Exam 2 PSYC 245

skewed right According to the verbal description, most haircuts cost less than $20, but there is a long right tail extending up to almost $60. Thus, the distribution of haircut prices is skewed right.

In a certain town, most haircuts are between $10 and $20, but a few salons cater to high‑end clients and charge $30 to $60. The distribution of haircut prices is -symmetric -skewed right -skewed left - bimodal.

44.22 To use the equation of the least‑squares line for prediction, substitute the value of the explanatory variable and calculate the response variable. In this case, the explanatory variable is the femur length and the response variable is the humerus length. The predicted humerus length for a fossil with a femur 40 cm long is humerus length=−3.66+(1.197)(40)=44.22 cm

Use the equation of the least‑squares line humerus length=−3.66+(1.197×femur length) to predict the humerus length for a fossil with a femur 40 cm long. (Give your answer as an exact number.)

skewed left

What is the shape of the exam score distribution? symmetric skewed right skewed left none of these

bar graph.

You have the average SAT score of entering freshman for five universities. The best graphical display for these data would be a -bar graph -line graph - pie chart. -side‑by‑side bar graph.

Exactly 1

One characteristic of a density curve is that there is a specific total area under the curve. What is this area equal to? - It depends on what is being measured - It depends on whether the distribution is Normal - Approximately 1. - Exactly 1.

A stemplot would have too many leaves, so a histogram would make it easier to view the distribution of the data.

Explain why we prefer a histogram to a stemplot for describing the returns on 1528 common stocks. - A stemplot would have too many leaves, so a histogram would make it easier to view the distribution of the data. - Histograms are the only graphical display of distributions, so a stemplot is not an option. - Because the data set is large, a stemplot would actually be better than a histogram. - A histogram is preferred to a stemplot because a histogram retains the original data in the graph.

mean=4300000 median=1800000

Find the values of the mean and the median. Round your answers to the nearest hundred thousand dollars.

the bars should have equal width.

For a bar graph to be accurate the bars should be vertical rather than horizontal. the bars should have equal height. the bars must touch each other. the bars should have equal width.

23 To find the number of stocks that lost money, add the frequency of the bars to the left of zero. There are seven bars to the left of zero. From right to left, the frequencies are approximately 11, 5, 3, 1, 1, 1, 1. Adding, 11%+5%+3%+1%+1%+1%+1%=23% This means 23% of all the stocks lost money.

A return less than zero means that owners of the stock lost money. About what percentage of all stocks lost money? Approximately. ?% of the stocks lost money.

"Car color" is categorical, "miles per day" is quantitative.

A survey was conducted and respondents were asked what color car they drive and how many miles they travel per day. Select the correct variable type for "car color" and "miles per day." "Car color" is quantitative, "miles per day" is categorical. "Car color" and "miles per day" both are categorical. "Car color" is categorical, "miles per day" is quantitative. "Car color" and "miles per day" both are quantitative..

More males than females are high school graduates or less.

Compare educational attainment for males and females. - About half of all people are high school graduates or less. - As the degree becomes more advanced, the greater the difference in attainment for males and females. - More males than females are high school graduates or less. - The relationships would be easier to see in a pie chart.

randomized comparative experiments.

The best evidence that changes in one variable cause changes in another comes from - the data for which the square of the correlation is near 1. - a plausible theory for causation. - the fact that higher values of the explanatory variable are associated with stronger responses. - randomized comparative experiments.

The correlation 𝑟=−0.86 is stronger because the absolute value of −0.86 is closer to 1 than is the absolute value of 0.704.

The correlation between percentage voting Democrat in 1980 and percentage voting Democrat in 1984 is 𝑟=0.704. The correlation between percentage of high school seniors taking the SAT and average SAT Mathematics score in the states is 𝑟=−0.86. Which of these two correlations indicates a stronger straight‑line relationship? Explain your answer. - The correlation 𝑟=−0.86 is stronger because the absolute value of −0.86 is closer to 1 than is the absolute value of 0.704. - The correlation 𝑟=−0.86 is stronger because −0.86 is smaller than 0.704. - The correlation 𝑟=0.704 is stronger because 0.704 is larger than −0.86. - The correlation 𝑟=0.704 is stronger because the absolute value of 0.704 is closer to 0 than the absolute value of −0.86.

lower bound=64.2 upper bound=74.2 The distribution of heights of men is approximately Normal, so you can use the 68-95-99.7 rule. According to this rule, the middle 95% of any Normal distribution lies within two standard deviations of the mean. The mean is 69.2 inches and the standard deviation is 2.5 inches. Two standard deviations is 5 inches here, so the middle 95% of men's heights are between 69.2−5=64.2 inches and 69.2+5=74.2 inches.

