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.

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:

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

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:

So,

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:

answer: 21$

step-by-step explanation:

$42 / $14 = 3   (ratio 3: 1)

$7 x 3 = $21

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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