homework 2 normal distribution and z scores

Chapter 6: The Normal Distribution and The Central Limit Theorem

6.1 The Standard Normal Distribution

Learning objectives.

By the end of this section, the student should be able to:

• Calculate and interpret z-scores.
• Recognize characteristics of the standard normal distribution.
• Apply the empirical rule to solve real-world applications.

The normal distribution has two parameters (two numerical descriptive measures): the mean ( μ ) and the standard deviation ( σ ). If X is a quantity to be measured that has a normal distribution with mean ( μ ) and standard deviation ( σ ), we designate this by writing X~N(μ, σ).

The probability density function of this curve is as follows:

f ( x ) = [latex]\frac{1}{\sigma \cdot \sqrt{2\cdot \pi }} \cdot {\text{ e}}^{-\frac{1}{2}\cdot {\left(\frac{x-\mu }{\sigma }\right)}^{2}}[/latex]

• -∞ < X < ∞
• -∞ < μ < ∞

As you can see, the normal pdf is a rather complicated function. Do not memorize it . It is not necessary. But this could be a problem since the normal distribution is so widely used. However we will see some ways we can work around this.

The cumulative distribution function is P ( X < x ). It is calculated either by a calculator or a computer, or it is looked up in a table. Technology has made the tables virtually obsolete.

The curve is symmetrical about a vertical line drawn through the mean, μ . In theory, the mean is the same as the median, because the graph is symmetric about μ . As the notation indicates, the normal distribution depends only on the mean and the standard deviation. Since the area under the curve must equal one, a change in the standard deviation, σ , causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on σ . A change in μ causes the graph to shift to the left or right. This means there are an infinite number of normal probability distributions. One of special interest is called the standard normal distribution .

The standard normal distribution is a normal distribution of standardized values called z -scores . A z -score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:

x = μ + ( z )( σ ) = 5 + (3)(2) = 11

The z -score is three.

The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = [latex]\frac{x-\mu }{\sigma }[/latex] produces the distribution Z ~ N (0, 1). The value x comes from a normal distribution with mean μ and standard deviation σ .

If X is a normally distributed random variable and X ~ N(μ, σ) , then the z -score is:

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.

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

Then: [latex]z=\frac{x–\mu }{\sigma }=\frac{17–5}{6}=2[/latex]

This means that x = 17 is two standard deviations (2 σ ) above or to the right of the mean μ = 5. The standard deviation is σ = 6.

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

Now suppose x = 1. Then: z = [latex]\frac{x–\mu }{\sigma }[/latex] = [latex]\frac{1–5}{6}[/latex] = –0.67 (rounded to two decimal places)

This means that x = 1 is 0.67 standard deviations (–0.67 σ ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)

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 .

Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. 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 ten 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?).

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.

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 ?

z = [latex]\frac{y-\mu }{\sigma }[/latex] = [latex]\frac{4-2}{1}[/latex] = 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 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 .

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

[latex]z=\frac{1–12}{3}\approx –3.67[/latex]

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 to 2010. Then X ~ N (170, 6.28).

a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 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?).

z = –0.32, z = 0.32, left, μ = 170

b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 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.

x = 177.98, 1.27, right

1. Suppose a 15- to 18-year-old male from Chile was 176 cm tall from 2009 to 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 from 2009 to 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.

Solve the equation z = [latex]\frac{x-\mu }{\sigma }[/latex] for x . x = μ + ( z )( σ )

• z = [latex]\frac{176-170}{6.28}[/latex] ≈ 0.96, This z -score tells you that x = 176 cm is 0.96 standard deviations to the right of the mean 170 cm.
• X = 157.44 cm, The z -score( z = –2) tells you that the male’s height is two standard deviations to the left of the mean

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 to 1985. Then Y ~ N (172.36, 6.34).

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 to 2010. 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?

The z -score for x = 160.58 is z = –1.5. The z -score for y = 162.85 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.

