Let X be a normal random variable with mean 3 and variance 4. (a) Find the probability P(2 < X < 6). (b) Find the value c such that P(X > c) = 0.33. (c) Find E[X2 ]. Hint: You can integrate with the density function, but it is quicker to relate E[X2 ] to the mean and variance.

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