# Normal distribution problem...

• March 29th 2010, 05:16 PM
FUTUREPASTRYCHEF
Normal distribution problem...
A normal distribution has a mean of 50 and a standard deviation of 4. Determine the value below which 95% of the observations will occur. In the book it says that the answer is X=56.60, but I have no clue how they got that.
• March 29th 2010, 07:01 PM
Anonymous1
Normalize.

$P(X< x)= P(\frac{X-\mu}{\sigma} < z)$

Then via $z-score$ tables find when $P(\frac{X-\mu}{\sigma} < z) \approx .95$ and use the corresponding $x$ value
• March 29th 2010, 07:20 PM
amul28
it goes like dis...
u know mean (μ) =50 and s.d (σ) =4
1-∝=95%

now according to normal dist"-

Z=[(x-μ)/σ ]~N(0,1)

using porbabily p(Z<1.645)=0.95

p((x-50)/4<1.645)=0.95

by simplifing p(x<56.58)=0.95
therefore x=56.58 when value below 95%of observations occur.

P.S: 1.645 is a table value if u take it as 1.65 u get answer as 56.60

hope it would help u..
thank u
amul