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 x 1 = 325 and x 2 = 366.21. Interpret each z -score. What can you say about x 1 = 325 and x 2 = 366.21?

The z -score for x 1 = 325 is z 1 = –1.14.

The z -score for x 2 = 366.21 is z 2 = –1.14.

Student 2 scored closer to the mean than Student 1, and since they both had negative z -scores, Student 2 had the better score.

Fill in the blanks.

Jerome averages 16 points a game with a standard deviation of four points. X ~ N (16,4). Suppose Jerome scores ten 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?).

1.5, left, 16

The Empirical Rule

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 says the following:

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

Notice that almost all the x values lie within three standard deviations of the mean. The empirical rule is also known as the 68-95-99.7 rule.

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

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

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 in 1984 to 1985. Then Y ~ N (172.36, 6.34).

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

• About 68% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively.
• About 95% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively.
• About 99.7% of the y values lie between what two values? These values are ________________. The z -scores are ________________, respectively.
• About 68% of the values lie between the values 41 and 63. The z -scores are –1 and 1, respectively.
• About 95% of the values lie between the values 30 and 74. The z -scores are –2 and 2, respectively.
• About 99.7% of the values lie between the values 19 and 85. The z -scores are –3 and 3, respectively.

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

between 20 and 30.

“Blood Pressure of Males and Females.” StatCrunch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013).

“The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013).

“2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013).

“Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013).

Data from the San Jose Mercury News .

Data from The World Almanac and Book of Facts .

“List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013).

Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013) .

Media Attributions

• Private: Figure 5.10 © Significant Statistics by John Morgan Russell is licensed under a CC BY-SA (Attribution ShareAlike) license
• Private: Figure 5.11 © Significant Statistics by John Morgan Russell is licensed under a CC BY-SA (Attribution ShareAlike) license

A function that defines a continuous random variable, and the likelihood of an outcome

a continuous random variable (RV) with a mean of 0 and a standard deviation of 1 which z-scores follow: X ~ N(0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N(0, 1).

the linear transformation of the form z = [latex]\frac{x\text{ }–\text{ }\mu }{\sigma }[/latex]; if this transformation is applied to any normal distribution X ~ N(μ, σ) the result is the standard normal distribution Z ~ N(0,1). If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ, the result is called the z-score of x. The z-score allows us to compare data that are normally distributed but scaled differently.

Roughly 68% of values are within 1 standard deviation of the mean, roughly 95% of values are within 2 standard deviations of the mean, and 99.7% of values are within 3 standard deviations of the mean

Introductory Statistics Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.

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6: The Normal Distribution

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In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them. The normal distribution has two parameters (two numerical descriptive measures), the mean (μ) and the standard deviation (σ).

• 6.0: Prelude to The Normal Distribution The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them. The normal distribution has two parameters (two numerical descriptive measures), the mean (μ) and the standard deviation (σ).
• 6.1E: The Standard Normal Distribution (Exercises)
• 6.2: Using the Normal Distribution The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean μ and the standard deviation σ. A special normal distribution, called the standard normal distribution is the distribution of z-scores. Its mean is zero, and its standard deviation is one.
• 6.3: Normal Distribution - Lap Times (Worksheet) A statistics Worksheet: The student will compare and contrast empirical data and a theoretical distribution to determine if Terry Vogel's lap times fit a continuous distribution.
• 6.4: Normal Distribution - Pinkie Length (Worksheet) A statistics Worksheet: The student will compare empirical data and a theoretical distribution to determine if data from the experiment follow a continuous distribution.
• 6.E: The Normal Distribution (Exercises) These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 4.

• Z-score introduction

Calculating z-scores

• Comparing with z-scores
• Z-scores-problem
• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Module 6: Normal Distribution

6.2: using the normal distribution, learning outcomes.

• Recognize the normal probability distribution and apply it appropriately.
• Compare normal probabilities by converting to the standard normal distribution.

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.

Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. Additionally, t his link houses a tool that allows you to explore the normal distribution with varying means and standard deviations as well as associated probabilities.  The following video explains how to use the tool.

To calculate the probability without the use of technology, use the probability tables provided  here . The tables include instructions for how to use them.

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

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

1 − 0.012 = 0.988

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

• Find the probability that a randomly selected student scored more than 65 on the exam.
• Find the probability that a randomly selected student scored less than 85.
• 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 ).
• 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 ).

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

• Go into 2nd DISTR . After pressing 2nd DISTR , press 2: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 enter 10^99 instead. The number 10 99 is way out in the right tail of the normal curve. We are calculating the area between 65 and 10 99 . In some instances, the lower number of the area might be –1E99 (= –10 99 ). The number –10 99 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
• 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 are 2nd Distr 3:invNorm (.6554) ENTER The answer is 0.3999 which rounds to 0.4.
• 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%).

• 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

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.

normalcdf(10 99 ,65,68,3) = 0.1587

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.

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

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.

normalcdf(66,70,68,3) = 0.4950

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.

• Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
• Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
• Find the 80th percentile of this distribution, and interpret it in a complete sentence.
• normalcdf(23,64.7,36.9,13.9) = 0.8186
• normalcdf(–1099,50.8,36.9,13.9) = 0.8413
• 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.

Use the information in Example 3 to answer the following questions.

• Find the 30th percentile, and interpret it in a complete sentence.
• 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.

Let X = a smart phone user whose age is 13 to 55+. X ~ N (36.9, 13.9)

• To find the 30th percentile, find k such that P ( x < k ) = 0.30.invNorm(0.30, 36.9, 13.9) = 29.6 yearsThirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.

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).

• Calculate the interquartile range ( IQR ).
• Forty percent of the ages that range from 13 to 55+ are at least what age?
• IQR = Q 3 – Q 1 Calculate Q 3 = 75th percentile and Q 1 = 25thpercentile. invNorm(0.75,36.9,13.9) = Q 3 = 46.2754invNorm(0.25,36.9,13.9) = Q 1 = 27.5246 IQR = Q 3 – Q 1 = 18.7508
• 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.

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.

• Calculate the first- and third-quartile scores for this exam.
• The middle 50% of the exam scores are between what two values?
• Q 1 = 25th percentile = invNorm(0.25,81,15) = 70.9 Q 3 = 75th percentile = invNorm(0.75,81,15) = 91.9
• The middle 50% of the scores are between 70.9 and 91.1.

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.

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

• 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
• 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.

Using the information from Example 5, answer the following:

• The middle 45% of mandarin oranges from this farm are between ______ and ______.
• Find the 16th percentile and interpret it in a complete sentence.
• The middle area = 0.40, so each tail has an area of 0.30.1 – 0.40 = 0.60The tails of the graph of the normal distribution each have an area of 0.30.Find k1 , the 30th percentile and k2 , the 70th percentile (0.40 + 0.30 = 0.70). k1 = invNorm(0.30,5.85,0.24) = 5.72 cm k2 = invNorm(0.70,5.85,0.24) = 5.98 cm
• normalcdf(5,1099,5.85,0.24) = 0.9998

“Naegele’s rule.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Naegele’s_rule (accessed May 14, 2013).

“403: NUMMI.” Chicago Public Media & Ira Glass, 2013. Available online at http://www.thisamericanlife.org/radio-archives/episode/403/nummi (accessed May 14, 2013).

“Scratch-Off Lottery Ticket Playing Tips.” WinAtTheLottery.com, 2013. Available online at http://www.winatthelottery.com/public/department40.cfm (accessed May 14, 2013).

“Smart Phone Users, By The Numbers.” Visual.ly, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed May 14, 2013).

“Facebook Statistics.” Statistics Brain. Available online at http://www.statisticbrain.com/facebook-statistics/(accessed May 14, 2013).

Concept Review

The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ . A special normal distribution, called the standard normal distribution is the distribution of z -scores. Its mean is zero, and its standard deviation is one.

