyle {(-a-8)/5}\leq {Z} \leq{(a-8)/5}$)=0.4
Since it is now standardized,

P($\displaystyle {(-a-8)/5}\leq {Z} \leq{(a-8)/5}$)
=1-2P($\displaystyle {Z}\geq{(a-8)/5})$=0.4
P($\displaystyle {Z}\geq{(a-8)/5})$=0.3
P($\displaystyle {Z}\leq{(a-8)/5})$=0.7

So I used Ti84 InvNorm Function to get $\displaystyle {(a-8)/5}$=0.5244
Which means $\displaystyle {a}$ is 10.622. However, the ans is 6.75.

Anyone can point out my mistake?

Thanks.
Your mistake is in believing that your answer is wrong and that 6.75 is correct.

Substitute a = 10.622 and use your TI-89 to confirm your answer.