Question

Assume that X N(12:6; 5:8). Find the following probabilities (note: 5.8 is a variance,
not a standard deviation)
a) P(X > 13:1)
b) P(X < 11:2)
c) P(X 12:3) (Hint: The answer is the same as that to P(X < 12:3).)
d) P(X < 7:8)

e) P(X > 9:6)
e) P(X > 9:6)

Answer 1

Solution :

Given that,

mean = = 12:6

variance = ² = 5.8

standard deviation = = 2.4083

a ) P (x > 13:1 )

= 1 - P (x < 13:1 )

= 1 - P ( x -  / ) < ( 13:1 - 12:6 / 2.4083)

= 1 - P ( z < 0.5 / 2.4083 )

= 1 - P ( z < 0.21)

Using z table

= 1 - 0.5832

= 0.4168

Probability = 0.4168

b ) P( x < 11:2 )

P ( x - / ) < ( 11:2 - 12:6 / 2.4083)

P ( z < -1.4 / 2.4083 )

P ( z < -0.58)

= 0.2810

Probability = 0.2810

c ) P( x < 12:3 )

P ( x - / ) < ( 12:3 - 12:6 / 2.4083)

P ( z < -0.3 / 2.4083 )

P ( z < -0.12 )

= 0.4522

Probability = 0.4522

d ) P( x < 7:8 )

P ( x - / ) < (7:8 - 12:6 / 2.4083)

P ( z < - 4.8 / 2.4083 )

P ( z < -1.99 )

= 0.0233

Probability = 0.0233

e ) P (x > 9:6 )

= 1 - P (x < 9:6 )

= 1 - P ( x -  / ) < ( 9:6 - 12:6 / 2.4083)

= 1 - P ( z < - 3 / 2.4083 )

= 1 - P ( z < -1.24)

Using z table

= 1 - 0.1075

= 0.8925

Probability = 0.8925