If the Test on an Art History Exam Were Normally Distrubuted With a Mean of 76

Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

NOTE

To calculate the probability, use the probability tables provided in Figure G1 without the use of technology. The tables include instructions for how to use them.

The probability is represented by the expanse under the normal curve. To observe the probability, calculate the z-score and look up the z-score in the z-table under the z-cavalcade. About z-tables show the area nether the normal curve to the left of z. Others evidence the mean to z area. The method used will be indicated on the tabular array.

We will discuss the z-tabular array that represents the area nether the normal bend to the left of z. Once you have located the z-score, locate the corresponding area. This volition be the surface area nether the normal bend, to the left of the z-score. This expanse can be used to find the area to the right of the z-score, or past subtracting from 1 or the total surface area under the normal curve. These areas can likewise be used to decide the area between two z-scores.

Example half dozen.7

If the surface area to the left is 0.0228, and so the surface area to the right is 1 – 0.0228 = 0.9772.

Endeavour Information technology 6.vii

If the area to the left of x is 0.012, then what is the expanse to the correct?

Example 6.8

The last examination scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

a. Detect the probability that a randomly selected student scored more than 65 on the exam.

Solution vi.eight

a. Permit X = a score on the final exam. 10 ~ N(63, 5), where μ = 63 and σ = 5.

Draw a graph.

Calculate the z-score:

$$z=\frac{x-\mu }{\sigma }=\frac{\mathrm{}65-63}{5}=\frac{2}{5}=.40$$

The z-tabular array shows that the area to the left of z is 0.6554. Subtracting this expanse from 1 gives 0.3446.

Then, discover P(x > 65).

$$P\text{(}x\text{ > 65) = 0}\text{.3446}$$

Figure 6.5

The probability that whatever student selected at random scores more than 65 is 0.3446.

Using the TI-83, 83+, 84, 84+ Calculator

Go into 2nd DISTR.

Later on pressing 2nd DISTR, press 2:normalcdf.

The syntax for the instructions is every bit follows:

normalcdf(lower value, upper value, mean, standard departure) For this trouble: normalcdf(65,1E99,63,5) = 0.3446. You lot get 1E99 (= ten⁹⁹) by pressing 1, the EE fundamental—a second cardinal—then 99. Or, you can enter 10^99 instead. The number 10⁹⁹ is way out in the correct tail of the normal bend. We are calculating the surface area betwixt 65 and 10⁹⁹. In some instances, the lower number of the area might be –1E99 (= –10⁹⁹). The number –10⁹⁹ is way out in the left tail of the normal curve. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values nether the curve to fall beneath it. This is an capricious value and one that works well, for our purpose.

Historical Note

The TI probability program calculates a z-score and then the probability from the z-score. Before engineering, the z-score was looked up in a standard normal probability table, also known as a Z-tabular array—the math involved to find probability is cumbersome. In this example, a standard normal table with area to the left of the z-score was used. Yous calculate the z-score and look up the area to the left. The probability is the area to the right.

Using the TI-83, 83+, 84, 84+ Computer

Calculate the z-score

*Press second Distr

*Press 3:invNorm(

*Enter the area to the left of

followed past )

*Printing ENTER.

For this Case, the steps are

2d Distr

3:invNorm(.6554) ENTER

The answer is 0.3999, which rounds to 0.four.

b. Discover the probability that a randomly selected student scored less than 85.

Solution 6.8

b. Draw a graph.

Then notice P(x

Using a reckoner or calculator, observe P(x

normalcdf(0,85,63,v) = ane (rounds to one)

The probability that ane pupil scores less than 85 is approximately one, or 100 percent.

c. Find the 90^th percentile—that is, find the score chiliad that has ninety percent of the scores below k and ten percent of the scores above 1000.

Solution 6.viii

c. Find the 90^th percentile. For each problem or part of a problem, draw a new graph. Depict the ten-axis. Shade the surface area that corresponds to the xc^thursday percentile. This time, we are looking for a score that corresponds to a given surface area under the curve.

Let k = the ninety^th percentile. The variable k is located on the x-axis. P(x 1000) is the surface area to the left of yard. The 90^th percentile grand separates the exam scores into those that are the same or lower than thousand and those that are the same or college. Ninety per centum of the test scores are the aforementioned or lower than m, and 10 percent are the same or higher. The variable k is frequently chosen a critical value.

