Question

1) The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

a) What is the probability that a sheet selected at random will be less than 29.75 inches long?

2) The weight of a product is normally distributed with a mean of four ounces and a variance of .25 ounces.

a) What is the probability that a randomly selected unit from a recently manufactured batch weighs no more than 3.5 ounces?

3) A normal distribution has µ = 30 and σ² = 25.

a) Find the raw score corresponding to z = -2

b) Find the raw score corresponding to z = 1.3

Answer 1

1)

Here, μ = 30.05, σ = 0.2 and x = 29.75. We need to compute P(X <= 29.75). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (29.75 - 30.05)/0.2 = -1.5

Therefore,
P(X <= 29.75) = P(z <= (29.75 - 30.05)/0.2)
= P(z <= -1.5)
= 0.0668

2)

Here, μ = 4, σ = 0.5 and x = 3.5. We need to compute P(X <= 3.5). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3.5 - 4)/0.5 = -1

Therefore,
P(X <= 3.5) = P(z <= (3.5 - 4)/0.5)
= P(z <= -1)
= 0.1587

c)

mean = 30 . s = 5

z = (x - mean)/s

-2 =(x - 30)/5
x = 5 * -2 + 30
x = 20

b)

z = (x - mean)/s

1.3 =(x - 30)/5
x = 5 * 1.3 + 30
x = 36.5