Data Analysis, Probability, and Statistics - HiSET

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Question

An ad campaign was run at 1200 fast food franchise locations. Two weeks later, a random sample of revenue of 80 locations was taken to measure the effectiveness of the ad campaign. Of the 80 locations, 60 locations reported increased sales.

Which of the following statements can be made with the greatest certainty?

Answer

From the question, we know that 60 out of a random sample of 80 locations experienced increased sales in the last two weeks; however, because this data is only from a random sampling of locations, it is difficult to say anything absolute about the population (all of the locations).

While 75% of the random sample did experience increased sales, we cannot know for certain what portion of the entire population experienced increased sales. What we can conclude from this data is that it is likely that the percentage of restaurants that experienced increased sales is roughly around 75%.

Therefore, the statement that can be made with the most certainty is that "It is very likely that more than half of all franchise locations experienced increased sales in the last two weeks."

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Question

Determine the linear regression line for the following paired data:

$\begin{matrix} \underline{x }&\underline{y} \ 1 & 3.5 \ 2 & 4.7\ 3 & 6.0\ 4 & 7.8 \end{matrix}$

Answer

The formula for a regression line for a set of paired data is the line of the equation

$y =b_{1}x+ b_{0}$ ,

where

$b_{1} = \frac{n (\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}$

and

$b_{0} = \overline{y} - b_{1}\overline{x}$ .

, the number of pairs.

$\sum x$ is the sum of the values:

$\sum x = 1 + 2 + 3 + 4 = 10$

$\sum y$ is the sum of the values:

$\sum y = 3.5 + 4.7 + 6.0 + 7.8 = 22.0$

$\sum x^{2}$ is the sum of the squares of the values:

$\sum x^{2} = 1^{2} + 2^{2} + 3 ^{2} + 4 ^{2} = 1 + 4 + 9 + 16 = 30$

$\sum xy$ is the sum of the products:

$\sum xy = 1(3.5 )+ 2(4.7)+ 3(6.0)+ 4 (7.8) = 3.5 + 9.4 + 18.0 + 31.2 = 62.1$

Substituting:

$b_{1} = \frac{4 (62.1)-(10)( 22.0)}{4(30)-(10)^{2}}$

$= \frac{248.4-220}{120-100}$

$= \frac{28.4}{20}$

$\overline{x}$ and $\overline{y}$ are the arithmetic means of the and values, respectively - the sums divided by the number of values:

$\overline{x} = \frac{\sum x}{n} = \frac{10}{4} = 2.5$

$\overline{y} = \frac{\sum y}{n} = \frac{22}{4} = 5.5$

$b_{0} = \overline{y} - b_{1}\overline{x}$

$b_{0} =5.5 - 1.42 (2.5)$

The equation of the linear regression line is .

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Question

Determine the linear regression line for the following paired data:

$\begin{matrix} \underline{x }&\underline{y} \ 1 &10.0 \ 2 &8.7\ 3 & 7.1 \ 4 & 5.0 \end{matrix}$

Answer

The formula for a regression line for a set of paired data is the line of the equation

$y =b_{1}x+ b_{0}$ ,

where

$b_{1} = \frac{n (\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}$

and

$b_{0} = \overline{y} - b_{1}\overline{x}$ .

, the number of pairs.

$\sum x$ is the sum of the values:

$\sum x = 1 + 2 + 3 + 4 = 10$

$\sum y$ is the sum of the values:

$\sum y= 10.0+8.7+ 7.1 +5.0 = 30.8$

$\sum x^{2}$ is the sum of the squares of the values:

$\sum x^{2} = 1^{2} + 2^{2} + 3 ^{2} + 4 ^{2} = 1 + 4 + 9 + 16 = 30$

$\sum xy$ is the sum of the products:

$\sum xy = 1(10.0 )+ 2(8.7)+ 3(7.1)+ 4 (5.0) = 10.0+ 17.4+21.3 + 20.0= 68.7$

Substituting:

