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luda_lava [24]
3 years ago
9

In this exercise, we estimate the rate at which the total personal income is rising in the Richmond-Petersburg, Virginia, metrop

olitan area. In 1999, the population in this area was 961400, and the population was increasing at roughly 9200 people per year. The average annual income was 30593 dollars per capita, and this average was increasing at about 1400 dollars per year. Use the Product Rule and these figures to estimate the rate at which the total personal income was rising in the Richmond-Petersburg area in 1999. Explain the meaning of each term in the Product Rule.
Mathematics
1 answer:
babymother [125]3 years ago
8 0

Answer:

The rate at which the total income is increasing is$1627415600

Step-by-step explanation:

From the given data

Population in the area=P=961400

The rate of increase in population=dP/dt=9200

The Average Income=I=$30593

The rate of increase in Average Income=dI/dt=$1400

Now the total income is given as

Total Income=TI=PI

So the rate of increase of total personal income is given as \frac{d}{dt}(PI)

From the product rule of derivatives

\frac{d}{dt}(P*I)=I*\frac{dP}{dt}+P*\frac{dI}{dt}

Here

I is the Average Income which is given as $30593

dI/dt is the rate of increase of income which is given as $1400.

P is the population of the area which is given as 961400

dP/dt=9200 is the rate of increase of population which is given as 9200

By substituting the values,

\frac{d}{dt}(P*I)=I*\frac{dP}{dt}+P*\frac{dI}{dt}\\\frac{d}{dt}(P*I)=30593*9200+961400*1400\\\frac{d}{dt}(P*I)=\$1627415600

So the rate at which the total income is increasing is$1627415600

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Using the normal distribution, we have that:

  • The distribution of X is X \approx (57,22).
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  • 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
  • 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
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In this problem, the parameters are given as follows:

\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358

Hence:

  • The distribution of X is X \approx (57,22).
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The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:

X = 58:

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Z = \frac{58 - 57}{22}

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

Z = \frac{X - \mu}{\sigma}

Z = \frac{55 - 57}{22}

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

Z = \frac{X - \mu}{s}

Z = \frac{58 - 57}{5.3358}

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Z = 0.19 has a p-value of 0.5753.

X = 55:

Z = \frac{X - \mu}{s}

Z = \frac{55 - 57}{5.3358}

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

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