6.1 The Standard Normal Distribution

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

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 …

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.

b. Suppose a person gained three pounds, a negative weight loss.…

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.

c. Suppose the random variables X and Y have the following norma…

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 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-…

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).

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 hei…

Use the information in the example 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 200…

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 know…

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 s…

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?

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 betwee…

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…

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 = ____________.

X ~ N(1, 2)

σ = _______

X ~ N(–4, 1)

What is the median?

X ~ N(–2, 1)

μ = _______

What does standardizing a normal distribution do to the mean?

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

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

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

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

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

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

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

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?

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.

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.

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

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

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

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

About what percent of x values lie between the first and second standard deviations from the mean, 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.

Define the random variable X in words. X = _______________.

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 10 days to recover?

1. 1.5
2. 0.2
3. 2.2
4. 7.3

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 basketball players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with…

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,…

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.

In 2005, 1,475,623 students heading to college took the SAT exam. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard de…