Formula Review

Normal Distribution: X ~ N ( µ , σ ) where µ is the mean and σ is the standard deviation.

Standard Normal Distribution: Z ~ N (0, 1).

Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation)

Calculator function for the k th percentile: k = invNorm (area to the left of k , mean, standard deviation)

• OpenStax, Statistics,Using the Normal Distribution. Authored by : . Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:41/Introductory_Statistics . License : CC BY: Attribution
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• How to use the online Normal Distribution Calculator. Authored by : gatorpj. Provided by : https://www.youtube.com/watch?v=rOs-jlJvcyM. License : All Rights Reserved . License Terms : Standard YouTube License

Z-score Calculator

Use this calculator to compute the z-score of a normal distribution.

Z-score and Probability Converter

Please provide any one value to convert between z-score and probability. This is the equivalent of referencing a z-table.

Probability between Two Z-scores

Use this calculator to find the probability (area P in the diagram) between two z-scores.

Related Standard Deviation Calculator

What is z-score?

The z-score, also referred to as standard score, z-value, and normal score, among other things, is a dimensionless quantity that is used to indicate the signed, fractional, number of standard deviations by which an event is above the mean value being measured. Values above the mean have positive z-scores, while values below the mean have negative z-scores.

The z-score can be calculated by subtracting the population mean from the raw score, or data point in question (a test score, height, age, etc.), then dividing the difference by the population standard deviation:

where x is the raw score, μ is the population mean, and σ is the population standard deviation. For a sample, the formula is similar, except that the sample mean and population standard deviation are used instead of the population mean and population standard deviation.

The z-score has numerous applications and can be used to perform a z-test, calculate prediction intervals, process control applications, comparison of scores on different scales, and more.

A z-table, also known as a standard normal table or unit normal table, is a table that consists of standardized values that are used to determine the probability that a given statistic is below, above, or between the standard normal distribution. A z-score of 0 indicates that the given point is identical to the mean. On the graph of the standard normal distribution, z = 0 is therefore the center of the curve. A positive z-value indicates that the point lies to the right of the mean, and a negative z-value indicates that the point lies left of the mean. There are a few different types of z-tables.

The values in the table below represent the area between z = 0 and the given z-score.

How to read the z-table

In the table above,

• the column headings define the z-score to the hundredth's place.
• the row headings define the z-score to the tenth's place.
• each value in the table is the area between z = 0 and the z-score of the given value, which represents the probability that a data point will lie within the referenced region in the standard normal distribution.

For example, referencing the right-tail z-table above, a data point with a z-score of 1.12 corresponds to an area of 0.36864 (row 13, column 4). This means that for a normally distributed population, there is a 36.864% chance, a data point will have a z-score between 0 and 1.12.

Because there are various z-tables, it is important to pay attention to the given z-table to know what area is being referenced.

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11.2E: The Standard Normal Distribution (Exercises)

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Exercise 6.2.7

A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable \(X\) in words. \(X =\) ____________.

ounces of water in a bottle

Exercise 6.2.8

A normal distribution has a mean of 61 and a standard deviation of 15. What is the median?

Exercise 6.2.9

\(X \sim N(1, 2)\)

\(\sigma =\) _______

Exercise 6.2.10

A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable \(X\) in words. \(X =\) ______________.

Exercise 6.2.11

\(X \sim N(-4, 1)\)

What is the median?

Exercise 6.2.12

\(X \sim N(3, 5)\)

Exercise 6.2.13

\(X \sim N(-2, 1)\)

\(\mu =\) _______

Exercise 6.2.14

What does a \(z\)-score measure?

Exercise 6.2.15

What does standardizing a normal distribution do to the mean?

The mean becomes zero.

Exercise 6.2.16

Is \(X \sim N(0, 1)\) a standardized normal distribution? Why or why not?

Exercise 6.2.17

What is the \(z\)-score of \(x = 12\), if it is two standard deviations to the right of the mean?