The distribution of heights of men is approximately Normal with mean 69.2 inches and standard deviation 2.5 inches. Between which heights do the middle 95% of men fall? (Give your answers to one decimal place.)

standard score: 1.12 standard score=observation−mean/ standard deviation Thus, the standard score of a height of 72 inches is 72−69.2 / 2.5= 1.12

The distribution of heights of young men is approximately Normal with mean 69.2 inches and standard deviation 2.5 inches. What is the standard score of a height of 72 inches (6 feet)? (Give your answer to two decimal places.)

100 percent

The heights of the bars on a relative frequency histogram displaying the lengths of rivers will add to the sum of all the river lengths. the midpoint of the distribution. the sample size. 100 percent.

is the line that makes the sum of the squared vertical distances of the data points from the line as small as possible.

The least‑squares regression line - is the line that makes the sum of the vertical distances of the data points from the line as small as possible. - is the line that makes the sum of the vertical distances of the data points from the line as large as possible. - is the line that makes the sum of the squared vertical distances of the data points from the line as small as possible. - is the line that makes the sum of the squared vertical distances of the data points from the line as large as possible.

lower bound= 218 upper bound 314 Almost all (99.7%) of the observations in a Normal distribution lie within three standard deviations of the mean. Since one standard deviation is 16) days, three standard deviations is 16⋅3=48 days. Therefore, almost all pregnancies are between 218 days (that is 266−48) and 314 days (that is 266+48).

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Use the 68-95-99.7 rule to answer the questions. Almost all (99.7%) pregnancies fall in what range of lengths? (Give your answers as exact numbers.)

more than 298 days long According to the 68-95-99.7 rule, the central 95% of a Normal distribution lies within two standard deviations of the mean. Since one standard deviation is 16 days, two standard deviations is 32 days here. So the central 95% of pregnancies are between 234 days (that is 266−32) and 298 days (that is 266+32). The other 5% of pregnancies have length outside the range from 234 to 298 days. Because the Normal distributions are symmetric, half of these pregnancies are on the long side. The longest 2.5% of pregnancies are more than 298 days long.

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Use the 68-95-99.7 rule to answer the questions. How long are the longest 2.5% of all pregnancies? (Give your answer as an exact number.)

less than 250 days long According to the 68-95-99.7 rule, the central 68% of a Normal distribution lies within one standard deviation of the mean. The standard deviation is 16 days here, so the central 68% of pregnancies are between 250 days (that is 266−16) and 282days (that is 266+16). The other 32% of pregnancies have a pregnancy length outside the range from 250 to 282 days. Because Normal distributions are symmetric, half of these pregnancies are on the short side. The shortest 16% of pregnancies are less than 250 days long.

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Use the 68-95-99.7 rule to answer the questions. How short are the shortest 16% of all pregnancies? (Give your answer as an exact number.)

the square of the correlation.

The quantity that tells us what fraction of the variation in the responses is explained by the straight line tie between the response and explanatory variables is - the correlation. - the square root of the correlation. - the square of the correlation. - the absolute value of the correlation.

39.7 Since the values in the table are in numerical order from smallest to largest, select the largest stem from the bottom of the table. Then select the largest leaf from that line. 39 7 As the key notes, the digits in the stem represent the percent, and the digits after the stem represent the digit after the decimal. Thus, this value is read as 39.7%.

p=?

Yes, it would also be correct to use a pie chart because the four categories represent an entire population when combined.

- Yes, it would also be correct to use a pie chart because the four categories represent an entire population when combined. - No, it would not be correct to use a pie chart because pie charts should only be used for percentages. - No, it would not be correct to use a pie chart because there aren't enough categories for a pie chart to be useful. - Yes, it would also be correct to use a pie chart because there are only four categories, which would make it easy to interpret.

The smallest total return is approximately -60%. The largest total return is approximately 100% The bar that is farthest to the left has endpoints of −70% and −60% and a center of −65%. Therefore, −65% is a good estimate for the smallest value, although it could be as low as −70%. The bar that is farthest to the right has endpoints of 100% and 110% and a center of 105%. Therefore, 105% is a good estimate for the largest value, although it could be as high as 110%. Sometimes, it can be difficult to determine the endpoints of the bars of a histogram. This histogram has tick marks and labels every 20% and there are two bars between each label. Therefore, the width of the bars of the histogram are all 10%. Since the leftmost bar has a right endpoint of −60%, its other endpoint is −60%−10%=−70%. The height of the leftmost bar is 1, so 1% of stocks that year lost between 60 and 70 percent of their value from the beginning of the year. Any number between −70% and −60% could be the smallest number. It is reasonble to estimate the smallest number with the midpoint of the endpoints of the leftmost bar, which is −60%+(−70%)2=−65% Similarly, the rightmost bar has endpoints of 100% and 110%, so 105% is a good estimate for the largest total return over the course of the year.

