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3 years ago

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Of the 200 students surveyed at the local high school, 126 say that they prefer earlier start times for school. What is the mean

of the sampling distribution of the proportion of the students who prefer later start times for school

1 answer:

ra1l [238]3 years ago

5 0

Answer:

The mean of sampling distribution of the proportion of the students who prefer later start times for school

μₓ = p = 0.63

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given sample size 'n' =200

given data Of the 200 students surveyed at the local high school, 126 say that they prefer earlier start times for school

sample proportion

$p = \frac{x}{n} = \frac{126}{200} = 0.63$

The mean of sampling distribution of the proportion of the students who prefer later start times for school

μₓ = p = 0.63

Standard deviation of the proportion

$S.D = \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.63(1-0.63)}{200} } = 0.034$

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Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species

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Answer:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

Step-by-step explanation:

The logistic equation is the following one:

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that $P(0) = 80, K = 2000$ .

The number of fish tripled in the first year. This means that $P(1) = 3P(0) = 3(80) = 240$ .

Using the equation for P(1), that is, P(t) when $t = 1$ , we find the value of r.

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

$240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}$

$280*(2000 + 80(e^{r} - 1)) = 160000e^{r}$

$280*(2000 + 80e^{r} - 80) = 160000e^{r}$

$280*(1920 + 80e^{r}) = 160000e^{r}$

$537600 + 22400e^{r} = 160000e^{r}$

$137600e^{r} = 537600$

$e^{r} = \frac{537600}{137600}$

$e^{r} = 3.91$

Applying ln to both sides.

$\ln{e^{r}} = \ln{3.91}$

$r = 1.36$

This means that the expression for the size of the population after t years is:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

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