Question

Many typical household cleaners contain a toxic chemical. Assume the percent of this toxic chemical in one glass cleaner is approximately normal. The mean percent is 3, the standard deviation of the percent is 1.

1 Randomly select one bottle of the glass cleaner, find the probability that the percentage of the toxic chemical is more than 3.255.

2.Randomly select 15 bottles of this glass cleaner, find the probability exactly 5 bottles which contain the toxic chemical more than 3.255.

3.Find the 65th percentile of the distribution of toxic chemical percent (of the glass cleaner), p65.

4. Assume random samples with sample size of 16, find the probability that the sample mean percentage (of the toxic chemical) is less than 2.75.

Answer 1

1)

Here, μ = 3, σ = 1 and x = 3.255. We need to compute P(X >= 3.255). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3.255 - 3)/1 = 0.26

Therefore,
P(X >= 3.255) = P(z <= (3.255 - 3)/1)
= P(z >= 0.26)
= 1 - 0.6026 = 0.3974

2)

Here, μ = 3, σ = 0.2582 and x = 3.255. We need to compute P(X >= 3.255). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3.255 - 3)/0.2582 = 0.99

Therefore,
P(X >= 3.255) = P(z <= (3.255 - 3)/0.2582)
= P(z >= 0.99)
= 1 - 0.8389 = 0.1611

3)

z value at 65% = 0.39

z = (x - mean)/s
0.39 = (x - 3)/1
x = 1 * 0.39 + 3
x = 3.39

4)

Here, μ = 3, σ = 0.25 and x = 2.75. We need to compute P(X <= 2.75). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (2.75 - 3)/0.25 = -1

Therefore,
P(X <= 2.75) = P(z <= (2.75 - 3)/0.25)
= P(z <= -1)
= 0.1587