Creating Box Plots Z Scores and Normal Distributions Review for Ap Stats

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, chosen z-scores, equally opposed to raw scores (the actual data values). A z-score indicates the number of standard deviation a score falls in a higher place or below the mean. Z-scores allow for comparing of scores, occurring in different data sets, with unlike 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 unlike samples that accept different ways and standard deviations. Z-scores can exist looked up in a Z-Table of Standard Normal Distribution, in social club to find the area under the standard normal bend, between a score and the mean, between ii scores, or above or below a score. The standard normal distribution allows u.s.a. 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 space number of data sets, no matter what the mean or standard deviation.

A z-score is calculated as $z=\frac{\mathrm{ten}-\mu }{\sigma }$ . The score itself tin be institute past using algebra and solving for ten. Multiplying both sides of the equation past σ gives: $\left(z\right)\left(\sigma \right)=\mathrm{10}-\mu$ . Adding μ to both sides of the equation gives $\mu +\left(z\right)\left(\sigma \right)=x$ .

Suppose we accept a data fix with a mean of five and standard departure of ii. Nosotros want to determine the number of standard deviations the score of eleven falls above the mean. We can detect this reply (or z-score) by writing

$$z=\frac{11-\mathrm{v}}{2}=3$$

or

$$\mathrm{v}+\left(z\right)\left(2\right)=11\text{,}$$

we can solve for z.

$$\begin{array}{l}2z=6\\ z=3\end{array}$$

Nosotros take determined that the score of 11 falls 3 standard deviations to a higher place the mean of 5.

With a standard normal distribution, we signal the distribution by writing Z ~ N(0, 1) which shows the normal distribution has a hateful of 0 and standard deviation of i. This notation only indicates that a standard normal distribution is existence used.

Z-Scores

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

$$z=\frac{x–\mu }{\sigma }\text{.}$$

The z-score tells you how many standard deviations the value x is above, to the right of, or beneath, to the left of, the hateful, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the hateful have negative z-scores. If 10 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 to a higher place for a normal distribution, volition be given.

Example vi.1

Suppose Ten ~ N(v, vi). This equation says that X is a normally distributed random variable with mean μ = v and standard deviation σ = six. Suppose x = 17. So,

$$z=\frac{x–\mu }{\sigma }=\frac{17–5}{6}=2\text{.}$$

This means that ten = 17 is 2 standard deviations (2σ) above, or to the right, of the hateful μ = five.

Notice that v + (ii)(6) = 17. The blueprint is μ + zσ = x.

Now suppose x = 1. So, z = $\frac{x–\mu }{\sigma }$ = $\frac{\mathrm{one}–5}{6}$ = –0.67, rounded to ii decimal places.

This means that x = i is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = v. 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 to a higher place 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, 10 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 hateful, in either direction.

Try It vi.one

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

Example 6.2

Some doctors believe that a person tin 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 Ten = the amount of weight lost, in pounds, by a person in a month. Apply a standard deviation of two pounds. Ten ~ N(v, 2). Fill in the blanks.

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

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

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

Try Information technology 6.2

Make full in the blanks.

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

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

• Almost 68 percentage of the x values lie between –1σ and +oneσ of the mean µ (within one standard divergence of the mean).
• Well-nigh 95 percent of the x values lie between –2σ and +twoσ of the hateful µ (within two standard deviations of the mean).
• About 99.7 per centum of the ten values lie between –threeσ and +threeσ of the hateful µ (within three standard deviations of the mean). Notice that almost all the ten values lie inside three standard deviations of the hateful.
• The z-scores for +oneσ and –oneσ are +1 and –1, respectively.
• The z-scores for +2σ and –twoσ are +ii and –2, respectively.
• The z-scores for +threeσ and –3σ are +3 and –3, respectively.

So, in other words, this is that about 68 percent of the values lie betwixt z-scores of –1 and one, nearly 95% of the values prevarication between z-scores of –2 and ii, and about 99.7 percent of the values lie between z-scores of -three and 3. These facts can be checked, by looking up the hateful 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.vii rule.

