The Normal Distribution

The standardized normal distribution is a type of normal distribution, with a mean of 0 and standard deviation of 1. It represents a distribution of standardized scores, called z-scores, as opposed to raw scores (the actual data values). A z-score indicates the number of standard deviation a score falls above or below the mean. Z-scores allow for comparison of scores, occurring in different data sets, with different means and standard deviations. It would not make sense to compare apples and oranges. Likewise, it does not make sense to compare scores from two different samples that have different means and standard deviations. Z-scores can be looked up in a Z-Table of Standard Normal Distribution, in order to find the area under the standard normal curve, between a score and the mean, between two scores, or above or below a score. The standard normal distribution allows us to interpret standardized scores and provides us with one table that we may use, in order to compute areas under the normal curve, for an infinite number of data sets, no matter what the mean or standard deviation.

A z-score is calculated as $z=\dfrac{x−μ}{σ}$. The score itself can be found by using algebra and solving for x. Multiplying both sides of the equation by σ gives: $(z)(σ)=x−μ$. Adding μ to both sides of the equation gives $μ+(z)(σ)=x$.

Suppose we have a data set with a mean of 5 and standard deviation of 2. We want to determine the number of standard deviations the score of 11 falls above the mean. We can find this answer (or z-score) by writing

$z=\dfrac{11−5}{2}=3$

or

$5+(z)(2)=11$,

we can solve for z.

$2z=6$

$z=3$

We have determined that the score of 11 falls 3 standard deviations above the mean of 5.

With a standard normal distribution, we indicate the distribution by writing Z ~ N(0, 1) which shows the normal distribution has a mean of 0 and standard deviation of 1. This notation simply indicates that a standard normal distribution is being used.

Z-Scores

As described previously, if X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is

$z=\dfrac{x – μ}{σ}$.

The z-score tells you how many standard deviations the value x is above, to the right of, or below, to the left of, the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.

When determining the z-score for an x-value, for a normal distribution, with a given mean and standard deviation, the notation above for a normal distribution, will be given.

Example 6.1

Suppose X ~ N(5, 6). This equation says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then,

$z=\dfrac{x–μ}{σ}=\dfrac{17–5}{6}=2$.

This means that x = 17 is two standard deviations (2σ) above, or to the right, of the mean μ = 5.

Notice that 5 + (2)(6) = 17. The pattern is μ + zσ = x.

Now suppose x = 1. Then, $z = \dfrac{x–μ}{σ} = \dfrac{1–5}{6} = –0.67$, rounded to two decimal places.

This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. This z-score shows that x = 1 is less than 1 standard deviation below the mean of 5. Therefore, the score doesn't fall very far below the mean.

Summarizing, when z is positive, x is above or to the right of μ, and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative, x is less than μ. The absolute value of z indicates how far the score is from the mean, in either direction.

Try It 6.1

What is the z-score of x, when x = 1 and X ~ N(12, 3)?

Example 6.2

Some doctors believe that a person can lose five pounds, on average, in a month by reducing his or her fat intake and by consistently exercising. Suppose weight loss has a normal distribution. Let X = the amount of weight lost, in pounds, by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.

a. Suppose a person lost 10 pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

Solution 1

a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.

b. Suppose a person gained three pounds, a negative weight loss. Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.

Solution 2

b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean.

c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. This was previously shown. If y = 4, what is z?

Solution 3

c. $z =\dfrac{y−μ}{σ} = \dfrac{4−2}{1}= 2$, where µ = 2 and σ = 1.

The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two of their own standard deviations to the right of their respective means.

The z-score allows us to compare data that are scaled differently. To better understand the concept, suppose X ~ N(5, 6) represents weight gains for one group of people who are trying to gain weight in a six-week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.

Try It 6.2

Fill in the blanks.

Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16, 4). Suppose Jerome scores 10 points in a game. The z-score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______ (right or left) of the mean______ (What is the mean?).

The Empirical Rule

If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following:

• About 68 percent of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean).
• About 95 percent of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean).
• About 99.7 percent of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean.
• The z-scores for +1σ and –1σ are +1 and –1, respectively.
• The z-scores for +2σ and –2σ are +2 and –2, respectively.
• The z-scores for +3σ and –3σ are +3 and –3, respectively.
  

So, in other words, this is that about 68 percent of the values lie between z-scores of –1 and 1, about 95% of the values lie between z-scores of –2 and 2, and about 99.7 percent of the values lie between z-scores of -3 and 3. These facts can be checked, by looking up the mean to z area in a z-table for each positive z-score and multiplying by 2.

The empirical rule is also known as the 68–95–99.7 rule.

Figure 6.3

Example 6.3

The mean height of 15-to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010. Then X ~ N(170, 6.28).

a. Suppose a 15-to 18-year-old male from Chile was 168 cm tall in 2009–2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

Solution 1

a. –0.32, 0.32, left, 170

b. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = 1.27. What is the male's height? The z-score (z = 1.27) tells you that the male's height is ________ standard deviations to the __________ (right or left) of the mean.

