Question

An orange juice producer buys only one kind of oranges. The amount of juice squeezed from each of these oranges is approximately normally distributed with a mean of 4.2 ounces and a population standard deviation of 1 ounce. If a sample of 100 oranges is selected:

(a) What is the probability that the average juice squeezed is less than 4.15 ounces?

(b) What is the probability that the average juice squeezed is more than 4.3 ounces?

(c) What is the probability that the average juice squeezed is between 4.15 ounces and 4.3 ounces?

(d) Do we need the Central Limit Theorem to solve (a) and (b)? Why or why not? Explain.

Answer 1

a)

Here, μ = 4.2, σ = 1/sqrt(100) = 0.1 and x = 4.15. We need to compute P(X <= 4.15). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (4.15 - 4.2)/0.1 = -0.5

Therefore,
P(X <= 4.15) = P(z <= (4.15 - 4.2)/0.1)
= P(z <= -0.5)
= 0.3085

b)
Here, μ = 4.2, σ = 0.1 and x = 4.3. We need to compute P(X >= 4.3). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (4.3 - 4.2)/0.1 = 1

Therefore,
P(X >= 4.3) = P(z <= (4.3 - 4.2)/0.1)
= P(z >= 1)
= 1 - 0.8413 = 0.1587

c)

Here, μ = 4.2, σ = 0.1, x1 = 4.15 and x2 = 4.3. We need to compute P(4.15<= X <= 4.3). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (4.15 - 4.2)/0.1 = -0.5
z2 = (4.3 - 4.2)/0.1 = 1

Therefore, we get
P(4.15 <= X <= 4.3) = P((4.3 - 4.2)/0.1) <= z <= (4.3 - 4.2)/0.1)
= P(-0.5 <= z <= 1) = P(z <= 1) - P(z <= -0.5)
= 0.8413 - 0.3085
= 0.5328

d)

yes