Figure 6.3

Example 6.3

The mean elevation of 15-to 18-year-old males from Republic of 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 fifteen-to 18-year-old male from Republic of chile in 2009–2010. Then X ~ Due north(170, vi.28).

a. Suppose a fifteen-to xviii-year-old male person from Chile was 168 cm tall in 2009–2010. The z-score when 10 = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the hateful _____ (What is the mean?).

b. Suppose that the top of a 15-to 18-yr-erstwhile male from Republic of chile in 2009–2010 has a z-score of z = 1.27. What is the male'south pinnacle? The z-score (z = 1.27) tells you that the male's summit is ________ standard deviations to the __________ (correct or left) of the hateful.

Endeavour Information technology 6.3

Use the information in Example 6.three to answer the following questions:

1. Suppose a 15-to 18-twelvemonth-onetime male person from Chile was 176 cm tall from 2009–2010. The z-score when x = 176 cm is z = _______. This z-score tells you that 10 = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the tiptop of a fifteen-to 18-year-old male from Republic of 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'due south height is ________ standard deviations to the __________ (right or left) of the hateful.

Example 6.4

From 1984 to 1985, the hateful height of 15-to 18-year-erstwhile males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Permit Y = the height of fifteen-to eighteen-year-former males from 1984–1985, and y = the tiptop of ane male from this group. Then Y ~ Due north(172.36, vi.34).

The hateful height of 15-to 18-yr-quondam males from Chile in 2009–2010 was 170 cm with a standard departure of vi.28 cm. Male person 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 ten = the height of ane male person from this group. So Ten ~ Due north(170, 6.28).

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

Effort It half dozen.iv

In 2012, 1,664,479 students took the Saturday exam. The distribution of scores in the verbal section of the Sabbatum had a mean µ = 496 and a standard deviation σ = 114. Permit X = a Sabbatum exam verbal section score in 2012. And so, 10 ~ N(496, 114).

Find the z-scores for 10 ₁ = 325 and x ₂ = 366.21. Interpret each z-score. What tin can you say about ten _one = 325 and x ₂ = 366.21, as they compare to their corresponding means and standard deviations?

Example half-dozen.5

Suppose x has a normal distribution with hateful 50 and standard departure 6.

• About 68 percent of the 10 values lie within ane standard deviation of the mean. Therefore, well-nigh 68 percentage of the x values lie between –1σ = (–1)(6) = –half-dozen and 1σ = (1)(6) = half dozen of the mean 50. The values 50 – 6 = 44 and l + vi = 56 are inside one standard deviation from the hateful 50. The z-scores are –ane and +1 for 44 and 56, respectively.
• About 95 pct of the x values lie within 2 standard deviations of the hateful. Therefore, about 95 percent of the x values lie between –2σ = (–ii)(6) = –12 and 2σ = (two)(six) = 12. The values l – 12 = 38 and fifty + 12 = 62 are inside two standard deviations from the mean 50. The z-scores are –2 and +ii for 38 and 62, respectively.
• About 99.7 percent of the x values lie inside three standard deviations of the hateful. Therefore, about 95 percent of the x values lie betwixt –3σ = (–3)(6) = –eighteen and 3σ = (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + eighteen = 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.five

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

Example six.6

From 1984–1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard divergence was 6.34 cm. Let Y = the pinnacle of 15-to 18-year-old males in 1984–1985. So Y ~ North(172.36, half-dozen.34).

1. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. Almost 95 pct of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
3. About 99.seven pct of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.

Try Information technology six.6

The scores on a higher archway exam accept an approximate normal distribution with hateful, µ = 52 points and a standard departure, σ = xi points.

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

mendozathadell.blogspot.com

Source: https://openstax.org/books/statistics/pages/6-1-the-standard-normal-distribution

Share this post