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 t
he 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
1 answer:
Answer:
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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