Solution 2

b. 177.98 cm, 1.27, right

Try It 6.3

Use the information in Example 6.3 to answer the following questions:

1. Suppose a 15-to 18-year-old male from Chile was 176 cm tall from 2009–2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = –2. What is the male's height? The z-score (z = –2) tells you that the male's height is ________ standard deviations to the __________ (right or left) of the mean.

Example 6.4

From 1984 to 1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males from 1984–1985, and y = the height of one male from this group. Then Y ~ N(172.36, 6.34).

The mean height of 15-to 18-year-old males from Chile in 2009–2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010, and x = the height of one male from this group. Then X ~ N(170, 6.28).

Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations?

Solution 1

The z-score for x = 160.58 cm is z = –1.5.
The z-score for y = 162.85 cm is z = –1.5.
Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.

Try It 6.4

In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then, X ~ N(496, 114).

Find the z-scores for x₁ = 325 and x₂ = 366.21. Interpret each z-score. What can you say about x₁ = 325 and x₂ = 366.21, as they compare to their respective means and standard deviations?

Example 6.5

Suppose x has a normal distribution with mean 50 and standard deviation 6.

• About 68 percent of the x values lie within one standard deviation of the mean. Therefore, about 68 percent of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively.
• About 95 percent of the x values lie within two standard deviations of the mean. Therefore, about 95 percent of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively.
• About 99.7 percent of the x values lie within three standard deviations of the mean. Therefore, about 95 percent of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations from the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively.

Try It 6.5

Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68 percent of the values lie?

Example 6.6

From 1984–1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males in 1984–1985. Then Y ~ N(172.36, 6.34).

1. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
3. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.

Solution 1

1. About 68 percent of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1.
2. About 95 percent of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2.
3. About 99.7 percent of the values lie between 1153.34 cm and 191.38 cm. The z-scores are –3 and 3.
   
   

Try It 6.6

The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points.

1. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
3. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.

Source: OpenStax, https://openstax.org/books/statistics/pages/6-introduction
This work is licensed under a Creative Commons Attribution 4.0 License.

The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to 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.

Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61.

Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

NOTE

To calculate the probability, use the probability tables provided in Appendix H Tables without the use of technology. The tables include instructions for how to use them.

The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. Others show the mean to z area. The method used will be indicated on the table.

We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores.

Example 6.7

If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.

Try It 6.7

If the area to the left of x is 0.012, then what is the area to the right?

Example 6.8

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

a. Find the probability that a randomly selected student scored more than 65 on the exam.

Solution 1

a. Let X= a score on the final exam. X~N(63, 5), where μ= 63 and σ= 5.

Draw a graph.

Calculate the z-score:

$z=\dfrac{x−μ}{σ}=\dfrac{65−63}{5}=\dfrac{2}{5}=.40$

The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446.

Then, find P(x ≥ 65).

$P(x ≥ 65) = 0.3446$

Figure 6.5

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

Using the TI-83, 83+, 84, 84+ Calculator

Go into 2nd DISTR. After pressing 2nd DISTR, press 2:normalcdf.

The syntax for the instructions is as follows:

normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf (65,1E99,63,5) = 0.3446. You get 1E99 (= 10⁹⁹) by pressing 1 , the EE key - a 2nd key - and then 99. Or, you can enter 10^99 instead. The number 10⁹⁹is 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 –10⁹⁹is way out in the left tail of the normal curve. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose.

Historical Note

The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table, also known as a Z-table - the math involved to find probability is cumbersome. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right.

Using the TI-83, 83+, 84, 84+ Calculator

Calculate thez-score

*Press 2nd Distr

*Press 3:invNorm(

*Enter the area to the left of z followed by )
*Press ENTER.

For this Example, the steps are


2nd Distr3:invNorm(.6554) ENTER

The answer is 0.3999, which rounds to 0.4.

b. Find the probability that a randomly selected student scored less than 85.

Solution 2

b. 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 percent.

c. Find the 90^thpercentile, - that is, find the score k that has 90 percent of the scores below k and 10 percent of the scores above k.

Solution 3

c. Find the 90^thpercentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90^thpercentile. This time, we are looking for a score that corresponds to a given area under the curve.

Let k= the 90^thpercentile.The variable k is located on the x-axis.P(x < k) is the area to the left of k. The 90^thpercentilekseparates 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 thank, and 10 percent are the same or higher. The variable k is often called a critical value.

We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation, into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9. Thus, we can write the following:

$1.28 = \dfrac{x-63}{5}$

Multiplying each side of the equation by 5 gives

$6.4 = x - 63$

Adding 63 to both sides of the equation gives

$69.4 = x$.

Thus, our score, k, is 69.4.