Approximately what were the smallest and largest total returns? (This describes the variability of the distribution.) The smallest total return is approximately ?%. The largest total return is approximately ?%

proportion =0.044 Let 𝑥 be the explanatory variable and 𝑦 be the response variable. The square of the correlation, 𝑟^2, is the proportion of the variation in the values of 𝑦 that is explained by the least‑squares regression of 𝑦 on 𝑥. In this case, 𝑥 is beer prices and 𝑦 is hot dog prices. The value of the correlation is 0.21. Thus, the proportion of the variation in hot dog prices that is explained by the least‑squares regression of hot dog prices on beer prices (per ounce) equals 𝑟^2=0.21^2≈0.044

The correlation between the prices is 𝑟=0.21. What proportion of the variation in hot dog prices is explained by the least‑squares regression of hot dog prices on beer prices (per ounce)? (Give your answer to three decimal places.)

About half of U.S. households in 2017 had incomes higher than $68,145 and about half had incomes less than $68,145.

You read that the median income of U.S. households in 2017 was $68,145. Select the correct explanation for the term "the median income. - "The total amount of household income produced in the U.S. in 2017 was $68,145 times the number of households. - The total amount of household income produced in the U.S. in 2017 divided by the number of households is $68,145. - More U.S. households in 2017 made $68,145 than any other household income. - About half of U.S. households in 2017 had incomes higher than $68,145 and about half had incomes less than $68,145.

1.3% To calculate the percentages in 1950, take the count in each age range and divide by the total for that year, 151.1. For example, the percent in the Under 10 age group is 29.3/151.1=0.1939 or 19.39%. Continue in this way for all the age groups in 1950. To calculate the percentages for 2050, do the same thing, but divide each count for each age group in that year by the total for that year, which is 310.6. For example, the percent in the Under 10 age group is 56.2/ 310.6=0.1809 or 18.09%. Continue in this way for all the age groups in 2050. By comparing the percents in the 1950 column to the 2050 column, the only age group that was larger in 1950 than 2050 by percent is the under 10 years age group. The difference between the percents in the two years is 19.39%−18.09%=1.30% Thus, the difference 𝑑 is 1.3%.

Because the total population in 2050 is much larger than the 1950 population, comparing percentages in each age group is clearer than comparing counts. Make a table of the percentage of the total population in each age group for both 1950 and 2050. Only one age group in 1950 represents a larger percentage of the population than it does in 2050. For that age group, determine the difference of the percentages between 1950 and 2050. Report your answer to the tenth of a percent. difference in percentages?

Ty Cobb's standard score: 4.15 Ted Williams' standard score: 4.26 George Brett's standard score: 4.07 Standard scores have a mean of 0 and a standard deviation of 1. Once each batting average is converted to a standard score, the standard scores can be compared to each other. The further a standard score is from zero, the more unusual the data point is. The player who was the best batter when compared to his peers will have the largest standard score. The standard score for any observation is standard score=observation−mean/ standard deviation For Ty Cobb, use the mean and standard deviation from the 1910s and substitute into the standard score formula. standard score for Ty Cobb=0.420−0.266 / 0.0371=4.15094 ≈4.15 For Ted Williams, use the mean and standard deviation from the 1940s and substitute into the standard score formula. standard score for Ted Williams=0.406−0.267/ 0.0326= 4.26380≈ 4.26 For George Brett, use the mean and standard deviation from the 1970s and substitute into the standard score formula. standard score for George Brett=0.390−0.261/ 0.0317= 4.06940≈ 4.07 Ted Williams deviated the most from his peers with a standard score of 4.26, then Ty Cobb with a standard score of 4.15,then George Brett with a standard score of 4.07.

Compute the standard scores for the batting averages of Cobb, Williams, and Brett to compare how far each stood above his peers. Enter your answers to two decimal places. Ty Cobb's standard score: Ted Williams' standard score: George Brett's standard score:

The graph is symmetric.

Describe the overall shape of the distribution of total returns. The graph is skewed right. The graph is symmetric. The graph is skewed left. The graph is uniform.

after 20 years= 6000 Equation = y=1000+300x To calculate how much Fred will have in his mattress, you can use the graph or substitute 𝑥=20 into the equation. 1000+250⋅20=6000 So, after 20 years, Fred will have $6000 in his mattress. To determine the new equation for Fred's savings, use the definitions of the intercept and slope of the line. Consider the old equation 𝑦=1000+250𝑥. The number 250 is the slope of the line, the amount by which 𝑦 changes when 𝑥 increases by one unit. The number 1000 is the intercept, the value of 𝑦 when 𝑥=0. If Fred had added $300 instead of $250 each year to his initial $1000, the new equation would have the new slope, 300, and the same intercept as the old equation. Thus, the equation that describes Fred's savings after 𝑥 years if he had added $300 instead of $250 each year to his initial $1000 is 𝑦=1000+300𝑥

