what proportion of the scores are above 700? round the answer to four decimal places.

The Normal Distribution

30 The Standard Normal Distribution

The standard normal distribution is a normal distribution of standardized values chosen z-scores. A z-score is measured in units of the standard divergence.

The hateful for the standard normal distribution is zero, and the standard departure is one. What this does is dramatically simplify the mathematical adding of probabilities. Take a moment and substitute zero and i in the advisable places in the higher up formula and you can meet that the equation collapses into one that tin can exist much more easily solved using integral calculus. The transformation z = produces the distribution Z ~ Due north(0, 1). The value x in the given equation comes from a known normal distribution with known hateful μ and known standard deviation σ. The z-score tells how many standard deviations a item x is abroad from the hateful.

Z-Scores

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

The z-score tells you lot how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. Values of 10 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, so x has a z-score of zero.

Suppose Ten ~ N(v, half-dozen). This says that X is a normally distributed random variable with hateful μ = five and standard deviation σ = 6. Suppose x = 17. And so:

This ways that x = 17 is two standard deviations (2σ) above or to the right of the hateful μ = 5.

Now suppose 10 = one. And then: z = = = –0.67 (rounded to ii decimal places)

This means that x = ane is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5.

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

• About 68% of the ten values lie between –1σ and +1σ of the mean µ (inside ane standard difference of the hateful).
• About 95% of the x values lie betwixt –2σ and +2σ of the mean µ (within 2 standard deviations of the mean).
• Nearly 99.vii% of the x values lie between –3σ and +3σ of the mean µ (within 3 standard deviations of the mean). Observe that most all the ten values lie within 3 standard deviations of the hateful.
• The z-scores for +1σ and –oneσ are +i and –1, respectively.
• The z-scores for +twoσ and –iiσ are +2 and –2, respectively.
• The z-scores for +threeσ and –3σ are +three and –3 respectively.

Suppose ten has a normal distribution with hateful fifty and standard difference six.

• Virtually 68% of the x values lie within one standard divergence of the mean. Therefore, near 68% of the ten values prevarication between –iσ = (–1)(6) = –6 and oneσ = (1)(six) = 6 of the mean 50. The values 50 – half dozen = 44 and 50 + 6 = 56 are within one standard difference from the hateful fifty. The z-scores are –ane and +1 for 44 and 56, respectively.
• About 95% of the x values prevarication within 2 standard deviations of the mean. Therefore, well-nigh 95% of the x values prevarication betwixt –2σ = (–2)(half dozen) = –12 and iiσ = (2)(6) = 12. The values 50 – 12 = 38 and l + 12 = 62 are within two standard deviations from the mean fifty. The z-scores are –2 and +two for 38 and 62, respectively.
• Well-nigh 99.7% of the x values lie within three standard deviations of the mean. Therefore, well-nigh 95% of the x values lie between –threeσ = (–3)(six) = –xviii and iiiσ = (three)(half-dozen) = 18 of the mean l. The values 50 – 18 = 32 and 50 + xviii = 68 are within iii standard deviations from the hateful 50. The z-scores are –3 and +three for 32 and 68, respectively.

Affiliate Review

A z-score is a standardized value. Its distribution is the standard normal, Z ~ N(0, 1). The mean of the z-scores is zip and the standard departure is one. If z is the z-score for a value ten from the normal distribution Northward(µ, σ) then z tells yous how many standard deviations x is to a higher place (greater than) or below (less than) µ.

Formula Review

Z ~ Northward(0, 1)

z = a standardized value (z-score)

mean = 0; standard difference = ane

To find the k ^thursday percentile of X when the z-scores is known:
k = μ + (z)σ

z-score: z = or z =

Z = the random variable for z-scores

Z ~ North(0, 1)

A bottle of h2o contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable 10 in words. X = ____________.

ounces of water in a bottle

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

<!– <solution id="fs-idm63161360″> 61 –>

A visitor manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard divergence of 0.two cm. Define the random variable X in words. 10 = ______________.

<!– <solution id="fs-idp15866016″> diameter of a rubber ball –>

X ~ N(–4, ane)

What is the median?

–4

<!– <solution id="fs-idm120454704″> v –>

What does a z-score measure?