$k=69.4$

Figure 6.6

The 90^thpercentile is 69.4. This means that 90 percent of the test scores fall at or below 69.4 and 10 percent fall at or above. To get this answer on the calculator, follow this next step:

Using the TI-83, 83+, 84, 84+ Calculator

invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)
For this problem, invNorm(0.90,63,5) = 69.4

d. Find the 70^thpercentile, - that is, find the score k such that 70 percent of scores are below k and 30 percent of the scores are above k.

Solution 4

d. Find the 70^thpercentile.

Draw a new graph and label it appropriately.k= 65.6

The 70^thpercentile is 65.6. This means that 70 percent of the test scores fall at or below 65.5 and 30 percent fall at or above.

invNorm(0.70,63,5) = 65.6

Try It 6.8

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.

Find the probability that a randomly selected golfer scored less than 65.

Example 6.9

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.

a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

Solution 1

a. 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).

First, calculate the z-scores for each x-value.

$z=\dfrac{1.8−2}{0.5}=\dfrac{−0.2}{0.5}=−0.40$

$z=\dfrac{2.75−2}{0.5}=\dfrac{0.75}{0.5}=1.5$

Now, use the Z-table to locate the area under the normal curve to the left of each of these z-scores.

The area to the left of the z-score of −0.40 is 0.3446. The area to the left of the z-score of 1.5 is 0.9332. The area between these scores will be the difference in the two areas, or, which equals 0.5886.

Figure 6.7

normalcdf(1.8,2.75,2,0.5) = 0.5886

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.

b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

Solution 2

b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25^thpercentile, k, where P(x < k) = 0.25.

Figure 6.8

invNorm(0.25,2,0.5) = 1.66

We use invNorm because we are looking for the k-value.

The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

Try It 6.9

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.

Example 6.10

In the United States smartphone users between the ages of 13 and 55+ between the ages of 13 and 55+ approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.

Solution 1

a. normalcdf(23,64.7,36.9,13.9) = 0.8186

The z-scores are calculated as

$z=\dfrac{23−36.9}{13.9}=\dfrac{−13.9}{13.9}=−1$

$z=\dfrac{64.7−36.9}{13.9}=\dfrac{27.8}{13.9}=2$

The Z-table shows the area to the left of a z-score with an absolute value of 1 to be 0.1587. It shows the area to the left of a z-score of 2 to be 0.9772. The difference in the two areas is 0.8185.

This is slightly different than the area given by the calculator, due to rounding.

b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.

Solution 2

b. normalcdf(–10⁹⁹,50.8,36.9,13.9) = 0.8413

c. Find the 80^thpercentile of this distribution, and interpret it in a complete sentence.

Solution 3

• invNorm(0.80,36.9,13.9) = 48.6
• The 80^thpercentile is 48.6 years.
• 80 percent of the smartphone users in the age range 13–55+ are 48.6 years old or less.
  
  

Try It 6.10

Use the information in Example 6.10 to answer the following questions:

1. Find the 30^thpercentile, and interpret it in a complete sentence.
2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old?
   
   

Example 6.11

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.

a. Calculate the interquartile range (IQR).

Solution 1

a.

• IQR=Q₃–Q₁
• CalculateQ₃= 75^thpercentile andQ₁= 25^thpercentile.
• Recall that we can use invNorm to find the k-value. We can use this to find the quartile values.
• invNorm(0.75,36.9,13.9) =Q₃= 46.2754
• invNorm(0.25,36.9,13.9) =Q₁= 27.5246
• IQR=Q₃–Q₁= 18.8

b. Forty percent of the ages that range from 13 to 55+ are at least what age?

Solution 2

b.

• Find k whereP(x ≥ k) = 0.40.At least translates to greater than or equal to.
• 0.40 = the area to the right
• The 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= 40.4.
• Forty percent of the ages that range from 13 to 55+ are at least 40.4 years.
  
  

Try It 6.11

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ= 81 points and standard deviation σ= 15 points.

1. Calculate the first- and third-quartile scores for this exam.
2. The middle 50 percent of the exam scores are between what two values?
   
   

Example 6.12

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.

a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.

Solution 1

a. normalcdf(6,10^99,5.85,0.24) = 0.2660

Figure 6.9

b. The middle 20 percent of mandarin oranges from this farm have diameters between ______ and ______.

Solution 2

b.

• 1 – 0.20 = 0.80. Outside of the middle 20 percent will be 80 percent of the values.
• The tails of the graph of the normal distribution each have an area of 0.40.
• Find k₁, the 40^thpercentile, and k₂, the 60^thpercentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution.
• k₁= invNorm(0.40,5.85,0.24) = 5.79 cm
• k₂= invNorm(0.60,5.85,0.24) = 5.91 cm

So, the middle 20 percent of mandarin oranges have diameters between 5.79 cm and 5.91 cm.

c. Find the 90^thpercentile for the diameters of mandarin oranges, and interpret it in a complete sentence.

Solution 3

c. 6.16, Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm.

Try It 6.12

Using the information from Example 6.12, answer the following:

1. The middle 45 percent of mandarin oranges from this farm are between ______ and ______.
2. Find the 16^thpercentile, and interpret it in a complete sentence.