Fred keeps his savings in his mattress. He began with $1000 from his mother and adds $250 each year. His total savings 𝑦 after 𝑥years are given by the equation 𝑦=1000+250𝑥 Draw a graph of this equation. (Choose two values of 𝑥, such as 0 and 10. Compute the corresponding values of 𝑦 from the equation. Plot these two points on graph paper and draw a straight line joining them.) After 20 years, how much will Fred have in his mattress? (Give your answer as an exact number.) If Fred had added $300 instead of $250 each year to his initial $1000, what is the equation that describes his savings after 𝑥 years? (Use 𝑦 for total savings and 𝑥 for years.)

p= 57.14 First calculate the standard score of 542. standard score=observation−mean / standard deviation 542−522 / 114=20 / 114=0.1754... The standard normal table provides scores to a single decimal place. The closest score is thus 0.2. Using the table of standard scores Standard Normal Table: Table B, look up the percentile corresponding to the standard score found, which is 57.93. Thus, the observation 542 represents approximately the 57.93rd percentile. In other words, about 57.93% of students that took the SAT ERW test received a score that was lower than the mean SAT Math score. Using software, the percentile corresponding to standard score is 56.96.

In 2018, the average performance of college‑bound seniors on the Evidence‑Based Reading and Writing (ERW) portion of the SAT followed a Normal distribution with mean 522 and standard deviation 114. The mean for the SAT Math portion was 542. What percentage 𝑝 of scores on the SAT ERW portion were lower than the SAT math mean? Give your answer to two decimal places.

points= 533 To determine the average Math SAT score based on the regression equation, replace the percentage taking variable with 61in the equation and round to the nearest whole number. average Math SAT score =602.4−(1.142⋅percentage taking) =602.4−(1.142⋅61)=532.738=533 Be careful to not confuse 61% with the proportion 0.61.

In Georgia, the percentage of high school seniors who took the SAT was 61%. Predict their average score to the nearest whole number. (The actual average score in Georgia was 515.) points:

The shape is skewed right with a center around 13. The minimum is 5.08 and the maximum is 42.00.

Make a graph to display the distribution of records for the states. Mark where your state lies in this distribution. Briefly describe the distribution in terms of shape, center, and variability. - The shape is roughly symmetric with a center around 13. The minimum is 5.08 and the maximum is 42.00. -The shape is skewed right with a center around 13. The minimum is 5.08 and the maximum is 42.00. - The shape is skewed left with a center around 13. The minimum is 5.08 and the maximum is 42.00. - The shape is skewed left with a center around 25. The minimum is 5.08 and the maximum is 42.00 - .The shape is skewed right with a center around 25. The minimum is 5.08 and the maximum is 42.00.

The sample was random, so not all samples will have the same number of people with IQ's less than 80. Since 0 and 2are pretty close together, it is not very unusual for a random sample to have no one with an IQ less than 80.

None of the 80 adults in our sample had scores this low. Your instructor wondered if students were surprised by that fact. Imagine that you are discussing this with a classmate, who says, "There must have been something wrong with the sample. 2.5%of 80 is 2, so I would expect to see two people in the sample with IQ's less than 80." What is the best response to the classmate? - It could be that something is wrong with the sample, but it could also be that the distribution of all IQ scores is not actually Normal. If the distribution is skewed, that could explain why no one with IQ scores less than 80 were found. - The sample was random, so not all samples will have the same number of people with IQ's less than 80. Since 0 and 2are pretty close together, it is not very unusual for a random sample to have no one with an IQ less than 80. - I agree that there must have been something wrong with the sample. A sample of size 80 is quite large, so it should accurately reflect the population unless it was biased. -It is very unlikely that anyone from the population has an IQ less than 80, so it is even more unlikely that a relatively small sample of size 80 would have anyone with an IQ less than 80. It would be unusual if the sample contained anyone with an IQ less than 80.

25th percentile =-0.67 75th percentile =0.67

The quartiles of any distribution are the 25th and 75th percentiles. Use Table B to determine the standard score for the first quartile (the 25th percentile). Use Table B to determine the standard score for the third quartile (the 75th percentile).

The shape of the distribution is symmetric, so the measures of center should all be similar.

The shape of the distribution suggests that the mean will be about the same as the median. Why? - The shape of the distribution is skewed, so the mean and the median will be similar. - The shape of the distribution is symmetric, so the measures of center should all be similar. - The mean and the median are always the same value since they both measure the center. - The mean and median are the same because there is an even number of values, so an average is used to calculate both values.

The slope indicates that as the percentage of students taking the exam increases by one, the average Math SAT score decreases by 1.142 points.