Nosotros know the hateful, standard deviation, and area under the normal curve. We demand to find the z-score that corresponds to the area of 0.ix then substitute information technology with the hateful and standard deviation into our z-score formula. The z-tabular array shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.ix. Thus, nosotros tin write the post-obit:

$$1.28=\frac{x-63\mathrm{}}{\mathrm{v}}$$

Multiplying each side of the equation by 5 gives

$$6.4=x-63$$

Calculation 63 to both sides of the equation gives

$$69.4=\mathrm{x.}$$

Thus, our score, k, is 69.iv.

$$k\text{ = 69}\text{.4}$$

Effigy 6.6

The 90^th percentile is 69.four. This means that ninety percentage of the examination scores fall at or below 69.4 and 10 percent fall at or above. To become this answer on the calculator, follow this next stride.

Using the TI-83, 83+, 84, 84+ Figurer

invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)

For this problem, invNorm(0.ninety,63,v) = 69.4

d. Find the lxx^th percentile—that is, find the score k such that 70 percent of scores are below k and xxx percent of the scores are above k.

Solution 6.viii

d. Discover the 70^th percentile.

Describe a new graph and label it appropriately. k = 65.six

The 70^th percentile is 65.6. This ways that seventy pct of the test scores fall at or beneath 65.5 and 30 percent fall at or above.

invNorm(0.70,63,v) = 65.six

Try It six.8

The golf scores for a school team were normally distributed with a hateful of 68 and a standard deviation of three.

Find the probability that a randomly selected golfer scored less than 65.

Example 6.9

A personal figurer is used for role piece of work at dwelling, research, advice, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per twenty-four hours. Presume the times for amusement are normally distributed and the standard deviation for the times is one-half an hr.

a. Find the probability that a household personal estimator is used for entertainment between i.8 and 2.75 hours per day.

Solution six.9

a. Permit X = the amount of time, in hours, a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.

Observe P(1.viii ten

First, calculate the z-scores for each 10-value.

$\begin{array}{l}z=\frac{\mathrm{ane.8}-2}{\mathrm{0.five}}=\frac{-0.2}{0.5}=-\mathrm{0.forty}\hfill \\ z=\frac{2.75-2}{0.5}=\frac{0.75}{0.5}=\mathrm{i.5}\hfill \end{array}$

Now, use the Z-table to locate the area under the normal bend to the left of each of these z-scores.

The area to the left of the z-score of −0.xl is 0.3446. The area to the left of the z-score of ane.5 is 0.9332. The area between these scores will be the difference in the 2 areas, or $0.9332-0.3446$, which equals 0.5886.

Figure vi.vii

normalcdf(i.8,2.75,2,0.5) = 0.5886

The probability that a household personal calculator is used between ane.8 and 2.75 hours per day for entertainment is 0.5886.

b. Notice the maximum number of hours per day that the lesser quartile of households uses a personal figurer for entertainment.

Solution 6.9

b. To find the maximum number of hours per day that the lesser quartile of households uses a personal reckoner for amusement, find the 25^th percentile, one thousand, where P(ten k) = 0.25.

Figure 6.8

invNorm(0.25,2,0.5) = ane.66

We utilise invNorm considering nosotros are looking for the k-value.

The maximum number of hours per twenty-four hours that the bottom quartile of households uses a personal computer for entertainment is i.66 hours.

Try It 6.9

The golf scores for a school team were unremarkably distributed with a mean of 68 and a standard deviation of three. Discover the probability that a golfer scored between 66 and 70.

Case 6.10

In the United States smartphone users between the ages of 13 and 55+ approximately follow a normal distribution with guess hateful and standard deviation of 36.9 years and 13.ix years, respectively.

a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.

Solution 6.10

a. normalcdf(23,64.vii,36.9,13.9) = 0.8186

The z-scores are calculated as

$\begin{array}{l}z=\frac{23-\mathrm{36.ix}}{\mathrm{thirteen.nine}}=\frac{-\mathrm{xiii.9}}{13.9}=-1\hfill \\ z=\frac{\mathrm{64.vii}-36.9}{\mathrm{13.nine}}=\frac{27.8}{\mathrm{xiii.9}}=2\hfill \end{array}$

The Z-table shows the expanse to the left of a z-score with an absolute value of i to exist 0.1587. It shows the expanse to the left of a z-score of 2 to be 0.9772. The difference in the 2 areas is 0.8185.

