Question

A variable is normally distributed with mean 9 and standard deviation 2.

a. Find the percentage of all possible values of the variable that lie between 4 and 12.

b. Find the percentage of all possible values of the variable that exceed 7.

c. Find the percentage of all possible values of the variable that are less than 8.

Answer 1

a)

Here, μ = 9, σ = 2, x1 = 4 and x2 = 12. We need to compute P(4<= X <= 12). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (4 - 9)/2 = -2.5
z2 = (12 - 9)/2 = 1.5

Therefore, we get
P(4 <= X <= 12) = P((12 - 9)/2) <= z <= (12 - 9)/2)
= P(-2.5 <= z <= 1.5) = P(z <= 1.5) - P(z <= -2.5)
= 0.9332 - 0.0062
= 0.9270

b)

Here, μ = 9, σ = 2 and x = 7. We need to compute P(X >= 7). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (7 - 9)/2 = -1

Therefore,
P(X >= 7) = P(z <= (7 - 9)/2)
= P(z >= -1)
= 1 - 0.1587 = 0.8413

c)

Here, μ = 9, σ = 2 and x = 8. We need to compute P(X <= 8). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (8 - 9)/2 = -0.5

Therefore,
P(X <= 8) = P(z <= (8 - 9)/2)
= P(z <= -0.5)
= 0.3085