<!– <solution id="fs-idm57247936″> The number of standard deviations a value is from the mean. –>

What does standardizing a normal distribution exercise to the mean?

The mean becomes null.

Is X ~ Northward(0, 1) a standardized normal distribution? Why or why not?

<!– <solution id="fs-idp48272224″> Yes considering the hateful is zero, and the standard deviation is one. –>

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

z = 2

What is the z-score of x = 9, if information technology is i.5 standard deviations to the left of the mean?

<!– <solution id="fs-idp6573504″> z = –1.v –>

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

z = ii.78

What is the z-score of x = 7, if information technology is 0.133 standard deviations to the left of the hateful?

<!– <solution id="fs-idp38051616″> z = –0.133 –>

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

x = 20

Suppose X ~ N(8, 1). What value of 10 has a z-score of –2.25?

<!– <solution id="fs-idp19875360″> x = five.75 –>

Suppose X ~ N(9, five). What value of 10 has a z-score of –0.5?

10 = half-dozen.5

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

<!– <solution id="fs-idp31923120″> x = –0.01 –>

Suppose X ~ Northward(4, 2). What value of 10 is one.5 standard deviations to the left of the mean?

x = i

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

<!– <solution id="fs-idm47755696″> ten = 8 –>

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

x = 1.97

Suppose X ~ North(–1, two). What is the z-score of 10 = two?

<!– <solution id="fs-idm119642576″> z = 1.5 –>

Suppose X ~ N(12, 6). What is the z-score of 10 = 2?

z = –1.67

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

<!– <solution id="fs-idm56843792″> z = 0 –>

Suppose a normal distribution has a hateful of half dozen and a standard deviation of 1.5. What is the z-score of x = v.5?

z ≈ –0.33

In a normal distribution, x = 5 and z = –ane.25. This tells you lot that x = 5 is ____ standard deviations to the ____ (right or left) of the mean.

<!– <solution id="fs-idm62539280″> 1.25, left –>

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

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

<!– <solution id="fs-idm102069296″> half-dozen, right –>

In a normal distribution, 10 = –five and z = –3.xiv. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean.

three.xiv, left

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

<!– <solution id="fs-idm98531584″> one.7, left –>

About what percentage of 10 values from a normal distribution lie within 1 standard deviation (left and right) of the mean of that distribution?

about 68%

Nearly what per centum of the x values from a normal distribution lie within ii standard deviations (left and correct) of the mean of that distribution?

<!– <solution id="fs-idm47813520″> about 95.45% –>

About what percent of 10 values lie between the 2nd and third standard deviations (both sides)?

about iv%

Suppose X ~ N(xv, iii). Between what x values does 68.27% of the information lie? The range of x values is centered at the mean of the distribution (i.e., xv).

<!– <solution id="fs-idm33513040″> between 12 and eighteen –>

Suppose X ~ Northward(–3, ane). Between what ten values does 95.45% of the data lie? The range of 10 values is centered at the mean of the distribution(i.east., –three).

between –5 and –1

Suppose X ~ Due north(–3, one). Between what x values does 34.14% of the data prevarication?

<!– <solution id="fs-idm68579024″> betwixt –4 and –iii or between –3 and –ii –>

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

almost 50%

About what percentage of ten values lie between the mean and one standard deviation?

<!– <solution id="fs-idp3556992″> about 34.xiv% –>

Near what percentage of ten values lie between the first and second standard deviations from the mean (both sides)?

most 27%

About what percent of x values lie betwween the kickoff and third standard deviations(both sides)?

<!– <solution id="fs-idp56159920″> almost 34.46% –>

Use the following data to answer the adjacent ii exercises: The life of Sunshine CD players is normally distributed with mean of iv.1 years and a standard difference of i.3 years. A CD thespian is guaranteed for 3 years. We are interested in the length of time a CD player lasts.

Define the random variable 10 in words. X = _______________.

The lifetime of a Sunshine CD player measured in years.

Ten ~ _____(_____,_____)

<!– <solution id="fs-idp89883200″> Ten ~ N(4.ane, 1.3) –>

Homework

Use the post-obit data to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of five.three days and a standard deviation of 2.1 days.

What is the median recovery time?

1. 2.vii
2. 5.three
3. seven.4
4. ii.one

<!– <solution id="fs-idm68359392″> b –>

What is the z-score for a patient who takes ten days to recover?

