Assume X is normally distributed with a mean of 5 and a standard deviation of 2. Determine the value for x that solves each of the following. a. P(X > x) = 0.5
b. P(X > x) = 0.95
c. P(x < X < 9) = 0.2
d. P(3 < X < x) = 0.95
e. P(-x < X - 5 < x) =0.99

U = 5

S = 4

1.) P(X>x) = 0.5

Prob = 1-0.5 = 0.5

We have z = 0, that is the z score with the probability of 0.5

X = u + z(s)

= 5+0*4

= 5

2.) 1-0.95 = 0.05

Z score having this probability

Z = -1.64

X = 5-1.64*4

= 5-6.56

= -1.56

3.) P(z<1.0) - p(X<x) = 0.2

0.841345-0.2 = .641345

We find the z score given this probability

Z= 0.36

X = 5+0.36*4

= 5+1.44

= 6.44

4.) P(X<x)-P(Z<-.5)

0.95 = p(X<x)-0.308538

p(X<x) = 0.308538 + 0.95

= 1.258538

There is no x value here, given that the probability is more than 1.

5. 1-0.99/2 = 0.005

We get the z score value

= -2.58

U - 5 = 5-5 = 0

-x = 0-2.58(4)

X = 10.32

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it is contrapositive

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