Question

According to a recent​ study, some experts believe that 15​% of all freshwater fish in a particular country have such high levels of mercury that they are dangerous to eat. Suppose a fish market has 150 fish we consider randomly sampled from the population of edible freshwater fish. Use the Central Limit Theorem​ (and the Empirical​ Rule) to find the approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15. You can use the Central Limit Theorem because the fish were randomly​ sampled; the population is more than 10 times 150​; and n times p is 22.5​, and n times​ (1 minus​ p) is 127.5​, and both are more than 10.

Answer 1

Answer:

The approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15 is 0.95.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 $\mu_{\hat p}=0.15$

The standard deviation of this sampling distribution of sample proportion is:

 $\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

As the sample size is large, i.e. n = 150 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=0.15\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.15(1-0.15)}{150}}=0.0292$

So, $\hat p\sim N(0.15, 0.0292^{2})$

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68%, 95% and 99.7% of the Normal distribution lie within one, two and three standard deviations of the mean, respectively.

Then,

                                  P (µ-σ < X < µ+σ) ≈ 0.68

                                  P (µ-2σ <X < µ+2σ) ≈ 0.95

                                  P (µ-3σ <X < µ+3σ) ≈ 0.997

Then the approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15 is 0.95.

That is:

$P(\mu_{\hat p}-2\sigma_{\hat p}$