This is slightly dissimilar than the surface area given by the calculator, due to rounding.

b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at almost l.viii years former.

Solution 6.x

b. normalcdf(–ten⁹⁹,50.viii,36.9,thirteen.ix) = 0.8413

c. Find the lxxx^th percentile of this distribution, and interpret it in a consummate sentence.

Solution 6.ten

c.

• invNorm(0.eighty,36.9,13.9) = 48.half-dozen
• The fourscore^th percentile is 48.6 years.
• 80 percent of the smartphone users in the historic period range 13–55+ are 48.vi years quondam or less.

Try It half-dozen.10

Utilize the information in Example 6.10 to answer the following questions:

1. Find the 30^th percentile, and translate information technology in a complete judgement.
2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old?

Instance 6.11

In the United States smartphone users between the ages of 13 and 55+ approximately follow a normal distribution with approximate mean and standard difference of 36.9 years and xiii.9 years, respectively. Using this information, respond the following questions—circular answers to one decimal place:

a. Calculate the interquartile range (IQR).

Solution 6.11

a.

• IQR = Q ₃ – Q _one
• Summate Q _three = 75^th percentile and Q ₁ = 25^th percentile.
• Recall that we tin can use invNorm to find the k-value. We can utilise this to find the quartile values.
• invNorm(0.75,36.9,13.9) = Q _three = 46.2754
• invNorm(0.25,36.9,xiii.9) = Q ₁ = 27.5246
• IQR = Q ₃ – Q _i = 18.viii

b. 40 percent of the ages that range from 13 to 55+ are at to the lowest degree what age?

Solution 6.11

b.

• Observe grand where P(x ≥ k) = 0.40. At least translates to greater than or equal to.
• 0.40 = the expanse to the right
• The expanse to the left = i – 0.40 = 0.sixty.
• The area to the left of k = 0.60
• invNorm(0.60,36.ix,13.9) = 40.4215
• thou = 40.iv.
• Forty per centum of the ages that range from 13 to 55+ are at least twoscore.4 years.

Try It half-dozen.xi

Two grand students took an test. The scores on the examination have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = xv points.

1. Calculate the kickoff- and 3rd-quartile scores for this exam.
2. The heart fifty percent of the exam scores are between what 2 values?

Case six.12

A citrus farmer who grows standard mandarin oranges finds that the diameters of mandarin oranges harvested on his subcontract follow a normal distribution with a mean bore of 5.85 cm and a standard deviation of 0.24 cm.

a. Find the probability that a randomly selected standard mandarin orangish from this farm has a diameter larger than half-dozen.0 cm. Sketch the graph.

Solution 6.12

a. normalcdf(half-dozen,10^99,5.85,0.24) = 0.2660

Figure 6.9

b. The middle 20 percentage of standard mandarin oranges from this farm have diameters between ______ and ______.

Solution 6.12

b.

• 1 – 0.20 = 0.fourscore. Outside of the center xx percent will be 80 per centum of the values.
• The tails of the graph of the normal distribution each have an area of 0.twoscore.
• Find k ₁, the forty^th percentile, and 1000 ₂, the 60^thursday percentile (0.40 + 0.20 = 0.sixty). This leaves the centre 20 percent, in the heart of the distribution.
• k ₁ = invNorm(0.forty,5.85,0.24) = five.79 cm
• k _two = invNorm(0.60,5.85,0.24) = 5.91 cm

So, the eye twenty pct of mandarin oranges have diameters betwixt 5.79 cm and 5.91 cm.

c. Detect the ninety^th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.

Solution vi.12

c. 6.sixteen: Ninety percent of the diameter of the standard mandarin oranges is at nearly 6.16 cm.

Attempt It 6.12

Using the information from Example half dozen.12, answer the post-obit:

1. The eye 45 pct of mandarin oranges from this subcontract are between ______ and ______.
2. Find the sixteen^th percentile, and interpret it in a complete sentence.

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Source: https://www.texasgateway.org/resource/62-using-normal-distribution

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