Answer:
a) the probability of (2 < X < 6) is 0.6247
b) the value of c is 3.878
c) the value of E[ x² ] is 13
Step-by-step explanation:
Given that;
mean μ = 3
variance = 4
standard deviation s = √variance = √4 = 2
(a) Find the probability P(2 < X < 6)
P(2 < X < 6) = p( (x - μ / s ) < z < (x - μ / s ) )
= p( (2 - 3 / 2 ) < z < (6 - 3 / 2 ) )
= p( -0.5 < z < 1.5)
from z-score table, 1.5; z = 0.9332 and -0.5; z = 0.3085
so
P(2 < X < 6) = 0.9332 - 0.3085 = 0.6247
Therefore, the probability of (2 < X < 6) is 0.6247
b) Find the value c such that P(X > c) = 0.33
with p-value = 0.33, the corresponding z -score to the right is 0.439
we know that;
z = x - μ / s
we substitute
0.439 = x - 3 / 2
x - 3 = 2 × 0.439
x - 3 = 0.878
x = 0.878 + 3
x = 3.878
Therefore, the value of c is 3.878
c) Find E[ x² ].
Variance = E[ x² ] - [ mean ]²
E[ x² ] = Variance + [ mean ]²
we substitute
E[ x² ] = 4 + [ 3 ]²
E[ x² ] = 4 + 9
E[ x² ] = 13
Therefore, the value of E[ x² ] is 13
