Question

Problem 6.
1. If X ~ N(9; 4), find Pr(|X - 2 |< 4).
2. If X ~ N(0; 1), find Pr(|X + 3|> 5).
3. If X ~ N(-2; 9), find the number c such that Pr(|X + 2| < c) = 0:5.

Answer 1

1)

X ~ N(9,4)

That is

= 9, = sqrt(4) = 2

We convert this to standard normal as

P( X < x) = P( Z < x - / )

We have to calculate,

P( | X - 2| < 4) = ?

P( | X - 2| < 4) = P( -4 < X - 2< 4)

= P( -4 +2 < X < 4 + 2)

= P( -2 < X < 6)

= P( X < 6) - P( X < -2)

= P( Z < 6 - 9 / 2) - P( Z < -2-9 / 2)

= P( Z < -1.5) - P( Z < -5.5)

= ( 1 - P( Z < 1.5) ) - ( 1 - P( Z < 5.5) )

= ( 1 - 0.9332) - ( 1 - 1)

= 0.0668

b)

X ~ N(0 , 1)

= 0 , = 1

We convert this to standard normal as

P( X < x) = P( Z < x - / )

We have to calculate,

P( | X + 3| > 5) = ?

P( | X + 3| > 5) = P( -5 < X + 3 < 5)

= P( -5 - 3 < X < 5 - 3 )

= P( -8 < X < 2)

= P( X < 2) - P( X < -8)

= P( Z < 2 - 0 / 1) - P( Z < -8 - 0 / 1)

= P( Z < 2) - P( Z < -8)

= 0.9772 - 0

= 0.9772

c)

X ~ N(-2 , 9)

= -2 , = 3

We convert this to standard normal as

P( X < x) = P( Z < x - / )

We have to calculate c such that

P( | X + 2| < c) = 0.5

P( -c < X + 2 < c ) = 0.5

P( -c -2 < X < c - 2) = 0.5

P( X < c - 2) - P( X < -c-2) = 0.5

P( Z < c - 2 +2 / 3) - P( Z < -c - 2 + 2 / 3) = 0.5

P( Z < c /3) - P( Z < -c / 3) = 0.5

P( Z < c / 3) - ( 1 - P( Z < c / 3) ) = 0.5

P( Z < c/3) - 1 + P( Z < c / 3) = 0.5

2 * P( Z < c / 3) = 1.5

P( Z < c / 3) = 0.75

From the Z table, z-score for the probability of 0.75 is 0.6745

(c ) / 3 = 0.6745

c = 2.023

c = 2 (Rounded to nearest integer)