Question

Suppose x is a normally distributed random variable with μ = 50 and Ϭ = 3. Find a value of the random variable, call it x0, such that a. P(x ≤ x0) = 0.8413 b. P(x > x0) = 0.25 c. P(x > x0) = 0.95 d. P(41 ≤ x < x0) = 0.8630 e. 10% of the values of x are less than x0. f. 1% of the values of x are greater than x0.

Answer 1

Answer:

Step-by-step explanation:

Given that X is N(50,3)

So whenever we want score for x, we can convert this to Z and use std normal distribution table.

The formula for conversion is

$Z = \frac{x-50}{3}$

Using this we can find out the corresponding scores

a) $P(X\leq X_0) = 0.8413$

Z value = 1 and hence

$X_o = 50+1(3) = 53$

b) $P(x > x0) = 0.25$

$z=0.5-0.0987=0.4013\\X0= 50+3(0.4013)\\= 51.2039$

c) $P(x > x0) = 0.95\\z=2\\X_0 = 50+6 =56$

d) $P(41 ≤ x < x0) = 0.8630 \\P(-3 ≤ Z<Z_0) = 0.8630\\Z_0 = 1.10\\X_0 = 50+3.3 = 53.3$

e) 10th percentile is $50-1.28(3) =46.16$

f) This is 99th percentile

=$50+2.33(3)\\= 56.69$