Exercise 1 Find the Probability That One Review Will Take Her 35 to 425 Hours Anwers

Learning Outcomes

• Recognize the standard normal probability distribution and apply information technology appropriately

The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard difference. For example, if the mean of a normal distribution is 5 and the standard deviation is two, the value 11 is iii standard deviations to a higher place (or to the right of) the mean. The calculation is every bit follows:

x = μ + (z)(σ) = 5 + (three)(two) = eleven

The z-score is three.

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

The following two videos requite a description of what it means to have a data set that is "normally" distributed.

Z-Scores

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

[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]

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

Example

Suppose X ~ North(five, six). This says that x is a unremarkably distributed random variable with mean μ = 5 and standard deviation σ = six. Suppose ten = 17. Then:

[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]= [latex]\displaystyle{z}=\frac{{17-5}}{{half-dozen}}={two}[/latex]

This means that ten = 17 is ii standard deviations (2σ) to a higher place or to the right of the mean μ = 5. The standard deviation is σ = 6.

Find that: 5 + (2)(6) = 17 (The pattern is μ + zσ = ten)

Now suppose x = i. Then: [latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex] = [latex]\displaystyle {z}=\frac{{one-5}}{{6}} = -{0.67}[/latex]

(rounded to 2 decimal places)

This ways that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. Observe that: 5 + (–0.67)(half-dozen) is approximately equal to 1 (This has the pattern μ + (–0.67)σ = 1)

Summarizing, when z is positive, x is in a higher place or to the right of μ and when zis 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 ten is less than μ.

try information technology

What is the z-score of x, when x = 1 and10 ~ North(12,3)?

[latex]\displaystyle {z}=\frac{{i-12}}{{3}} = -{iii.67} [/latex]

Example

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. Allow 10 = the amount of weight lost(in pounds) by a person in a month. Use a standard difference of two pounds. X ~ N(five, two). Fill in the blanks.

1. Suppose a person lost x pounds in a calendar month. The z-score when ten = 10 pounds is z = 2.v (verify). This z-score tells you thatx = ten is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the hateful?).
2. Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z-score tells you that x = –iii is ________ standard deviations to the __________ (right or left) of the mean.

Solution:

1. This z-score tells you that x = x is ii.5 standard deviations to the correct of the hateful five.
2. z= –4. This z-score tells you that ten = –3 is four standard deviations to the left of the hateful.

Suppose the random variables X and Y have the following normal distributions: Ten ~ Due north(5, 6) and Y ~ N(2, ane). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z? [latex]\displaystyle {z}=\frac{{y - \mu}}{{\sigma}} = \frac{{4-2}}{{1}}[/latex].

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 usa to compare data that are scaled differently. To understand the concept, suppose 10 ~ N(v, vi) represents weight gains for ane 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 2d group of people. A negative weight gain would be a weight loss. Since x = 17 and y= 4 are each ii standard deviations to the correct of their means, they represent the aforementioned, standardized weight gain relative to their means

Endeavour It

Fill in the blanks.

Jerome averages 16 points a game with a standard deviation of four points. Ten ~N(16,4). Suppose Jerome scores ten points in a game. The z–score when x = 10 is –1.v. This score tells you that x = x is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).

1.v, left, sixteen

The Empirical Rule

If X is a random variable and has a normal distribution with hateful µ and standard difference σ, then the Empirical Rule says the post-obit:

• About 68% of the x values lie between –oneσ and +1σ of the hateful µ (within ane standard deviation of the mean).
• Near 95% of the 10 values prevarication betwixt –twoσ and +2σ of the hateful µ (within two standard deviations of the mean).
• Nigh 99.seven% of the x values lie between –3σ and +3σ of the hateful µ(within three standard deviations of the mean). Notice that well-nigh all the10 values lie inside three standard deviations of the mean.
• The z-scores for +1σ and –1σ are +i and –i, respectively.
• The z-scores for +twoσ and –2σ are +2 and –2, respectively.
• The z-scores for +threeσ and –3σ are +iii and –3 respectively.

The empirical dominion is also known as the 68-95-99.vii rule.

Example

The hateful pinnacle of fifteen to 18-year-erstwhile 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 Ten = the height of a fifteen to xviii-twelvemonth-quondam male person from Chile in 2009 to 2010. Then 10 ~ North(170, half dozen.28).

a. Suppose a fifteen to 18-year-old male from Republic of chile was 168 cm alpine from 2009 to 2010. The z-score when x = 168 cm is z = _______. This z-score tells you that ten = 168 is ________ standard deviations to the ________ (right or left) of the hateful _____ (What is the mean?).

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

Solution:

a. –0.32, 0.32, left, 170

b. 177.98, ane.27, right

try it

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

1. Suppose a 15 to eighteen-year-one-time male person from Chile was 176 cm alpine from 2009 to 2010. The z-score when x = 176 cm is z = _______. This z-score tells y'all that 10 = 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 eighteen-year-one-time male person from Republic of chile from 2009 to 2010 has a z-score of z = –2. What is the male'southward height? The z-score (z = –2) tells you lot that the male person's height is ________ standard deviations to the __________ (right or left) of the mean.

