Use the function, f(x)= 2x^2+3x-2 to find f(2).

34kurt

Answer:

$144 - 120x + 25 {x}^{2} \\ 144 - 95 {x}^{3}$

7 0

3 years ago

A random sample of 1500 self-employed individuals was chosen and asked about the number of hours they work per week. When the da

bija089 [108]

Answer:

1. Approximately 68% work between 39 and 55 hours; 2. Approximately 47.5%; 3. 16% of self-employed individuals work less than 39 hours per week.

Step-by-step explanation:

We can solve this question using the <em>68-95-99.7 rule</em> which states that, in a normal distribution, approximately 68% (more exactly, 68.27%) of the data are about <em>one standard deviation</em> <em>above</em> and <em>below</em> the population mean, 95% (or more exactly, 95.45%) are<em> two standard deviations</em> <em>above</em> and <em>below</em> the population mean, and finally 99.7% (or more exactly, 99.73%) of the data are <em>three standard deviations above</em> and <em>below</em> the population mean.

We can convert or transform any value given in the question to the equivalent <em>z-score</em> so that we can determine how many standard deviations from the mean are any of these values. For this, we can use the z-score formula:

$\\ z = \frac{x - \mu}{\sigma}$

Where

<em>x</em> is the value given to determine how far is from the population mean.

$\\ \mu$ is the <em>population mean</em>, and

$\\ \sigma$ is the <em>population standard deviation</em>.

In this case

$\\ \mu = 47\;hours$

$\\ \sigma = 8\;hours$

We need to remember that we can use the <em>standard normal table</em> that has all the probabilities for any <em>z-score</em>, that is, <em>standardized values</em> for any normal distribution.

<h3>Solutions</h3><h3>1. Percent of 68% of the self-employed individuals work</h3>

We need to convert any value given to the equivalent z-score and proceed as we described before. So, for any range of values, we have:

For P(x>55 hours) and P(x<39 hours)

$\\ z = \frac{x - \mu}{\sigma}$

x = 55:

$\\ z = \frac{55 - 47}{8} = 1$ (a value <em>above</em> the mean).

x = 39:

$\\ z = \frac{39 - 47}{8} = -1$ (a value <em>below</em> the mean).

As we can see, from the 68-95-99.7 rule, values between x = 39 and x= 47 represent approximately 68% (or more exactly, 68.27%) of the data because they are one standard deviation above (z = 1) and one standard deviation below the mean (z = -1).

Mathematically, P(39<x<55) = 68.27% (approximately 68%). And, this seems to be the answer to the question, since the sum of P(x>55) and P(x<39) represent the complement probability for P(39<x<55), that is, 1 - 0.6827 = 0.3173 (or 31.73%), and because they are at the extremes of the normal distribution at symmetrically both sides, then, the probability for P(x<39) = (0.3173/2 = 0.1586) and P(x>55) = (0.3173/2 = 0.1586).

For the rest of the values, we have:

P(31<x<63)

For x = 31:

$\\ z = \frac{31 - 47}{8} = -2$ (below the mean)

For x = 63:

$\\ z = \frac{63 - 47}{8} = 2$ (above the mean)

Thus, P(31<x<63) (or equivalently, P(-2<z<2)) are between <em>two standard deviations</em> above and below the population mean, so this represents, according to the 68-95-99.7 rule, approximately 95% of the cases.

P(23<x<71)

For x = 23:

$\\ z = \frac{23 - 47}{8} = -3$ (below the mean)

For x = 71:

$\\ z = \frac{71 - 47}{8} = 3$ (above the mean)

Thus, P(23<x<71) are between three standard deviations above and below the population mean, and this represents, according to the 68-95-99.7 rule, approximately 99.7% of the cases.

P(39<x<55)

We already confirm that these values are between one standard deviation above and below the mean, so this represents, approximately, 68% of the cases, according to the 68-95-99.7 rule.

P(x>47)

We know that the population's mean is 47 hours, and for any normal distribution, values above and below the mean are 50% of the cases. So, P(x>47) = 50% of the cases.

Therefore, "based on the collected data we can say that approximately 68% of the self-employed work" between 39 and 55 hours.

<h3>2. Percent of self-employed individuals that work between 47 and 63 hours</h3>

Using the 68-95-99.7 rule, we can use the following reasoning: because the value for x = 63 is two standard deviations above the mean and for values of two standard deviations above and below the mean this represents 95%, and we already know that the mean = 47, the values above the mean are only the half of 95% (95/2 = 47.5%).

Then, the percent of self-employed individuals that work between 47 and 63 hours per week is 47.5%.

<h3>3. 16% of self-employed individuals work less than how many hours per week</h3>

We know from the 68-95-99.7 rule that approximately 68% of the cases are between one standard deviation below and above the mean. The complement of the values at both extremes of the normal distribution are one half of the complement below and above the mean. So, the complement of 68% is 1 - 0.68 = 0.32. Then, one half (0.32/2 = 0.16) must be below the mean and the other above the mean, at the extreme sides of the normal distribution. Then, those values for P(z<-1) represents a probability of 16%.

Using the formula for z-scores:

$\\ z = \frac{x - \mu}{\sigma}$

$\\ -1 = \frac{x - 47}{8}$