Question

Suppose that the volume in a soda can is normally distributed with mean 350 mL and standard deviation 6 mL.

a) Determine the probability that a single randomly selected can has a volume between 345 and 355 mL.

Now consider a random sample of 6 cans.

b) Is the CLT for means valid? (yes/no)

c) Calculate the SD of the sampling distribution. (4 decimal places)

d) Determine the probability that the average volume in this six pack is between 345 and 355 mL.

Answer 1

Here, μ = 350, σ = 6, x1 = 345 and x2 = 355. We need to compute P(345<= X <= 355). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (345 - 350)/6 = -0.83
z2 = (355 - 350)/6 = 0.83

Therefore, we get
P(345 <= X <= 355) = P((355 - 350)/6) <= z <= (355 - 350)/6)
= P(-0.83 <= z <= 0.83) = P(z <= 0.83) - P(z <= -0.83)
= 0.7967 - 0.2033
= 0.5934

b)

yes

c)

SD = s/sqrt(n)
= 6/sqrt(6)
= 2.4495

d)

Here, μ = 350, σ = 2.4495, x1 = 345 and x2 = 355. We need to compute P(345<= X <= 355). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (345 - 350)/2.4495 = -2.04
z2 = (355 - 350)/2.4495 = 2.04

Therefore, we get
P(345 <= X <= 355) = P((355 - 350)/2.4495) <= z <= (355 - 350)/2.4495)
= P(-2.04 <= z <= 2.04) = P(z <= 2.04) - P(z <= -2.04)
= 0.9793 - 0.0207
= 0.9586