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Lorico [155]
2 years ago
6

In 2019, 15.9% of Broadway actors were acting in their first role on Broadway. Suppose we took a survey of 38 Broadway actors an

d found that 18.4% of the actors we surveyed were first-timers. What are the mean and standard deviation for the sampling distribution of pÌ p^? 1. Mean: 0.159, Standard Deviation: 0.0592. Mean: 0.159, Standard Deviation: 0.36573. Mean: 0.184, Standard Deviation: 0.0634. Mean: 0.184, Standard Deviation: 0.0595. Mean: 0.159, Standard Deviation: 0.063In 2019 the CDC reported that 14.0% of US adults are smokers. Suppose you take a random sample of 30 smokers and find that the proportion of them who are current smokers is 16.7%.What is the mean and the standard deviation of the sampling distribution of pÌ p^ ?1. mean = 0.140, standard deviation = 0.0682. mean = 0.167, standard deviation = 0.0633. mean = 0.140, standard deviation = 0.0634. mean = 0.167, standard deviation = 0.068
Mathematics
1 answer:
SpyIntel [72]2 years ago
3 0

Answer:

For the Broadway actors acting in their first role on Broadway, mean: 0.184, Standard Deviation: 0.063.

For the proportion of smokers, mean = 0.167, standard deviation = 0.068

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Suppose we took a survey of 38 Broadway actors and found that 18.4% of the actors we surveyed were first-timers.

This means that p = 0.184, n = 38

What are the mean and standard deviation for the sampling distribution of p^?

Mean:

\mu = p = 0.184

Standard deviation:

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.184*0.816}{38}} = 0.063

Suppose you take a random sample of 30 smokers and find that the proportion of them who are current smokers is 16.7%.

This means that n = 30, p = 0.167

What is the mean and the standard deviation of the sampling distribution of p^ ?

Mean:

\mu = p = 0.167

Standard deviation:

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.167*0.833}{30}} = 0.068

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The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg
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Answer:

Approximately 101\; \rm ft (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

  • Let \rm O denote the observer.
  • Let \rm A denote the top of the tree.
  • Let \rm R denote the base of the tree.
  • Let \rm B denote the point where line \rm AR (a vertical line) and the horizontal line going through \rm O meets. \angle \rm B\hat{A}R = 90^\circ.

Angles:

  • Angle of elevation of the base of the tree as it appears to the observer: \angle \rm B\hat{O}R = 51^\circ.
  • Angle of elevation of the top of the tree as it appears to the observer: \angle \rm B\hat{O}A = 57^\circ.

Let the length of segment \rm BR (vertical distance between the base of the tree and the base of the hill) be x\; \rm ft.

The question is asking for the length of segment \rm AB. Notice that the length of this segment is \mathrm{AB} = (x + 20)\; \rm ft.

The length of segment \rm OB could be represented in two ways:

  • In right triangle \rm \triangle OBR as the side adjacent to \angle \rm B\hat{O}R = 51^\circ.
  • In right triangle \rm \triangle OBA as the side adjacent to \angle \rm B\hat{O}A = 57^\circ.

For example, in right triangle \rm \triangle OBR, the length of the side opposite to \angle \rm B\hat{O}R = 51^\circ is segment \rm BR. The length of that segment is x\; \rm ft.

\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}.

Rearrange to find an expression for the length of \rm OB (in \rm ft) in terms of x:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}.

Similarly, in right triangle \rm \triangle OBA:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}.

Equate the right-hand side of these two equations:

0.810\, x \approx 0.649\, (x + 20).

Solve for x:

x \approx 81\; \rm ft.

Hence, the height of the top of this tree relative to the base of the hill would be (x + 20)\; {\rm ft}\approx 101\; \rm ft.

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