Due to the pollution from the industry around an apple farm, the apples grown there may be contaminated by heavy metals. It is believed that the amount of heavy metals in an apple of the farm follows the Normal distribution.
N(16,16) which has a mean μ = 16 units and σ = √16 = 4 units. An apple is said to be "acceptable" if the amount of heavy metals is less than 18 units.
First, what is the probability that a randomly selected apple is acceptable?
Second, if each apple is taken to be conducted in a test of heavy metals content (assuming each test is absolutely accurate), what is the probability that more than 2 tests are required for obtaining the first acceptable apple.
Third, now the pollution is getting worse and an investigation is conducted by randomly selecting 100 apples. The average amount of heavy metals is 20 units. We assume that the s.d. remains the same as the question says. Construct a 95% confidence interval for the mean amount of heavy metals μ.
P(X <=-k) = P(X>=k)
The normal distribution pdf
Z = (X - μ)/σ
P( (barX - μ)/(σ/√n) <= Zα/2) = β, where confidence level β = 1 - α
The Attempt at a Solution
Let X be the random variable meaning the amount of heavy metals found in an apple.
For the first question, I try to calculate P(X<=18). By converting it to standard normal distribution using
Z = (X - μ)/σ. But I have a question on whether I should calculate P(X<18). If so, does it mean I have to minus P(X=18) from P(X<=18)? Or are they the same? Or I should us the continuity correction factor?
For the second question, I have thought about calculating that P(more than 2 tests) = 1 - P(1 test) - P(2 tests).
Does it mean that I use the prob. I got in part a i: P(1st rejected) = 1 - P(1st acceptable) = 1 - P?
For the third question, I just plug in the formula P( (barX - μ)/(σ/√n) <= Zα/2) = β to find the intervals?