Exercise 6.2.18

What is the \(z\)-score of \(x = 9\), if it is 1.5 standard deviations to the left of the mean?

Exercise 6.2.19

What is the \(z\)-score of \(x = -2\), if it is 2.78 standard deviations to the right of the mean?

\(z = 2.78\)

Exercise 6.2.20

What is the \(z\)-score of \(x = 7\), if it is 0.133 standard deviations to the left of the mean?

Exercise 6.2.21

Suppose \(X \sim N(2, 6)\). What value of x has a z -score of three?

Exercise 6.2.22

Suppose \(X \sim N(8, 1)\). What value of \(x\) has a \(z\)-score of –2.25?

Exercise 6.2.23

Suppose \(X \sim N(9, 5)\). What value of \(x\) has a \(z\)-score of –0.5?

\(x = 6.5\)

Exercise 6.2.24

Suppose \(X \sim N(2, 3)\). What value of \(x\) has a \(z\)-score of –0.67?

Exercise 6.2.25

Suppose \(X \sim N(4, 2)\). What value of \(x\) is 1.5 standard deviations to the left of the mean?

Exercise 6.2.26

Suppose \(X \sim N(4, 2)\). What value of \(x\) is two standard deviations to the right of the mean?

Exercise 6.2.27

Suppose \(X \sim N(8, 9)\). What value of \(x\) is 0.67 standard deviations to the left of the mean?

\(x = 1.97\)

Exercise 6.2.28

Suppose \(X \sim N(-1, 12)\). What is the \(z\)-score of \(x = 2\)?

Exercise 6.2.29

Suppose \(X \sim N(12, 6)\). What is the \(z\)-score of \(x = 2\)?

\(z = –1.67\)

Exercise 6.2.30

Suppose \(X \sim N(9, 3)\). What is the \(z\)-score of \(x = 9\)?

Exercise 6.2.31

Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the \(z\)-score of \(x = 5.5\)?

\(z \approx –0.33\)

Exercise 6.2.32

In a normal distribution, \(x = 5\) and \(z = –1.25\). This tells you that \(x = 5\) is ____ standard deviations to the ____ (right or left) of the mean.

Exercise 6.2.33

In a normal distribution, \(x = 3\) and \(z = 0.67\). This tells you that \(x = 3\) is ____ standard deviations to the ____ (right or left) of the mean.

0.67, right

Exercise 6.2.34

In a normal distribution, \(x = –2\) and \(z = 6\). This tells you that \(z = –2\) is ____ standard deviations to the ____ (right or left) of the mean.

Exercise 6.2.35

In a normal distribution, \(x = –5\) and \(z = –3.14\). This tells you that \(x = –5\) is ____ standard deviations to the ____ (right or left) of the mean.

Exercise 6.2.36

In a normal distribution, \(x = 6\) and \(z = –1.7\). This tells you that \(x = 6\) is ____ standard deviations to the ____ (right or left) of the mean.

Exercise 6.2.37

About what percent of \(x\) values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution?

Exercise 6.2.38

About what percent of the \(x\) values from a normal distribution lie within two standard deviations (left and right) of the mean of that distribution?

Exercise 6.2.39

About what percent of \(x\) values lie between the second and third standard deviations (both sides)?

Exercise 6.2.40

Suppose \(X \sim N(15, 3)\). Between what \(x\) values does 68.27% of the data lie? The range of \(x\) values is centered at the mean of the distribution (i.e., 15).

Exercise 6.2.41

Suppose \(X \sim N(-3, 1)\). Between what \(x\) values does 95.45% of the data lie? The range of \(x\) values is centered at the mean of the distribution (i.e., –3).

between –5 and –1

Exercise 6.2.42

Suppose \(X \sim N(-3, 1)\). Between what \(x\) values does 34.14% of the data lie?

Exercise 6.2.43

About what percent of \(x\) values lie between the mean and three standard deviations?