$b_{1} = \frac{4 (68.7)-(10)( 30.8)}{4(30)-(10)^{2}}$

$= \frac{274.8 - 308}{120-100 }$

$= \frac{-33.2}{20 }$

$\overline{x}$ and $\overline{y}$ are the arithmetic means of the and values, respectively - the sums divided by the number of values:

$\overline{x} = \frac{\sum x}{n} = \frac{10}{4} = 2.5$

$\overline{y} = \frac{\sum y}{n} = \frac{30.8}{4} = 7.7$

$b_{0} = \overline{y} - b_{1}\overline{x}$

The linear regression line is the line of the equation .

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Question

Researchers performed a survey of the populations of several small islands. The data is represented by the following histogram.

What percentage of islands surveyed had a population of less than 140?

Answer

On the horizontal axis of a histogram, you have labels representing ranges of values. For example, the label "140–159" represents the range of islands with populations between 140 and 159.

The question asks for the percent of islands with populations under 140. The ranges that are smaller than 140 are the "120–139" range and the "100–119" range. Match the bars of each with the percents on the vertical (y) axis. The percents corresponding to these ranges are 35% and 30%, respectively.

Adding both of these percents yields the answer, 65%.

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Question

The yearly revenue over 10 years of a company is given by the following graph.

Using the graph, determine the approximate revenue in the year 2004.

Answer

The question asks for you to find the revenue in the year 2004. The horizontal axis is labeled "Year," so look there first to find where the label for 2004 is.

Tracing upwards, you will see that the value 2004 corresponds to a value of about 16 on the vertical axis. Since the label for the vertical axis is "Dollars in Millions," this means that the revenue in 2004 was 16 million dollars.

In numerical form, 16 million dollars is $16,000,000.

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Question

The approximate salaries of employees in a company are given by the histogram below:

Using the data presented in the histogram, determine what percentage of employees earn a weekly salary of $180,000 or more.

Answer

Each bar in this histogram represent the percent of employees with salary in the given range. For example, the bar above the label "150–159" means that 25% of the employees earn between $150,000 and $159,000.

Since we want the percent of employees who earn $180,000 or more, we need to add the values for the bars for "180–189", "190–199", and "200–210".

The bars show that 15% of employees earn between $180,000 and $189,000, 15% earn between $190,000 and $199,000, and 5% earn between $200,000 and $210,000.

Adding these percentages together yields 35%.

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Question

Which of these data sets would be best represented using a line graph?

Answer

The line graph is best used with data that occurs sequentially, especially sequentially over a period of time. The line graph is best at describing a general trend, such as an increase or decrease, over time.

The decrease in weight of an adult male over 12 months fits a line graph well because there is data that can be organized sequentially (chronologically, in this case) and because we would be looking for some sort of trend in the data, a decrease over time in this case.

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Question

The revenue of a small company over 10 years is portrayed by the following line graph.

Which of the following statements can best be made about the data?

Answer

The question asks you to identify a trend that best fits the data.

Over 10 years, the revenue of the company decreases and increases with no clear overall pattern or trend, so the answers claiming that there is a general increase or decrease are incorrect.

While the revenue was at a high point ($65,000) in the year 1994, it was not the highest point in the 10 years. The highest point was in 1998, with a revenue of $77,000.

While the revenue did increase from 1991 to 1992, it decreased by even more from 1991 to 1993. Therefore, from 1991 to 1993, there was actually a net decrease in revenue, not a net increase.

The correct answer is: "The company's revenue generally decreased from 1994 to 1996." Notice the data points matching the years 1994, 1995, and 1996 are linked by a line that has a downward slope. This indicates a general decrease in revenue.

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Question

An individual's income over 10 years is represented by the following line chart.

Approximate individual's highest annual income from 1994 to 1998.

Answer

Notice that the question narrows the time period you are examining to 1994–1998. While there are years with higher incomes than in this time period (namely, 1992 and 1999), we can only consider the years from 1994 to 1998.

