Question

Use an appropriate normal random variable.

1. Find the value of Z such that the area to the right of the Z is 0.72.

2. The middle 99% of the standard normal distribution is contained between -Z and Z. Find these values.

3. Suppose that the area that can be painted using a single can of spray paint is slightly variable and follows a normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. What is the probability that the area covered by a can of spray paint is between 21 and 26 square feet?

Answer 1

#1.
Look up the left tail z-table for area in the left tail = 1 - 0.72 = 0.28

z-value = -0.58

#2.
Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

z = 2.58
-z = -2.58

#3.
Here, μ = 25, σ = 3, x1 = 21 and x2 = 26. We need to compute P(21<= X <= 26). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (21 - 25)/3 = -1.33
z2 = (26 - 25)/3 = 0.33

Therefore, we get
P(21 <= X <= 26) = P((26 - 25)/3) <= z <= (26 - 25)/3)
= P(-1.33 <= z <= 0.33) = P(z <= 0.33) - P(z <= -1.33)
= 0.6293 - 0.0918
= 0.5375