1. i.5
2. 0.2
3. 2.ii
4. 7.3

c

The length of fourth dimension to discover it takes to find a parking space at ix A.M. follows a normal distribution with a hateful of 5 minutes and a standard divergence of ii minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true?

1. The data cannot follow the uniform distribution.
2. The data cannot follow the exponential distribution..
3. The information cannot follow the normal distribution.

1. I only
2. Ii merely
3. Three merely
4. I, Ii, and Three

<!– <solution id="fs-idp16013392″> b –>

The heights of the 430 National Basketball Association players were listed on team rosters at the first of the 2005–2006 flavor. The heights of basketball players have an approximate normal distribution with hateful, µ = 79 inches and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences.

1. 77 inches
2. 85 inches
3. If an NBA role player reported his height had a z-score of 3.five, would you believe him? Explain your answer.

1. Use the z-score formula. z = –0.5141. The peak of 77 inches is 0.5141 standard deviations below the mean. An NBA thespian whose pinnacle is 77 inches is shorter than boilerplate.
2. Apply the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose peak is 85 inches is taller than average.
3. Height = 79 + 3.5(3.89) = 92.615 inches, which is taller than 7 feet, 8 inches. There are very few NBA players this tall and then the answer is no, not likely.

The systolic claret pressure (given in millimeters) of males has an approximately normal distribution with hateful µ = 125 and standard difference σ = xiv. Systolic blood pressure for males follows a normal distribution.

1. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters.
2. If a male friend of yours said he thought his systolic claret pressure was two.five standard deviations below the mean, but that he believed his claret pressure was between 100 and 150 millimeters, what would you say to him?

<!– <solution id="eip-774″> Use the z-score formula. 100 – 125 fourteen ≈ –i.eight and 100 – 125 14 ≈ ane.viii I would tell him that 2.5 standard deviations beneath the mean would requite him a blood pressure reading of 90, which is below the range of 100 to 150. –>

Kyle's doctor told him that the z-score for his systolic blood pressure is i.75. Which of the following is the best estimation of this standardized score? The systolic blood pressure (given in millimeters) of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = xiv. If X = a systolic blood pressure score and so X ~ N (125, 14).

1. Which reply(s) is/are right?
   1. Kyle'southward systolic blood pressure is 175.
   2. Kyle'due south systolic blood force per unit area is 1.75 times the boilerplate claret force per unit area of men his age.
   3. Kyle'southward systolic claret force per unit area is 1.75 in a higher place the average systolic blood pressure of men his age.
   4. Kyles's systolic blood pressure level is ane.75 standard deviations above the average systolic claret pressure for men.
2. Calculate Kyle's blood pressure.

1. iv
2. Kyle's blood pressure is equal to 125 + (1.75)(14) = 149.five.

Height and weight are two measurements used to track a child's development. The Earth Health Arrangement measures child development by comparison the weights of children who are the same summit and the same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ North(10.2, 0.8). Calculate the z-scores that correspond to the following weights and translate them.

1. xi kg
2. 7.9 kg
3. 12.two kg

<!– <solution id="eip-182″> 11 – x.two 0.8 = one A child who weighs 11 kg is one standard deviation higher up the mean of 10.ii kg. vii.9 – 10.2 0.8 = –2.875 A child who weighs 7.9 kg is 2.875 standard deviations below the mean of 10.2 kg. 12.ii – 10.two 0.8 = 2.v A child who weighs 12.2 kg is ii.5 standard departure to a higher place the mean of 10.2 kg. –>

In 2005, 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 standard deviation σ = 115.

1. Calculate the z-score for an Sabbatum score of 720. Interpret it using a complete judgement.
2. What math Sat score is ane.5 standard deviations above the mean? What can you say well-nigh this SAT score?
3. For 2012, the Sabbatum math examination had a hateful of 514 and standard deviation 117. The ACT math test is an alternate to the Sabbatum and is approximately usually distributed with hateful 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a 2d person took the ACT math examination and scored 30, who did amend with respect to the test they took?

<!– <para id="fs-idp207361600″>Use the following information to answer the side by side 3 exercises: X ~ U(3, 13) –>

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Source: https://opentextbc.ca/introbusinessstatopenstax/chapter/the-standard-normal-distribution/

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