Answer to Question #188279 in Statistics and Probability for Tani Mog

Question

is normally distributed and the mean X is 12 S.D. is 4.

a.

Find out the probability of the following

(i) X ≥ 20, (ii) X ≤ 20 and (iii) 0 ≤ X ≤ 12

b. Find x’ when P(X>x’) = 0.24

c.

Find x

0’ and x1’ when P(x0’<X< x1’) = 0.50 and P(X> x1’) = 0.25

Answer 1

mean, $\mu=12, \sigma=4$

(a)

(i) $P(X ≥ 20)=P(z\ge \dfrac{20-12}{4})=P(z\ge 2)=0.02275$

(ii) $P(X ≤ 20) = P(z\le \dfrac{20-12}{4})=P(z\le 2)=0.97725$

(iii) $P(0 ≤ X ≤ 12)=P(\dfrac{0-12}{4}\le z\le \dfrac{12-12}{4})=P(-3\le z\le 0)= 0.49865$

(b) $P(X>x')=0.24$

from the normal distribution table $x'=0.706$

(c) $P(x_0’<X< x_1’) = 0.50 \text{ and }P(X> x_1’) = 0.25$

As $P(X>x')=0.25 \text{ then } x_1'=0.674$

Also $P(x_0'<X<x_1')=0.25$

Then, $x_0'= -0.674$