> ## Documentation Index
> Fetch the complete documentation index at: https://v1-learn.neoartd.my.id/llms.txt
> Use this file to discover all available pages before exploring further.

# Applications of the Normal Distribution

### Applications of the Normal Distribution - Example

#### Question 1

<Card>
  Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. Assume the amount generated is normally distributed.

  <br />

  If a household is selected at random, find the probability of its generating:

  <AccordionGroup>
    <Accordion title="More than 30.2 pounds per month.">
      * First find the z-value for 30.2.

      <br />

      * $~~~~~~~~~~z = \dfrac{X-\mu}{\sigma} = \dfrac{30.2-28}{2} = 1.1 $

      <br />

      * Thus, $P(z > 1.1) = 0.5 - 0.3643 = 0.1357$.
      * That is, the probability that a randomly selected household will generate more than 30.2 lbs. of newspapers is 0.1357 or 13.57%.

      <br />

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/7/images/screenshot-1715689675128.png" alt="Question 1" />
    </Accordion>

    <Accordion title="Between 27 and 31 pounds per month.">
      * First find the z-value for 27 and 31.

      <br />

      * $~~~~~~~~~~z_1 = \dfrac{X-\mu}{\sigma} = \dfrac{27-28}{2} = -0.5 $
      * $$
        $$
      * $~~~~~~~~~~z_2 = \dfrac{X-\mu}{\sigma} = \dfrac{31-28}{2} = 1.5 $

      <br />

      * Thus, $P(-0.5 < z < 1.5) = 0.1915 + 0.4332 = 0.6247$.

      <br />

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/7/images/screenshot-1715689884329.png" alt="Question 1" />
    </Accordion>
  </AccordionGroup>
</Card>

#### Question 2

<Card>
  The American Automobile Association
  reports that the average time it takes to
  respond to an emergency call is 25
  minutes. Assume the variable is
  approximately normally distributed and
  the standard deviation is 4.5 minutes.

  <br />

  If 80 calls are randomly selected

  <AccordionGroup>
    <Accordion title="approximately how many will be responded to in less than 15 minutes?">
      * First find the z-value for 15 is

      <br />

      * $~~~~~~~~~~z = \dfrac{X-\mu}{\sigma} = \dfrac{15-25}{4.5} = -2.22 $

      <br />

      * Thus, $P(z < -2.22) = 0.5000 - 0.4868 = 0.0132$.
      * The number of calls that will be made in less than 15 minutes = (80)(0.0132) = 1.056 $\approx$ 1.

      <br />

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/7/images/screenshot-1715690076828.png" alt="Question 1" />
    </Accordion>
  </AccordionGroup>
</Card>

#### Question 3

<Card>
  An exclusive college desires to accept
  only the top 10% of all graduating
  seniors based on the results of a
  national placement test. This test has a
  mean of 500 and a standard deviation
  of 100.

  <br />

  <AccordionGroup>
    <Accordion title="Find the cutoff score for the exam.">
      * Assume the variable is normally distributed.
      * Work backward to solve this problem.
      * Subtract 0.1 (10%) from 0.5 to get the area under the normal curve for accepted students.
      * Find the z value that corresponds to an area of 0.4000 by looking up 0.4000 in the area portion of Table E. Use the closest value, 0.3997.
      * Substitute in the formula $z = \dfrac{X-\mu}{\sigma}$ and solve for $X$.
      * The z-value for the cutoff score $(X) ~ \text{is} ~ z = \dfrac{X-\mu}{\sigma} = \dfrac{X-500}{100} = 1.28$.
      * Thus, $X = (1.28)(100) + 500 = 628$.
      * The score of 628 should be used as a cutoff score.

      <br />

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/7/images/screenshot-1715690372098.png" alt="Question 1" />
    </Accordion>
  </AccordionGroup>
</Card>

#### Question 4

<Card>
  <Note>
    To solve for X, use the following
    formula: $X = z \times \sigma + \mu$.
  </Note>

  For a medical study, a
  researcher wishes to select people in
  the middle 60% of the population based
  on blood pressure.

  <br />

  If the mean systolic blood
  pressure is 120 and the standard
  deviation is 8

  <AccordionGroup>
    <Accordion title="find the upper and lower readings that would qualify people to participate in the study.">
      * Note that two values are needed, one above the mean and one below the mean. The closest z values are 0.84 and - 0.84 respectively.
      * $X = z \times \sigma + \mu = (0.84)(8) + 120 = 126.72$ The other $X = (-0.84)(8) + 120 = 113.28$.
      * i.e. the middle 60% of BP readings is between 113.28 and 126.72.

      <br />

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/7/images/screenshot-1715691068465.png" alt="Question 1" />
    </Accordion>
  </AccordionGroup>
</Card>