Respuesta :
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:
[tex]\mu_{\hat p}=0.15[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
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:
[tex]\mu_{\hat p}=0.15\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.15(1-0.15)}{150}}=0.0292[/tex]
So, [tex]\hat p\sim N(0.15, 0.0292^{2})[/tex]
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:
[tex]P(\mu_{\hat p}-2\sigma_{\hat p}<\hat p<\mu_{\hat p}+2\sigma_{\hat p})=0.95\\\\P(0.15-2\cdot0.0292<\hat p<0.15+2\cdot0.0292)=0.95\\\\P(0.092<\hat p<0.208)=0.95[/tex]