What does the slope 𝑏=−1.142 tell us about the relationship between these variables? - The slope indicates that as the average Math SAT score increases by one, the percentage of students taking the exam decreases by 1.142 points. - The slope indicates that as the percentage of students taking the exam decreases by one, the average Math SAT score decreases by 1.142 points. - The slope indicates that as the percentage of students taking the exam increases by one, the average Math SAT score decreases by 1.142 points. - The slope indicates that as the average Math SAT score decreases by one, the percentage of students taking the exam decreases by 1.142 points.

having time intervals that are not equally spaced

Which of the following is not acceptable for a line graph? having equally spaced time intervals having seasonal variation having lines that cross having time intervals that are not equally spaced

minimum value:0.8 first quartile:3.8 median: 11 third quartile: 27.8 maximum: 139.4 To determine the five‑number summary, you should first order the data from the lowest to highest values. The minimum and maximum values are simply the minimum and maximum values in the data. These are 0.8 and 139.4, respectively. To determine the median, find the midpoint of the distribution such that half of the 𝑛 observations are smaller, and the other half are larger. Since there are 26 observations in the data, the median is the average of the two center observations in the ordered list. To find the center position of the ordered list, use the formula: position of median=𝑛+12=26+12=13.5 From the bottom of the list, the position of the median is the average of the 13th and 14th observations. The 13th observation is 10.1 and the 14th observation is 11.9. The median is then 10.1+11.92=11. The first quartile is the median of the observations whose position in the ordered list is to the left of the location of the overall median. The position of the first quartile is: position of first quartile=13+1 /2= 7 The seventh position, or first quartile, is 3.8. The third quartile is the median of the observations whose position in the ordered list is to the right of the location of the overall median. The position of the third quartile is: position of third quartile=26− 13+1 / 2= 26−7= 19 The 19th position, or third quartile, is 27.8. The five number summary is 0.8, 3.8, 11.0, 27.8, and 139.4.

minimum value: first quartile: median: third quartile: maximum:

It is easier to compare the heights of the bars on the bar graph than it is to compare the size of the angles on the pie chart.

- It is easier to see that the bars of the bar graph sum to 100% than it is to see that the wedges of the pie chart sum to 100%. - The categories on a bar graph can be ordered from greatest to least, but the categories on a pie chart cannot. - It is easier to compare the heights of the bars on the bar graph than it is to compare the size of the angles on the pie chart. - There are no advantages. Bar graphs and pie charts can always be used interchangeably.

The percentages do not represent parts of a whole.

According to the National Household Survey on Drug Use and Health, when asked in 2016, 23.5% of those aged 18 to 25 years used cigarettes in the past month, 5.2% used smokeless tobacco, 23.2% used illicit drugs, and 38.4% engaged in binge alcohol drinking. Explain why it is not correct to display these data in a pie chart - .The respondents are likely to be lying about illicit drug use and binge drinking. - The percentages do not represent parts of a whole. - The variable of interest is quantitative, not categorical. - There is no data value for not doing any of those things.

The distribution is right skewed. The modal class is children 10 and under. There is a sharp drop off between young children and teens.

Make a histogram of the 1950 age distribution (in percents). Describe the main features of the distribution. In particular, describe the percentage of children relative to the rest of the population. - The distribution is right skewed but is nearly uniform until around age 50. The sharpest drop off is at age 90. - The distribution is right skewed. The modal class is children 10 and under. There is a sharp drop off between young children and teens. - The distribution is left skewed. The modal class is age 70 to 79. The biggest jump is from the under 10 group to the teenage group. - The distribution is right skewed. The modal class is ages 20 to 29. There is almost no one alive older than age 90.

one can recover the actual observations from the display.

An advantage of a stemplot over a histogram is one can recover the actual observations from the display. the classes are chosen for you. they are good for really large data sets. they are horizontal.

The mean is likely greater than the median because the distribution is skewed to the right.

Construct a graph of the data. Based on the shape of the distribution, describe the position of the mean relative to the median. - The mean and the median are approximately equal because the distribution is symmetric. - The mean is likely greater than the median because the distribution is skewed to the right. - The mean is likely less than the median because the distribution is skewed to the left. - The mean is likely less than the median because the distribution is skewed to the right.

The shape is roughly symmetric. The center is around 35.9%. The variability is 32.5% to 39.7%.

Describe the shape, center, and variability of this distribution. - The shape is roughly symmetric. The center is around 35.9%. The variability is 32.5% to 39.7%. - The shape is skewed to the right. The center is around 36.8%. The variability, excluding outliers, is 33.4% to 38.7%. - The shape is roughly symmetric. The center is around 35.9%. The variability, excluding outliers, is 33.4% to 38.7%. - The shape is skewed to the right. The center is around 36.8%. The variability is 32.5% to 39.7%.