The highest point within this period is in the year 1998. Matching this data point with the values on the y-axis (the income values), you will see that this data point lies between $400,000 and $500,000. Looking more closely, you can even say that it is above halfway between the values, so it appears to be greater than $450,000.

Thus, the value that approximates the income in 1998 is $460,000.

Be careful to take into account the units on each axis. The y-axis measures income in thousands of dollars, so the answer is $460,000 and not $460.

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Question

If I have quarters, what is the probability that I will get no more than heads?

Answer

Step 1: If we have 3 quarters and each quarter has two outcomes, then there are total outcomes that can come out of all of the rolls of the quarters.

Those rolls are:

HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.

Step 2: Determine how many outcomes have no more than heads....

We cannot count HHH because it has heads.

We also cannot count TTT because it has no heads at all.

Step 3: We had options and we can't take options..so we have a total of options left.

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Question

In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?

Answer

Step 1: Find the probability of getting a green marble...

The probability of a green marble is $\dfrac {5}{40}$ .

Step 2: Find the probability of getting a blue marble (without replacement)

When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is less.

Probability of a blue marble is $\dfrac {15}{39}$ .

Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.

So, $\dfrac {5}{40}\times \dfrac {15}{39}=\dfrac {5\times 15}{40\times 39}=\dfrac {75}{1560}$

Step 4: Simplify...

$\dfrac {75}{1560}=\dfrac {\dfrac {75}{5}}{\dfrac {1560}{5}}=\dfrac {15}{312}=\dfrac {15/3}{312/3}=\dfrac {5}{104}$

$\dfrac {5}{104}$ is the answer.

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Question

What is the probability that we will roll a one on a six sided die and flip a coin to reveal "heads"?

Answer

In order to solve this standard we need to understand two primary components: probabilities and the property of independent events. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

$P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}$

An intersection is the likelihood that one event will occur with another.

We can calculate that the probability of rolling a one on a six-sided die is:

$P(rolling\ a\ one)=\frac{1}{6}$

Now, let's determine the probability of flipping a coin to reveal the "heads" side. A coin has two sides: heads or tails. The probability of rolling heads is as follows:

$P(heads)=\frac{1}{2}$

An intersection is the probability of two events occurring simultaneously; therefore, we need to multiply the probabilities.

$P(rolling\ a\ one\ and\ flipping\ heads)=\frac{1}{6}\times \frac{1}{2}=\frac{1}{12}$

Convert to a percentage.

$\frac{1}{12}=0.0833$

$0.0833\times100%=8.33%$

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Question

The spades and the aces are removed from a standard deck of 52. Give the probability that a card drawn at random will be a red three.

Answer

In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of

cards, to leave a modified deck of

cards.

None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is

$P(R3) = \frac{n(R3)}{n(C)} = \frac{2}{36} = \frac{1}{18}$ .

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Question

How many ways can eight people be seated at two tables if four people should be at each table, and it only matters at which table each person sits (ass opposed to which seat)?

Answer

The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is .

Since

$C(n, r) = \frac{n!}{(n-r)!r!}$ ,

then, setting ,

$C(8,4) = \frac{8!}{(8-4)!4!}$

$= \frac{8!}{4!4!}$

$= \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 }{1 \times 2 \times 3 \times 4 \times 1 \times 2 \times 3 \times 4}$

$= \frac{ 5 \times 6 \times 7 \times 8 }{ 1 \times 2 \times 3 \times 4}$

$= \frac{1680}{24}$

,

the correct number of arrangements.

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Question

A standardized test comprises five examinations: reading, writing, math, science, and social studies, and these examinations may be taken in any order; however, in one state, there is one restriction: neither the writing nor the math test may be taken last.

In how many different orders can the five tests be taken so that this restriction is met?

Answer

Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:

First, the test to be taken last must be one of three tests: reading, science, or social studies;

Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.

By the multiplication principle, this makes

$3 \times 4 \times 3 \times 2 \times 1 = 72$

possible schedules.

