Use a proportion to find the part. 15% of 15 is what number? brainly

Calculations of Probabilities

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

Notation

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

The probability is represented by the surface area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-tabular array under the z-column. Most z-tables prove the area under the normal curve to the left of z. Others show the hateful to z area. The method used will be indicated on the tabular array.

We will talk over the z-table that represents the area under the normal bend to the left of z. Once you lot have located the z-score, locate the respective area. This volition be the surface area under the normal curve, to the left of the z-score. This surface area can be used to find the area to the correct of the z-score, or by subtracting from 1 or the total expanse under the normal curve. These areas tin can also be used to determine the area between 2 z-scores.

Example 6.seven

If the surface area to the left is 0.0228, so the area to the correct is one – 0.0228 = 0.9772.

Try It half dozen.7

If the area to the left of x is 0.012, and then what is the area to the right?

Example 6.8

The terminal exam scores in a statistics class were normally distributed with a hateful of 63 and a standard deviation of five.

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

Solution 6.8

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

Draw a graph.

Calculate the z-score:

$$z=\frac{\mathrm{ten}-\mu }{\sigma }=\frac{\mathrm{}65-63}{\mathrm{v}}=\frac{2}{5}=\mathrm{.twoscore}$$

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

And so, discover P(10 > 65).

$$P\text{(}\mathrm{10}\text{ > 65) = 0}\text{.3446}$$

Figure 6.5

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

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

Go into 2nd DISTR.

Afterwards pressing 2nd DISTR, press 2:normalcdf.

The syntax for the instructions is as follows:

normalcdf(lower value, upper value, mean, standard deviation) For this trouble: normalcdf(65,1E99,63,v) = 0.3446. Y'all get 1E99 (= 10⁹⁹) past pressing i, the EE central—a 2nd cardinal—and then 99. Or, y'all tin can enter 10^99 instead. The number 10⁹⁹ is style out in the correct tail of the normal curve. We are calculating the area between 65 and x⁹⁹. In some instances, the lower number of the area might be –1E99 (= –ten⁹⁹). The number –10⁹⁹ is way out in the left tail of the normal curve. Nosotros chose the exponent of 99 because this produces such a big number that we can reasonably look all of the values under the bend to fall below it. This is an arbitrary value and 1 that works well, for our purpose.

Historical Note

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

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

Summate the z-score

*Press 2nd Distr

*Press iii:invNorm(

*Enter the expanse to the left of

followed by )

*Printing ENTER.

For this Instance, the steps are

2nd Distr

3:invNorm(.6554) ENTER

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

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

Solution 6.8

b. Depict a graph.

And so detect P(ten

Using a reckoner or calculator, notice P(x

normalcdf(0,85,63,5) = 1 (rounds to 1)

The probability that one student scores less than 85 is approximately one, or 100 per centum.

c. Find the 90^th percentile—that is, notice the score g that has xc percent of the scores below thousand and x pct of the scores above k.

Solution 6.viii

c. Discover the 90^thursday percentile. For each trouble or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90^thursday percentile. This time, we are looking for a score that corresponds to a given expanse nether the curve.

Let k = the xc^thursday percentile. The variable k is located on the 10-axis. P(x k) is the expanse to the left of chiliad. The 90^th percentile grand separates the exam scores into those that are the aforementioned or lower than k and those that are the same or higher. Ninety percentage of the test scores are the same or lower than k, and ten percent are the aforementioned or higher. The variable k is often called a critical value.

We know the hateful, standard departure, and surface area under the normal curve. We demand to discover the z-score that corresponds to the area of 0.ix and so substitute it with the mean and standard deviation into our z-score formula. The z-tabular array shows a z-score of approximately 1.28, for an surface area under the normal bend to the left of z (larger portion) of approximately 0.ix. Thus, nosotros tin can write the following:

$$1.28=\frac{x-63\mathrm{}}{5}$$

Multiplying each side of the equation by 5 gives

$$\mathrm{six.4}=x-63$$

Adding 63 to both sides of the equation gives

$$69.4=\mathrm{10.}$$

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

$$\mathrm{1000}\text{ = 69}\text{.4}$$

Figure 6.6

The 90^th percentile is 69.4. This means that ninety percent of the test scores fall at or beneath 69.iv and 10 pct autumn at or to a higher place. To become this answer on the calculator, follow this next step.

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

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

For this problem, invNorm(0.ninety,63,5) = 69.4

d. Notice the 70^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 vi.eight

d. Find the 70^th percentile.

Draw a new graph and label information technology appropriately. thou = 65.6

The 70^th percentile is 65.6. This ways that 70 percent of the test scores fall at or below 65.5 and xxx percent autumn at or above.

invNorm(0.70,63,5) = 65.half-dozen

Try It 6.8

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

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

Case 6.9

A personal calculator is used for role work at home, enquiry, advice, personal finances, instruction, 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 day. Presume the times for entertainment are normally distributed and the standard divergence for the times is half an hour.

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

Solution vi.9

a. Let X = the amount of time, in hours, a household personal calculator is used for entertainment. X ~ N(2, 0.v) where μ = 2 and σ = 0.five.

Find P(1.8 x

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

$\begin{array}{l}z=\frac{\mathrm{one.eight}-2}{0.5}=\frac{-0.2}{0.5}=-\mathrm{0.forty}\hfill \\ z=\frac{2.75-\mathrm{two}}{\mathrm{0.five}}=\frac{0.75}{\mathrm{0.v}}=1.5\hfill \end{array}$

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

The surface area to the left of the z-score of −0.twoscore is 0.3446. The expanse to the left of the z-score of i.v is 0.9332. The expanse between these scores volition be the difference in the two areas, or $0.9332-0.3446$, which equals 0.5886.

