Using the Normal Distribution

Example

The shaded area in the following graph indicates the area to the left of
x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x).

The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.

Example

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

1. Find the probability that a randomly selected student scored more than 65 on the exam.
2. Find the probability that a randomly selected student scored less than 85.
3. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).
4. Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).

Solution:

1. LetX = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5
   1. Draw a graph. Then, find P(x > 65). P(x > 65) = 0.3446
      

The probability that any student selected at random scores more than 65 is 0.3446.

1. 1. Go into 2nd DISTR. After pressing 2nd DISTR, press2:normalcdf. The syntax for the instructions are as follows:normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10
      ⁹⁹) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter10^99 instead. The number 10⁹⁹ is way out in the right tail of the normal curve. We are calculating the area between 65 and 10⁹⁹. In some instances, the lower number of the area might be –1E99 (= –10⁹⁹). The number –10⁹⁹ is way out in the left tail of the normal curve.[latex]\displaystyle{z}=\frac{{{65}-{63}}}{{5}}={0.4}[/latex]Area to the left is 0.6554.
      P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446
   2. Calculate the z-score:*Press
      2nd Distr
*Press 3:invNorm(*Enter the area to the left of z followed by )*Press ENTER.For this Example, the steps are2nd Distr3:invNorm(.6554) ENTERThe answer is 0.3999 which rounds to 0.4.
2. Draw a graph. Then find P(x < 85), and shade the graph. Using a computer or calculator, find P(x < 85) = 1.normalcdf(0,85,63,5) = 1 (rounds to one)The probability that one student scores less than 85 is approximately one (or 100%).
3. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile.
   Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value.k = 69.4
   The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)For this problem, invNorm(0.90,63,5) = 69.4
4. Find the 70th percentile. Draw a new graph and label it appropriately. k = 65.6The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.invNorm(0.70,63,5) = 65.6

Example

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.

1. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
2. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

Solution:

1. Let X= the amount of time (in hours) a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5. Find P(1.8 < x < 2.75).The probability for which you are looking is the area
   betweenx = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886
   normalcdf(1.8,2.75,2,0.5) = 0.5886The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
2. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25.
   invNorm(0.25,2,0.5) = 1.66The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

Example

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

1. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
2. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
3. Find the 80th percentile of this distribution, and interpret it in a complete sentence.

Solution:

1. normalcdf(23,64.7,36.9,13.9) = 0.8186
2. normalcdf(–1099,50.8,36.9,13.9) = 0.8413
3. invNorm(0.80,36.9,13.9) = 48.6 The 80th percentile is 48.6 years. 80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.

Example

There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

1. Calculate the interquartile range (IQR).
2. Forty percent of the ages that range from 13 to 55+ are at least what age?

Solution:

1. IQR = Q₃ – Q₁Calculate
   Q₃ = 75th percentile and Q₁ = 25thpercentile. invNorm(0.75,36.9,13.9) = Q₃ = 46.2754invNorm(0.25,36.9,13.9) = Q₁ = 27.5246
   IQR = Q₃ – Q₁ = 18.7508
2. Find k where P(x > k) = 0.40 (“At least” translates to “greater than or equal to.”) 0.40 = the area to the right. Area to the left = 1 – 0.40 = 0.60. The area to the left of k = 0.60.invNorm(0.60,36.9,13.9) = 40.4215. k = 40.42. Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.

Example

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.

1. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.
2. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.
3. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.

Solution:

1. normalcdf(6,10^99,5.85,0.24) = 0.2660
   
2. 1 – 0.20 = 0.80 The tails of the graph of the normal distribution each have an area of 0.40. Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60). k1 = invNorm(0.40,5.85,0.24) = 5.79 cm k2 = invNorm(0.60,5.85,0.24) = 5.91 cm
3. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.