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.

Respuesta :

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 [tex]\hat p[/tex], 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 [tex]\hat p[/tex] is:

[tex]\hat p\sim N(p, \frac{p(1-p)}{n} )[/tex]

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

Apply continuity correction:

[tex]P(\hat p\geq 0.85)=P(\hat p>0.85+0.50)[/tex]

                   [tex]=P(\hat p>0.90)\\[/tex]

                   [tex]=P(\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}>\frac{0.85-0.80}{\sqrt{\frac{0.80(1-0.80)}{100}}})[/tex]

                   [tex]=P(Z>1.25)\\=1-P(Z<1.25)\\=1-0.89435\\=0.10565\\\approx0.1057[/tex]

*Use a z-table for the probability.

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

The probability that 85% or more of the sampled mussels will be infected would be "0.1303".

Probability

According to the question,

Sample number, n = 100

Number of infected individuals, p = 80% or,

                                                         = 0.8

Now, μ = np

            = 100 × 0.8

            = 80

and, σ = √np (1 - p)

           = √100 (0.8) (0.2)

           = 4

The z-value will be:

→ z = [tex]\frac{x- \mu}{\sigma}[/tex]

     = [tex]\frac{84.5 - 80}{4}[/tex]

     = 1.125

hence, the probability be:

p (z > 1.125) = 1 - p (z [tex]\leq[/tex] 1.125)

                   = 1 - 0.8697

                   = 0.1303  

Thus the above response is correct.

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