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Applications of the Normal Distribution - Example

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Question 1

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.
If a household is selected at random, find the probability of its generating:
  • First find the z-value for 30.2.

  •           z=X−μσ=30.2−282=1.1~~~~~~~~~~z = \dfrac{X-\mu}{\sigma} = \dfrac{30.2-28}{2} = 1.1

  • Thus, P(z>1.1)=0.5−0.3643=0.1357P(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%.

Question 1
  • First find the z-value for 27 and 31.

  •           z1=X−μσ=27−282=−0.5~~~~~~~~~~z_1 = \dfrac{X-\mu}{\sigma} = \dfrac{27-28}{2} = -0.5
  •           z2=X−μσ=31−282=1.5~~~~~~~~~~z_2 = \dfrac{X-\mu}{\sigma} = \dfrac{31-28}{2} = 1.5

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

Question 1

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Question 2

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.
If 80 calls are randomly selected
  • First find the z-value for 15 is

  •           z=X−μσ=15−254.5=−2.22~~~~~~~~~~z = \dfrac{X-\mu}{\sigma} = \dfrac{15-25}{4.5} = -2.22

  • Thus, P(z<−2.22)=0.5000−0.4868=0.0132P(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.

Question 1

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Question 3

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.
  • 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=X−μσz = \dfrac{X-\mu}{\sigma} and solve for XX.
  • The z-value for the cutoff score (X) is z=X−μσ=X−500100=1.28(X) ~ \text{is} ~ z = \dfrac{X-\mu}{\sigma} = \dfrac{X-500}{100} = 1.28.
  • Thus, X=(1.28)(100)+500=628X = (1.28)(100) + 500 = 628.
  • The score of 628 should be used as a cutoff score.

Question 1

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Question 4

To solve for X, use the following formula: X=z×σ+μX = z \times \sigma + \mu.
For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure.
If the mean systolic blood pressure is 120 and the standard deviation is 8
  • 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×σ+μ=(0.84)(8)+120=126.72X = z \times \sigma + \mu = (0.84)(8) + 120 = 126.72 The other X=(−0.84)(8)+120=113.28X = (-0.84)(8) + 120 = 113.28.
  • i.e. the middle 60% of BP readings is between 113.28 and 126.72.

Question 1