written 5.0 years ago by | modified 2.1 years ago by |
If x is a normal variable with mean 10 and standard deviation 4, find:
(i) P(|x-14| $\lt$ 1)
(ii) P(5 $\le$ x $\le$18)
(iii) P(x $\le$ 12)
written 5.0 years ago by | modified 2.1 years ago by |
If x is a normal variable with mean 10 and standard deviation 4, find:
(i) P(|x-14| $\lt$ 1)
(ii) P(5 $\le$ x $\le$18)
(iii) P(x $\le$ 12)
written 5.0 years ago by |
Solution:
We have, mean=10 and standard deviation=4
$z=\cfrac{x-m}{\sigma}=\cfrac{x-10}{4}$
(i) When x=14, P(|x-14|<1),
$z=\cfrac{x-m}{\sigma}=\cfrac{14-10}{4}=1$
$\therefore$ P(|x-14| $\le$ 1)=P(|z| $\le$1)=area between (z=-1 and z=1)=2(area between z=0 to 1)=2(0.3414)=0.6826
(ii) P(5 $\le$ x $\le$ 18)
When x=5, $z=\cfrac{x-m}{\sigma}=\cfrac{5-10}{4}=-1.25$
When x=18, $z=\cfrac{x-m}{\sigma}=\cfrac{18-10}{4}=2$
$\therefore$ P(5 $\le$ x $\le$ 18)=P(-1.25 $\le$ x $\le$ 2)=area between z=-1.25 to z=2 =0.3944+0.4772 = 0.8716
(iii) When z=12, for P(x$\le$12)
$z=\cfrac{x-m}{\sigma}=\cfrac{12-10}{4}=0.5$
$\therefore$ P(x$\le$12) = P(z$\le$0.5) = (area between z=-$\infty$ to 0) + (area between z=0 to z=0.5) = 0.5+0.1915 = 0.6915