Question

The number of contaminants in samples of one cubic centimetre of polluted water has a Poisson distribution with a mean of 1.

a. What is the probability a sample will contain some (one or more) contaminants?

b. If four samples are independently selected from this water, find the probability that at least one sample will contain some (one or more) contaminants?

c. If 100 samples are selected instead, what is approximately the probability of seeing at least 60 samples with a contaminant? (Use the Normal table provided for your calculations)

Answer 1

Let 'X' be the success of having the contaminants.

per sample

E(X) =

P( X = x) =

a. What is the probability a sample will contain some (one or more) contaminants?

P(X >= 1 ) = 1 - P(X <1)

= 1 - P( X = 0)

= 1 -

= 1 - 0.3679

P( X >=1) = 0.6321

b. If four samples are independently selected from this water, find the probability that at least one sample will contain some (one or more) contaminants?

Above we had one trial that is one sample. Here we have 4 trials. These are independent so the probability that each one will have some contaminants will be same. So this can be treated as binomial where the success event is having contaminant.

P( X =x) =

P( At least one ) = P( X >= 1)

= 1 - P(X < 1)

= 1 - P(X =0)

= 1 -

P(At least one) = 0.9817

c. If 100 samples are selected instead, what is approximately the probability of seeing at least 60 samples with a contaminant?

Here we have n = 100 therefore np = 63.212

WE can use normal approximation because 'n > 30' and 'np > 10'.

Where

z-score =

But since we are approximating a discrete by continuous we need to do continuity corrrection.

P( AT least 60) = P( X >= 60)

= P( X > 59.5)

P( X> 59.5) = P( Z > -0.77)

= P Z < 0.77)

using normal distribution tables we have

P( X >= 60 ) = 0.7793

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