Sales and leases for both cars and trucks were at the highest in 1999 and 2000. Later years show an overall trend of gradually decreasing sales and leases until 2007 when the trend decreases sharply for both cars and trucks before beginning to rise again. Overall, trucks have much lower sales and leases than cars, except for the period 2002 to 2008 when truck sales and leases were only slightly higher than cars.

Do consumers prefer trucks, SUVs, and minivans to passenger cars? The data provided give the sales and leases of new cars and trucks (in thousands of vehicles) in the United States from 1996 to 2010. (The definition of "truck" includes SUVs and minivans.) Plot two line graphs on the same axes to compare the change in car and truck sales over time. Click to download the data in your preferred format. CSV Excel JMP Mac-Text Minitab PC-Text R SPSS TI CrunchIt! Describe the trend that you see in the line graphs. - Both cars and trucks show an overall trend of gradually decreasing sales and leases until 2007 when they decrease sharply for both cars and trucks. Trucks typically have much lower sales and leases than cars. - The highest quantity of sales and leases for both cars and trucks occurred in 1999 and 2000. Later years show an overall trend of gradually decreasing sales and leases until 2007 when the trend decreases sharply for both cars and trucks before beginning to rise again. Generally, cars have much lower sales and leases than trucks except for the period 2002 to 2008 when car sales and leases were only slightly higher than trucks. - Sales and leases for both cars and trucks were at the highest in 1999 and 2000. Later years show an overall trend of gradually decreasing sales and leases until 2007 when the trend decreases sharply for both cars and trucks before beginning to rise again. Overall, trucks have much lower sales and leases than cars, except for the period 2002 to 2008 when truck sales and leases were only slightly higher than cars. - The overall trend of truck sales and leases shows minimal change from 1996 to 2010, while car sales and leases decreased by approximately 3000. There is minimal fluctuation in sales and leases for both trucks and cars over the 14 years.

because the pattern observed in the data may be different outside the range of the data.

Extrapolation, or prediction outside the range of the data, is risky - unless the correlation is very close to 1. - because the pattern observed in the data may be different outside the range of the data. - because correlation does not necessarily imply causation. - unless the square of the correlation is very close to 1.

No, it is not possible. All the values in the stemplot are sorted by size. Only the values are included, not the states they belong to.

From an examination of the stemplot for residents aged 18 to 44, is it possible to determine the percentage of residents in that age range that lived in Arizona in 2010? Explain. - No, it is not possible. All the values in the stemplot are sorted by size. Only the values are included, not the states they belong to. - Yes, it is possible. The data from which the stemplot was built should have that information. - No, it is impossible to determine raw values since, like a histogram, a stemplot only shows the count of data in a range. - Yes, it is possible. Arizona is the third state alphabetically. Since the values in the list are ordered, the third value in the stemplot is the corresponding value.

The variability will change but not the shape or center. Hawaii was already an outlier and is now the new maximum, but that does not affect the shape or center.

In April 2018, it was reported that 49.69 inches of rain fell in a 24‑hour period on the island of Kauai, in Hawaii. If we change the precipitation amount for Hawaii from 38.00 inches to 49.69 inches, how does this impact the distribution of records for the states? In other words, would your description of shape, center, and variability need to be modified? Please select the best answer that explains the impact. - The shape, center, and variability will change because the change in the precipitation amount for Hawaii is very large. - The center and variability will change but not the shape. The shape is still right skewed, but the center is smaller and the maximum is larger. - The variability will change but not the shape or center. Hawaii was already an outlier and is now the new maximum, but that does not affect the shape or center. - Neither the shape, the center, nor the variability will change because the change in the precipitation amount for Hawaii is slight.

The distribution for the residents aged 65 and older is more variable because the data cover a larger range than for the distribution for residents aged 18 to 44.

Is the distribution for residents aged 18 to 44 years more or less variable than the distribution for residents aged 65 and older? Explain. - All the values in the distribution for ages 65 and older are less than all the values in the distribution for ages 18 to 44, and so there is less variability. - The distribution for the residents aged 65 and older is left‑skewed and so it cannot have the same variability as the distribution for residents aged 18 to 44. - The distribution for the residents aged 65 and older is more variable because the data cover a larger range than for the distribution for residents aged 18 to 44. - The distribution for the residents aged 65 and older has approximately the same variability because the range is the same for the distribution for residents aged 18 to 44. - The distribution for the residents aged 65 and older is less variable because the data cover a smaller range than for the distribution for residents aged 18 to 44.

The 2050 distribution is more uniform in the early years, and a much larger percentage survive beyond age 60 than was the case in 1950.