Solve the equation [latex]\displaystyle{z}=\frac{{10 - \mu}}{{\sigma}}[/latex] for x. x = μ + (z)(σ) for x. ten = μ + (z)(σ)

1. z<=[latex]\displaystyle\frac{{176-170}}{{0.96}}[/latex], This z-score tells you that x = 176 cm is 0.96 standard deviations to the right of the hateful 170 cm.
2. X = 157.44 cm, The z-score(z = –2) tells you that the male'southward height is two standard deviations to the left of the mean.

Example

From 1984 to 1985, the hateful height of 15 to eighteen-year-sometime males from Chile was 172.36 cm, and the standard deviation was half dozen.34 cm. Let Y = the pinnacle of fifteen to 18-yr-old males from 1984 to 1985. And so Y ~ N(172.36, six.34).

The hateful height of xv to 18-year-old males from Republic of chile from 2009 to 2010 was 170 cm with a standard deviation of vi.28 cm. Male heights are known to follow a normal distribution. Allow X = the height of a xv to 18-twelvemonth-onetime male person from Chile in 2009 to 2010. Then X ~ North(170, vi.28).

Notice the z-scores for x = 160.58 cm and y = 162.85 cm. Translate each z-score. What tin can you say virtually x = 160.58 cm and y = 162.85 cm?

Solution:

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.

try it

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

Find the z-scores for xone = 325 and 102 = 366.21. Translate each z-score. What can yous say about xone = 325 and x2 = 366.21?

The z-score for xane = 325 is zi = –1.xiv.

The z-score for 102 = 366.21 is z2 = –one.14.

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

Case

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

• About 68% of the ten values lie betwixt –aneσ = (–one)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + vi = 56 are inside one standard deviation of the hateful 50. The z-scores are –ane and +1 for 44 and 56, respectively.
• Nearly 95% of the ten values lie between –iiσ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are inside two standard deviations of the hateful 50. The z-scores are –2 and +2 for 38 and 62,respectively.
• Nigh 99.7% of the ten values prevarication between –3σ = (–3)(6) = –18 and 3σ= (3)(6) = 18 of the mean 50. The values fifty – eighteen = 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

try it

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

Solution:

Between 20 and 30.

Case

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

1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively.
3. About 99.7% of the y values prevarication between what two .7% of the values lie betwixt 153.34 and 191.38. The z-scores are –3 and iii.
   1. About 68% of the y values lie betwixt what 2 values? These values are ________________. Thez-scores are ________________, respectively.

endeavor information technology

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

.seven% of the values prevarication betwixt 153.34 and 191.38. The z-scores are –3 and 3.

1. About 68% of the y values lie between what two values? These values are ________________. Thez-scores are ________________, respectively.
2. Well-nigh 95% of the y values lie between what ii values? These values are ________________. Thez-scores are ________________, respectively.
3. Well-nigh 99.seven% of the y values lie betwixt what two values? These values are ________________. Thez-scores are ________________, respectively.

Solution:

1. About 68% of the values lie betwixt the values 41 and 63. Thez-scores are –1 and 1, respectively.
2. Near 95% of the values prevarication between the values 30 and 74. Thez-scores are –two and 2, respectively.
3. About 99.7% of the values lie between the values 19 and 85. Thez-scores are –3 and iii, respectively.

References

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

"The Employ 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://disharmonize.lshtm.ac.uk/page_125.htm (accessed May 14, 2013).

"2012 College-Spring Seniors Total Grouping Profile Report." CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/inquiry/TotalGroup-2012.pdf (accessed May 14, 2013).

"Digest of Didactics Statistics: Human activity score average and standard deviations by sex and race/ethnicity and percentage of ACT examination takers, by selected composite score ranges and planned fields of report: 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 xiv, 2013).

Data from the San Jose Mercury News.

Data from The Earth Almanac and Book of Facts.

"Listing 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. Bachelor online at www.nba.com (accessed May 14, 2013).

Concept Review

A z-score is a standardized value. Its distribution is the standard normal, Z ~N(0, i). The hateful of the z-scores is zip and the standard deviation is 1. If zis the z-score for a value x from the normal distribution N(µ, σ) then z tells you how many standard deviations x is above (greater than) or beneath (less than) µ.

Formula Review

Z ~ N(0, 1)

z = a standardized value (z-score)

mean = 0; standard divergence = 1

To find the Mth percentile of Ten when the z-scores is known:

g = μ + (z)σ

z-score:[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]

Z = the random variable for z-scores

Z ~ N(0, 1)

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Source: https://courses.lumenlearning.com/introstats1/chapter/the-standard-normal-distribution/

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