Who ever gets to this first, earns 100 pts! Please help!
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Answer:
The probability that 85% or more of the sampled mussels will be infected is 0.1057.
Step-by-step explanation:
Let <em>X</em> = number of mussels infected with an intestinal parasite.
The probability that a random selected mussel is infected with an intestinal parasite is, <em>p</em> = 0.80.
A random sample of <em>n</em> = 100 mussels from the population are examined by a marine biologist.
The random variable <em>X</em> follows a Binomial distribution with parameters n = 100 and p = 0.80.
But the sample selected is too large, i.e. <em>n</em> = 100 > 30.
So a Normal approximation to binomial can be applied to approximate the distribution of <em>, </em>the sample proportion of mussels infected with an intestinal parasite, if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
Check the conditions as follows:
n × p = 100 × 0.80 = 80 > 10
n × (1 - p) = 100 × (1 - 0.80) = 20 > 10
Thus, a Normal approximation to binomial can be applied.
So, the distribution of is:
Compute the probability that 85% or more of the sampled mussels will be infected as follows:
Apply continuity correction:
*Use a <em>z</em>-table for the probability.
Thus, the probability that 85% or more of the sampled mussels will be infected is 0.1057.
Answer:
that would be how to solve it
