Question

In a certain population of mussels (Mytilus edulis), 80% of the individuals are infected with an intestinal parasite. A marine biologist plans to examine 100 randomly chosen mussels from the population.
a) Find the probability that 85% or more of the sampled mussels will be infected using normal approximation to binomial distribution.

Answer 1

Answer:

The probability that 85% or more of the sampled mussels will be infected is 0.1057.

Step-by-step explanation:

Let X = number of mussels infected with an intestinal parasite.

The probability that a random selected mussel is infected with an intestinal parasite is, p = 0.80.

A random sample of n = 100 mussels from the population are examined by a marine biologist.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.80.

But the sample selected is too large, i.e. n = 100 > 30.

So a Normal approximation to binomial can be applied to approximate the distribution of $\hat p$, the sample proportion of mussels infected with an intestinal parasite, if the following conditions are satisfied:

1. np ≥ 10
2. 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 $\hat p$ is:

$\hat p\sim N(p, \frac{p(1-p)}{n} )$

Compute the probability that 85% or more of the sampled mussels will be infected as follows:

Apply continuity correction:

$P(\hat p\geq 0.85)=P(\hat p>0.85+0.50)$

                   $=P(\hat p>0.90)$

                   $=P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.85-0.80}{\sqrt{\frac{0.80(1-0.80)}{100}}})$

                   $=P(Z>1.25)\\=1-P(Z$

*Use a z-table for the probability.

Thus, the probability that 85% or more of the sampled mussels will be infected is 0.1057.