Make a histogram of the projected age distribution for the year 2050. Use the same scales as your histogram in the previous question for easy comparison. What are the most important changes in the U.S. age distribution projected for the 100-year period between 1950 and 2050? - The biggest difference is that the number of people age 10 to 19 is the largest group, whereas in 1950, this group was much larger than groups older or younger than it. - The biggest difference between the two distributions is the number of people in the 90 to 99 age group, which is many times more in 2050 than it was in 1950. - The 2050 distribution is more uniform in the early years, and a much larger percentage survive beyond age 60 than was the case in 1950. - The biggest difference is the birth rate is at a replacement rate (each age group remains about the same over time), whereas in the 1950s, this was the start of the Baby Boom and the population was increasing sharply.

The distribution is slightly skewed to the left. Because of the slight skew, the median, which is 46, is the best way to describe a typical year. Sixty home runs is not an outlier.

Make a stemplot of these data. Is the distribution roughly symmetric, clearly skewed, or neither? About how many home runs did Ruth hit in a typical year? Is his famous 60 home runs in 1927 an outlier? - The distribution is roughly symmetric. Because the distribution is symmetric, the midpoint, which is 41, is the best way to describe the number of home runs Babe Ruth hit in a typical year. Sixty is not an outlier because there are other years in which Ruth hit nearly 60 home runs. - The distribution is slightly skewed to the left. Because of the slight skew, the median, which is 46, is the best way to describe a typical year. Sixty home runs is not an outlier. - The distribution is roughly symmetric. A typical number of runs is 44 because the mean is 43.9. Sixty home runs is an outlier because it is an extremely rare for a baseball player to hit that many home runs in a year. - The distribution is slightly skewed to the right. Because the distribution is slightly skewed, the median, which is 46, is the best way to describe a typical year. Sixty home runs is not an outlier.

Mookie Betts' standard score: 2.67 Christian Yelich's standard score:2.45 To covert the batting average to a standard score, use the formula standard score=observation−mean/ standard deviation Mookie Betts has a batting average of 0.346 and is in the American League. Using the mean of 0.262 and standard deviation of 0.0315, the calculation is standard score for Mookie Betts=0.346−0.262/ 0.0315= 2.66667≈ 2.67 Christian Yelich has a batting average of 0.326 and is in the National League. Using the mean of 0.269 and standard deviation of 0.0233, the calculation is standard score for Christian Yelich=0.326−0.269/ 0.0233= 2.44635≈ 2.45

Mookie Betts' standard score: Christian Yelich's standard score:

The median is less than $240,000 because average income is inflated by the relatively few number of individuals who have much higher incomes than average.

Seattle Magazine reports that the average income of its readers is $240,000. Is the median wealth of these readers greater or less than $240,000? Why? - The median is greater than $240,000 because average income is inflated by the relatively few number of individuals who have much higher incomes than average. - The median is less than $240,000 because average income is inflated by the relatively few number of individuals who have much higher incomes than average. - The median is less than $240,000 because average income is deflated by the relatively few number of individuals who have much lower incomes than average. - The median is greater than $240,000 because average income is deflated by the relatively few number of individuals who have much lower incomes than average.

approximately 16% of adults have pulse rates higher than Bonnie's. According to the 68-95-99.7 rule, the central 68% of any Normal distribution lies within one standard deviation of the mean. The mean of the resting pulse rates distribution is 69 beats per minute and the standard deviation is 9.5 beats per minute. Bonnie's pulse rate is one standard deviation above the mean since 78.5=69+9.5. Note that 32% of healthy adults have pulse rates outside the interval within one standard deviation of the mean. Because the Normal distributions are symmetric, approximately 16% of adults have pulse rates higher than Bonnie's. Since 69+9.5 equals 78.5, Bonnie's pulse rate is one standard deviation above the mean, not two standard deviations above the mean. A standard score is the number of standard deviations an observation is above or below the mean. Sine Bonnie's pulse rate is one standard deviation above the mean, the standard score is 1 not 1.5.

Suppose that resting pulse rates for healthy adults are found to follow a Normal distribution, with a mean of 69 beats per minute and a standard deviation of 9.5 beats per minutes. If Bonnie has a pulse rate of 78.5 beats per minute, this means that - Bonnie's pulse rate is two standard deviations above the mean. - Bonnie's pulse rate, when converted to a standard score, would be 1.5 - .approximately 16% of adults have pulse rates higher than Bonnie's. - approximately 32% of adults have pulse rates higher than Bonnie's.

No, it will not be Normal. Even though the distribution of each one is Normal, the means are not in the same place and the partially overlapping distributions will have two peaks.

