1. standard normal

Let X follow a normal distribution
mean=80 and Var=100

the probability is 0.08 that X is in the symmetric interval about the mean between which two numbers?

I did:

P( a <x< b)= .08
a= mean - k
b= mean + k

P( -k/stand dev < Z< k/st dev) =.08
F(k/st dev)- F(-k/st dev) =.08
F(k/st dev) - [ 1- F(k/ st dev)]= 0.08
2F(k/st dev)=1.08 from which k= 1 and a= 79

but the result on the book is a= 62.5

2. Originally Posted by 0123
Let X follow a normal distribution
mean=80 and Var=100

the probability is 0.08 that X is in the symmetric interval about the mean between which two numbers?

I did:

P( a <x< b)= .08
a= mean - k
b= mean + k

P( -k/stand dev < Z< k/st dev) =.08
F(k/st dev)- F(-k/st dev) =.08
F(k/st dev) - [ 1- F(k/ st dev)]= 0.08
2F(k/st dev)=1.08 from which k= 1 and a= 79

but the result on the book is a= 62.5

$\displaystyle X$ is $\displaystyle N(80 , 100)$ and $\displaystyle Z = \frac{X - 80}{10}$

so if $\displaystyle P(80-k < X < 80 + k) = .08$, then
$\displaystyle P\left({\frac{-k}{10} < Z < \frac{k}{10}}\right) = P\left({|Z| < \frac{k}{10}}\right) = 0.08$

this implies that $\displaystyle P\left({Z < \frac{k}{10}}\right) = 0.08 + \left({0.5 - \frac{0.08}{2}}\right) = 0.54$
which implies that $\displaystyle \frac{k}{10} = .1004 \implies k=1.004$

indeed, you might be right..
try checking this.. Z table - Normal Distribution