Exercise 6.2.44

About what percent of \(x\) values lie between the mean and one standard deviation?

Exercise 6.2.45

About what percent of \(x\) values lie between the first and second standard deviations from the mean (both sides)?

Exercise 6.2.46

About what percent of \(x\) values lie between the first and third standard deviations(both sides)?

Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts.

Exercise 6.2.47

Define the random variable \(X\) in words. \(X =\) _______________.

The lifetime of a Sunshine CD player measured in years.

Exercise 6.2.48

\(X \sim\) _____(_____,_____)

6.2 Using the Normal Distribution

The shaded area in the following graph indicates the area to the right of x . This area is represented by the probability P ( X > x ). Normal tables provide the probability between the mean, zero for the standard normal distribution, and a specific value such as x 1 x 1 . This is the unshaded part of the graph from the mean to x 1 x 1 .

Because the normal distribution is symmetrical , if x 1 x 1 were the same distance to the left of the mean the area, probability, in the left tail, would be the same as the shaded area in the right tail. Also, bear in mind that because of the symmetry of this distribution, one-half of the probability is to the right of the mean and one-half is to the left of the mean.

Calculations of Probabilities

To find the probability for probability density functions with a continuous random variable we need to calculate the area under the function across the values of X we are interested in. For the normal distribution this seems a difficult task given the complexity of the formula. There is, however, a simply way to get what we want. Here again is the formula for the normal distribution:

Looking at the formula for the normal distribution it is not clear just how we are going to solve for the probability doing it the same way we did it with the previous probability functions. There we put the data into the formula and did the math.

To solve this puzzle we start knowing that the area under a probability density function is the probability.

This shows that the area between X 1 and X 2 is the probability as stated in the formula: P (X 1 ≤ x ≤ X 2 )

The mathematical tool needed to find the area under a curve is integral calculus. The integral of the normal probability density function between the two points x 1 and x 2 is the area under the curve between these two points and is the probability between these two points.

Doing these integrals is no fun and can be very time consuming. But now, remembering that there are an infinite number of normal distributions out there, we can consider the one with a mean of zero and a standard deviation of 1. This particular normal distribution is given the name Standard Normal Distribution. Putting these values into the formula it reduces to a very simple equation. We can now quite easily calculate all probabilities for any value of x, for this particular normal distribution, that has a mean of zero and a standard deviation of 1. These have been produced and are available here in the appendix to the text or everywhere on the web. They are presented in various ways. The table in this text is the most common presentation and is set up with probabilities for one-half the distribution beginning with zero, the mean, and moving outward. The shaded area in the graph at the top of the table in Statistical Tables represents the probability from zero to the specific Z value noted on the horizontal axis, Z.

The only problem is that even with this table, it would be a ridiculous coincidence that our data had a mean of zero and a standard deviation of one. The solution is to convert the distribution we have with its mean and standard deviation to this new Standard Normal Distribution. The Standard Normal has a random variable called Z.

Using the standard normal table, typically called the normal table, to find the probability of one standard deviation, go to the Z column, reading down to 1.0 and then read at column 0. That number, 0.3413 is the probability from zero to 1 standard deviation. At the top of the table is the shaded area in the distribution which is the probability for one standard deviation. The table has solved our integral calculus problem. But only if our data has a mean of zero and a standard deviation of 1.

However, the essential point here is, the probability for one standard deviation on one normal distribution is the same on every normal distribution. If the population data set has a mean of 10 and a standard deviation of 5 then the probability from 10 to 15, one standard deviation, is the same as from zero to 1, one standard deviation on the standard normal distribution. To compute probabilities, areas, for any normal distribution, we need only to convert the particular normal distribution to the standard normal distribution and look up the answer in the tables. As review, here again is the standardizing formula :

where Z is the value on the standard normal distribution, X is the value from a normal distribution one wishes to convert to the standard normal, μ and σ are, respectively, the mean and standard deviation of that population. Note that the equation uses μ and σ which denotes population parameters. This is still dealing with probability so we always are dealing with the population, with known parameter values and a known distribution. It is also important to note that because the normal distribution is symmetrical it does not matter if the z-score is positive or negative when calculating a probability. One standard deviation to the left (negative Z-score) covers the same area as one standard deviation to the right (positive Z-score). This fact is why the Standard Normal tables do not provide areas for the left side of the distribution. Because of this symmetry, the Z-score formula is sometimes written as:

Where the vertical lines in the equation means the absolute value of the number.

What the standardizing formula is really doing is computing the number of standard deviations X is from the mean of its own distribution. The standardizing formula and the concept of counting standard deviations from the mean is the secret of all that we will do in this statistics class. The reason this is true is that all of statistics boils down to variation, and the counting of standard deviations is a measure of variation.

This formula, in many disguises, will reappear over and over throughout this course.

Example 6.3

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

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

Draw a graph.

Then, find P ( x > 65).

P ( x > 65) = 0.3446

P ( x ≥ x 1 ) = P ( Z ≥ Z 1 ) = 0.3446

The probability that any student selected at random scores more than 65 is 0.3446. Here is how we found this answer.

The normal table provides probabilities from zero to the value Z 1 . For this problem the question can be written as: P(X ≥ 65) = P(Z ≥ Z 1 ), which is the area in the tail. To find this area the formula would be 0.5 – P(X ≤ 65). One half of the probability is above the mean value because this is a symmetrical distribution. The graph shows how to find the area in the tail by subtracting that portion from the mean, zero, to the Z 1 value. The final answer is: P(X ≥ 63) = P(Z ≥ 0.4) = 0.3446

z = 65  – 63 5 65  – 63 5 = 0.4

Area to the left of Z 1 to the mean of zero is 0.1554

P ( x > 65) = P ( z > 0.4) = 0.5 – 0.1554 = 0.3446

Z = x - μ σ = 85 - 63 5 = 4.4 Z = x - μ σ = 85 - 63 5 = 4.4 which is larger than the maximum value on the Standard Normal Table. Therefore, the probability that one student scores less than 85 is approximately one or 100%.

A score of 85 is 4.4 standard deviations from the mean of 63 which is beyond the range of the standard normal table. Therefore, the probability that one student scores less than 85 is approximately one (or 100%).

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 6.4

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.

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).

The probability for which you are looking is the area between x = 1.8 and x = 2.75. P (1.8 < x < 2.75) = 0.5886

P (1.8 ≤ x ≤ 2.75) = P ( Z i ≤ Z ≤ Z 2 )

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.

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 th percentile, k , where P ( x < k ) = 0.25.

f ( Z ) = 0.5 - 0.25 = 0.25 f ( Z ) = 0.5 - 0.25 = 0.25 , therefore Z ≈ -0.675 ≈ -0.675 ( or just 0.67 using the table ) Z = x - μ σ = x - 2 0.5 = -0.675 Z = x - μ σ = x - 2 0.5 = -0.675 , therefore x = -0.675 * 0.5 + 2 = 1.66 = -0.675 * 0.5 + 2 = 1.66 hours.

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

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 6.5

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.

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

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

Example 6.6

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.

P ( x ≥ 6) = P ( z ≥ 0.625) = 0.2670

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

f ( Z ) = 0.20 2 = 0.10 f ( Z ) = 0.20 2 = 0.10 , therefore Z ≈ ± 0.25 ≈ ± 0.25 Z = x - μ σ = x - 5.85 0.24 = ± 0.25 → ± 0.25 · 0.24 + 5.85 = ( 5.79 , 5.91 ) Z = x - μ σ = x - 5.85 0.24 = ± 0.25 → ± 0.25 · 0.24 + 5.85 = ( 5.79 , 5.91 )

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• Book title: Introductory Business Statistics
• Publication date: Nov 29, 2017
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• Section URL: https://openstax.org/books/introductory-business-statistics/pages/6-2-using-the-normal-distribution

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