Figure 6.seven

normalcdf(one.8,2.75,two,0.5) = 0.5886

The probability that a household personal calculator is used betwixt 1.8 and 2.75 hours per solar day for entertainment is 0.5886.

b. Observe the maximum number of hours per day that the bottom quartile of households uses a personal computer for amusement.

Solution half-dozen.9

b. To observe the maximum number of hours per twenty-four hour period that the lesser quartile of households uses a personal computer for entertainment, observe the 25^th percentile, thou, where P(10 chiliad) = 0.25.

Figure half dozen.8

invNorm(0.25,2,0.5) = 1.66

We employ invNorm because we are looking for the k-value.

The maximum number of hours per 24-hour interval that the lesser quartile of households uses a personal computer for entertainment is 1.66 hours.

Try It half dozen.nine

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

Example 6.10

In the United states of america smartphone users between the ages of 13 and 55+ approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

a. Make up one's mind the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.seven years former.

Solution half-dozen.10

a. normalcdf(23,64.7,36.nine,13.9) = 0.8186

The z-scores are calculated as

$\begin{array}{l}z=\frac{23-\mathrm{36.nine}}{13.9}=\frac{-\mathrm{xiii.9}}{13.9}=-1\hfill \\ z=\frac{64.7-36.9}{13.9}=\frac{27.8}{\mathrm{13.nine}}=\mathrm{ii}\hfill \end{array}$

The Z-table shows the area to the left of a z-score with an absolute value of 1 to exist 0.1587. It shows the surface area to the left of a z-score of 2 to be 0.9772. The difference in the two areas is 0.8185.

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

b. Decide the probability that a randomly selected smartphone user in the age range xiii to 55+ is at well-nigh l.8 years old.

Solution half dozen.10

b. normalcdf(–ten⁹⁹,50.viii,36.9,xiii.9) = 0.8413

c. Find the 80^th percentile of this distribution, and interpret it in a complete sentence.

Solution 6.10

c.

• invNorm(0.80,36.9,13.nine) = 48.6
• The 80^thursday percentile is 48.6 years.
• 80 percentage of the smartphone users in the age range 13–55+ are 48.6 years one-time or less.

Try Information technology 6.10

Employ the data in Example vi.10 to answer the post-obit questions:

1. Detect the xxx^th percentile, and interpret it in a complete sentence.
2. What is the probability that the age of a randomly selected smartphone user in the range xiii to 55+ is less than 27 years old?

Case 6.11

In the Us smartphone users between the ages of 13 and 55+ approximately follow a normal distribution with gauge mean and standard deviation of 36.9 years and 13.9 years, respectively. Using this data, answer the post-obit questions—round answers to one decimal place:

a. Summate the interquartile range (IQR).

Solution 6.11

a.

• IQR = Q ₃ – Q ₁
• Summate Q ₃ = 75^th percentile and Q ₁ = 25^th percentile.
• Remember that we can use invNorm to find the k-value. Nosotros tin can employ this to find the quartile values.
• invNorm(0.75,36.9,thirteen.ix) = Q ₃ = 46.2754
• invNorm(0.25,36.nine,13.9) = Q ₁ = 27.5246
• IQR = Q ₃ – Q ₁ = 18.8

b. Forty percent of the ages that range from 13 to 55+ are at least what age?

Solution 6.11

b.

• Find k where P(x ≥ k) = 0.xl. At least translates to greater than or equal to.
• 0.40 = the area to the right
• The area to the left = 1 – 0.40 = 0.threescore.
• The surface area to the left of k = 0.lx
• invNorm(0.60,36.9,13.9) = 40.4215
• k = 40.iv.
• Forty pct of the ages that range from xiii to 55+ are at least twoscore.4 years.

Attempt Information technology half-dozen.eleven

Two grand students took an exam. The scores on the exam have an guess normal distribution with a hateful μ = 81 points and standard difference σ = 15 points.

1. Calculate the first- and tertiary-quartile scores for this test.
2. The middle fifty percent of the exam scores are betwixt what two values?

Instance vi.12

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a hateful diameter of v.85 cm and a standard departure of 0.24 cm.

a. Find the probability that a randomly selected mandarin orange from this farm has a bore larger than 6.0 cm. Sketch the graph.

Solution 6.12

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

Figure vi.9

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

Solution 6.12

b.

• 1 – 0.20 = 0.eighty. Outside of the eye twenty percent will be 80 percent of the values.
• The tails of the graph of the normal distribution each have an expanse of 0.40.
• Find k ₁, the 40^th percentile, and k _two, the 60^th percentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution.
• one thousand ₁ = invNorm(0.40,5.85,0.24) = five.79 cm
• m ₂ = invNorm(0.lx,5.85,0.24) = 5.91 cm

So, the centre xx pct of mandarin oranges have diameters between 5.79 cm and 5.91 cm.

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

Solution 6.12

c. half-dozen.16: Ninety percent of the bore of the mandarin oranges is at most 6.16 cm.

Try It 6.12

Using the information from Example 6.12, answer the following:

1. The centre 45 percent of mandarin oranges from this farm are betwixt ______ and ______.
2. Observe the 16^thursday percentile, and interpret it in a complete judgement.

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