Question

Given a normal distribution with mean equal to 40 and a standard deviation of 6, find the value of x that has 14% of the area to the right.

Answer 1

$p(x>k=0.14$

i.e $p(x<k)=0.86$

So K correspond to P86

{usng excel}

$=$ NORMSINV $(0.86)=1.0803$

Hence Z $=1.0803$

$(K-40)16=1.0803$

$K=40+1.0803\times 6$

$K=46.4819$

$P86=46.4819$

$K=46.48$

So, the value of $x=46.48$ that has $14$% of the area to the right