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Question

What is the probability that a person will roll an even number on a six-sided fair die?

Answer

In order to solve this problem, we need to discuss probabilities. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

$P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}$

The die has six sides with the following values: one tow, three, four, five, and six. Of these values the die has three that are even: two, four, and six. We can write the following probability.

$P=\frac{3\textup{ even values}}{6\textup{ total possible outcomes}}$

Simplify.

$P=\frac{3}{6}$

$P=\frac{1}{2}$

Now, let's convert this into a percentage:

$\frac{1}{2}=0.5$

$0.5\times100%=50%$

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Question

Two fair dice are tossed. You are not shown the dice but you are told that exactly one die shows an even number. Give the probability that a total of seven was thrown.

Answer

Two fair dice can be thrown with 36 equally probable rolls. 18 rolls comprise one even number and one odd number. They are as follows - with the rolls that total seven in boldface:

$1\textup{-}2, 1\textup{-}4, \textbf{1-6}, 2\textup{-}1, 2\textup{-}3, \textbf{2-5}, 3\textup{-}2, \textbf{3-4}, 3\textup{-}6,$

$4\textup{-}1, \textbf{4-3}, 4\textup{-}5, \textbf{5-2}, 5\textup{-}4, 5\textup{-}6, \textbf{6-1}, 6\textup{-}3, 6\textup{-}5.$

6 of the 18 possible rolls have total 7, making the probability of a roll of 7, given that exactly one doe shows an even number, equal to $\frac{6}{18} = \frac{1}{3}$ .

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Question

Two fair dice are tossed. You are not shown the dice but you are told that at least one die shows an even number. Give the probability that a total of seven was thrown.

Answer

Two fair dice can be thrown with 36 equally probable rolls.9 rolls can be thrown that show two odd numbers: they are

$1\textup{-}1, 1\textup{-}3, 1\textup{-}5, 1\textup{-}1, 1\textup{-}3, 3\textup{-}5, 3\textup{-}1, 3\textup{-}3, 3\textup{-}5$

This leaves 27 rolls with at least one even number, 6 of which result in a sum of 7:

$1\textup{-}6, 2\textup{-}5, 3\textup{-}4, 4\textup{-}3, 5\textup{-}2, 6\textup{-}1$

This makes the probability of rolling a sum of 7

$\frac{6}{27} = \frac{2}{9}$ ,

the correct choice.

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Question

Find the mean of the following data set:

$10,\ 12,\ 3,\ 4,\ 7,\ 8,\ 9,\ 10,\ 12,\ 11,\ 13,\ 14,\ 16, \textup{ and } 18$

Answer

The mean is found by adding the values in a set and dividing that number by the total number of values.

$\textup{Mean}=\frac{10+12+3+4+7+8+9+10+12+11+13+14+16+18}{14}$

$\textup{Mean}=\frac{147}{14}$

$\textup{Mean}=10.5$

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Question

Consider the data set

$\left { 10, 12, 12, 12, 12, 13, 16, 17, 20 \right }$

Of the mean, the median, the mode, and the midrange of the data set, which two are equal?

Answer

The mean of a data set $\overline{x}$ is equal to the sum of the terms divided by the number of terms, which here is 9. Therefore,

$\overline{x} = \frac{ 10+ 12+ 12+ 12+12+ 13+ 16+ 17+ 20}{9} = \frac{129}{9} \approx 13.8$

The median of a data set with an odd number of terms is the term that falls in the middle when the terms are ordered. The terms are already ordered:

$\left { 10, 12, 12, 12, \textbf{12}, 13, 16, 17, 20 \right }$

so the element that falls in the middle is 12.

The mode of the data set is the term that occurs most frequently, which is 12.

The midrange is the mean of the least and greatest elements, which is

$\frac{10+20}{2} = \frac{30}{2} = 15$ .

Of the four - mean, median, mode, and midrange - the median and the mode are equal.

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