The mean height of men is about 69.2 inches. Women that age have a mean height of about 63.7 inches. Do you think that the distribution of heights for all adults is approximately Normal? Explain your answer. Yes, it will be at least approximately Normal since both the heights of men and the heights of women are normally distributed. No, it will not be Normal. Since men are taller than women, the distribution will be skewed right .No, it will not be Normal. Even though the distribution of each one is Normal, the means are not in the same place and the partially overlapping distributions will have two peaks. No, it will not be Normal. Since there are more women than men, the distribution will be skewed left.

roughly symmetric because team payrolls are capped as part of the collective bargaining agreement to provide equality among teams.

There are 30 teams in the National Basketball Association (NBA) and each team has a team payroll. The team payroll consists of the total amount of money available to pay all players on the team. As an example, in the 2018‑19 season, the LA Lakers had a team payroll of $107,020,840. Suppose you were able to look at the distribution of team payrolls for all 30 teams in the NBA during the 2018‑19 season. The distribution of team payrolls should be - clearly skewed to the right because there are relatively few players with very large salaries. - clearly skewed to the left because there are relatively few players with very small salaries. - roughly symmetric because team payrolls are capped as part of the collective bargaining agreement to provide equality among teams.

x=3 s= 1.095 x= 5 s=1.095

Use a calculator to find the mean 𝑥⎯⎯⎯ and standard deviation 𝑠 of the two sets of numbers. (Use decimal notation. Give your answers to three decimal places if necessary.) The first set is 4, 2, 4, 2, 4, 2. Add 2 to each of the numbers in the first data set. The data are 6, 4, 6, 4, 6, 4.

you want to look at the distribution of a quantitative variable.

Use a histogram when you want to look at the distribution of a quantitative variable. the number of observations is small. you want to show the actual observations. you want to look at the distribution of a categorical variable.

The reasonable range for prediction is from 0% to 100% of seniors taking the SAT.

Using least‑squares regression to do prediction outside the range of available data is risky. For what range of data is it reasonable to use the least‑squares regression line for predicting average SAT Math score from percentage taking? - The reasonable range for prediction is from 20% to 100% of seniors taking the SAT. - The reasonable range for prediction is from 0% to 20% of seniors taking the SAT. - The reasonable range for prediction is from 0% to 100% of seniors taking the SAT. - The reasonable range for prediction is from 20% to 80% of seniors taking the SAT.

is between 15 and 20, so 18 is a good estimate.

What is the approximate center of this distribution? -The approximate center of this distribution - cannot be determined from the histogram. - is equal to 15. - is between 15 and 20, so 18 is a good estimate. - is between 20 and 25, so 22 is a good estimate.

p= 2.5% Since the standard deviation of the distribution is 10 and the mean is 100, 80−100/ 10=−2 That is, 80 is 2 standard deviations below the mean. (Similarly, 120 is 2 standard deviations above the mean.) The 68-95-99.7 rule states that about 95% of the values in a Normal distribution are within 2 standard deviations of the mean. Therefore, 95% of IQ scores are between 80 and 120. Therefore, 5% of IQ scores are not between 80 and 100. The Normal distribution is symmetric, so half of 5%, or 2.5%, of IQ scores are less than 80 (and 2.5% are greater than 120.)

What percentage 𝑝 of all students have IQ scores below 80? Give your answer as a percentage to one decimal place.

The median is the balance point in a density curve.

Which of the following is an incorrect statement? - The mean of a skewed distribution is pulled toward the long tail. - In a symmetric density curve, the mean is equal to the median. - If a density curve is skewed to the right, the mean will be larger than the median. - The median is the balance point in a density curve.

the five‑number summary

Which of the following should you use to describe a distribution that is skewed? the five‑number summary the mean and standard deviation the mean, the first quartile, and the third quartile the median and standard deviation

Removing an outlier will decrease the standard deviation.

Which of the following statements is true of the standard deviation? It is the difference between the minimum and maximum values. Removing an outlier will decrease the standard deviation. It is the difference between the first and third quartile. Removing an outlier will increase the standard deviation.

minimum:107 first quartile:138.5 median:153 third quartile:180.5 maximum:195 The first quartile, 𝑄1, is the median of the 8 observations below the median, not including the median. Using the location formula for the median, you can find the location of 𝑄1. Because the number of observations below the median is even, the first quartile will be the average of the 4th and 5th observation. location of 𝑄1=𝑛+1/ 2= 9/ 2=4.5 The value of the first quartile is 𝑄1=138+139/ 2= 138.5 The third quartile, 𝑄3, is the median of the 𝑛=8 observations above the median, not including the median. Thus, the average of the 13th and 14th observation gives 180.5. 𝑄3=179+182/ 2= 180.5 The following represents the five‑number summary for the number of calories in a hot dog. Minimum=107, 𝑄1=138.5, 𝑀=153, 𝑄3=180.5, Maximum=195

minimum: first quartile: median: third quartile: maximum: