6.1 The Standard Normal Distribution
Converting a measurement into a z-score that counts standard deviations from the mean, comparing values from different bell curves, and the empirical rule.
Introductory Statistics 2e, 6.1 The Standard Normal Distribution
31 exercises
6.1 Exercise 2
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing their 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 …
6.1 Exercise 3
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing their 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.
b. Suppose a person gained three pounds (a negative weight loss).…
6.1 Exercise 4
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing their 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.
c. Suppose the random variables X and Y have the following normal…
6.1 Exercise 6
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 = _______.…
6.1 Exercise 8
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 …
6.1 Exercise 11
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).
- 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…
6.1 Exercise 13
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 = ____________.
6.1 Exercise 15
X ~ N(1, 2)
σ = _______
6.1 Exercise 17
X ~ N(–4, 1)
What is the median?
6.1 Exercise 19
X ~ N(–2, 1)
μ = _______
6.1 Exercise 21
What does standardizing a normal distribution do to the mean?
6.1 Exercise 23
What is the z-score of x = 12, if it is two standard deviations to the right of the mean?
6.1 Exercise 25
What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean?
6.1 Exercise 27
Suppose X ~ N(2, 6). What value of x has a z-score of three?
6.1 Exercise 29
Suppose X ~ N(9, 5). What value of x has a z-score of –0.5?
6.1 Exercise 31
Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean?
6.1 Exercise 33
Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean?
6.1 Exercise 35
Suppose X ~ N(12, 6). What is the z-score of x = 2?
6.1 Exercise 37
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?
6.1 Exercise 39
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.
6.1 Exercise 41
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.
6.1 Exercise 43
About what percent of x values from a normal distribution lie within one standard deviation (left and right) of the mean of that distribution?
6.1 Exercise 45
About what percent of x values lie between the second and third standard deviations (both sides)?
6.1 Exercise 47
Suppose X ~ 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).
6.1 Exercise 49
About what percent of x values lie between the mean and three standard deviations?
6.1 Exercise 51
About what percent of x values lie between the first and second standard deviations from the mean (both sides)?
6.1 Exercise 53
Use the following information to answer the next two exercises: The life of wearable fitness devices is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A wearable fitness device is guaranteed for three years. We are interested in the length of time a wearable fitness device lasts.
Define the random variable X in words. X = _______________.
6.1 Exercise 56
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the z-score for a patient who takes ten days to recover?
- 1.5
- 0.2
- 2.2
- 7.3
6.1 Exercise 58
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
The heights of the 430 National Basketball Association players were listed on team rosters at the start of a recent season. The heights of basketball players have an approximate normal di…
6.1 Exercise 60
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure…
6.1 Exercise 62
Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days.
During a certain year, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and st…
Adapted from Introductory Statistics 2e © 2023 Rice University. Licensed under CC BY 4.0. Source: Introductory Statistics 2e. Imported at revision 50ec917.