Question 1

The mean test score on a statistics test was 72 with a standard deviation of 10.How many standard deviations from the mean is a test score of 90?&#10;A)1.80 standard deviations above the mean&#10;B)1.80 standard deviations below the mean&#10;C)0.69 standard deviations above the mean&#10;D)0.80 standard deviations above the mean&#10;E)0.69 standard deviations below the mean

Accepted Answer

To find how many standard deviations a score is from the mean, subtract the mean from the score and divide by the standard deviation: (90 - 72) / 10 = 18 / 10 = 1.80 standard deviations above the mean.

Question 2

Suppose that the average amount of sugar a person eats per year is 7 kg with a standard deviation of 1.5 kg.How many standard deviations from the mean is the consumption of 11 kg of sugar?&#10;A)About 1.33 standard deviations above the mean&#10;B)About 1.33 standard deviations below the mean&#10;C)About 4.67 standard deviations above the mean&#10;D)About 2.67 standard deviations above the mean&#10;E)About 2.67 standard deviations below the mean

Accepted Answer

To find how many standard deviations from the mean a value is, subtract the mean from the value and divide by the standard deviation: (11 kg - 7 kg) / 1.5 kg = 4 / 1.5 = About 2.67 standard deviations above the mean.

Question 3

Adam played golf on Saturday and Sunday.He scored 82 both days.The scores of all golfers Saturday averaged 68 with a standard deviation of 17.The scores on Sunday averaged 81 with a standard deviation of 9.On which day did Adam do better compared with the other golfers? Explain.Note that the smaller the score the better in golf.

A)Sunday.A score of 82 Sunday is

\frac { 14 } { 17 }

standard deviations from the mean while a score of 82 Saturday is

\frac { 1 } { 9 }

standard deviations from the mean.
B)Sunday.A score of 82 Sunday is

\frac { 1 } { 9 }

standard deviations from the mean while a score of 82 Saturday is

\frac { 14 } { 17 }

standard deviations from the mean.
C)Saturday.A score of 82 Saturday is

\frac { 14 } { 17 }

standard deviations from the mean while a score of 82 Sunday is

\frac { 1 } { 9 }

standard deviations from the mean.
D)He did not do better either day,he scored the same.
E)Saturday.A score of 82 Saturday is

\frac { 1 } { 9 }

standard deviations from the mean while a score of 82 Sunday is

\frac { 14 } { 17 }

standard deviations from the mean.

Accepted Answer

We can use z-scores to compare Adam's score with the average scores of other golfers. For Saturday, Adam's z-score is (82-68)/17 = 0.82, which means he is 0.82 standard deviations above the mean. For Sunday, his z-score is (82-81)/9 = 0.11, which means he is only 0.11 standard deviations above the mean. Since a smaller z-score indicates a better performance compared to other golfers, we can conclude that Adam did better on Sunday. Choice B correctly states that Sunday was the better day for Adam.

Question 4

The setter on your school's volleyball team averages 58 assists per match with a standard deviation of 6.How many standard deviations from the mean is an outing with 79 assists?&#10;A)1.75 standard deviations below the mean&#10;B)1.36 standard deviations below the mean&#10;C)3.50 standard deviations below the mean&#10;D)3.50 standard deviations above the mean&#10;E)1.75 standard deviations above the mean

Accepted Answer

To find how many standard deviations from the mean an outing with 79 assists is, we use the formula: (X - μ) / σ, where X is the value in question (79 assists), μ is the mean (58 assists), and σ is the standard deviation (6). Plugging in the numbers: (79 - 58) / 6 = 21 / 6 = 3.5. Therefore, an outing with 79 assists is 3.50 standard deviations above the mean.

Question 5

The average number of days absent per student per year at West Valley School is 16 days with a standard deviation of 5 days.How many standard deviations from the mean is of 6 absent days?&#10;A)0.83 standard deviations above the mean&#10;B)1.83 standard deviations below the mean&#10;C)2.00 standard deviations above the mean&#10;D)2.00 standard deviations below the mean&#10;E)1.83 standard deviations above the mean

Accepted Answer

To find how many standard deviations from the mean 6 absent days is, we use the formula: (X - μ) / σ, where X is the value (6 days), μ is the mean (16 days), and σ is the standard deviation (5 days). So, (6 - 16) / 5 = -10 / 5 = -2. This means 6 absent days is 2 standard deviations below the mean.

Question 6

A basketball coach kept statistics for his team in free throw percentage and steals (among others).At the last game,Erin's free throw percentage was 79% and she had 4 steals.The team averaged 90% from the free throw line with a standard deviation of 6% and they averaged 7 steals with a standard deviation of 4.In which category did Erin do better compared with her team? Explain.

A)Steals.4 steals is -

\frac { 11 } { 6 }

standard deviations from the mean while 79% free throw average is -

\frac { 3 } { 4 }

standard deviations from the mean.
B)Free throw percentage.79% free throw average is -

\frac { 3 } { 4 }

standard deviations from the mean while 4 steals is -

\frac { 11 } { 6 }

standard deviations from the mean.
C)Free throw percentage.79% free throw average is -

\frac { 11 } { 6 }

standard deviations from the mean while 4 steals is -

\frac { 3 } { 4 }

standard deviations from the mean.
D)One can't compare the two categories,they are too different.
E)Steals.4 steals is -

\frac { 3 } { 4 }

standard deviations from the mean while 79% free throw average is -

\frac { 11 } { 6 }

standard deviations from the mean.

Accepted Answer

The answer of A basketball coach kept statistics for his...

Question 7

The average amount of sugar a person eats per year is 4 kg.A person who consumes 3.25 kg of sugar has a z-score of -0.75.Find the standard deviation.&#10;A)4 kg&#10;B)1 kg&#10;C)0.1 kg&#10;D)5 kg&#10;E)10 kg

Accepted Answer

Using the formula for z-score:
z = (x - μ) / σ

where:
x = 3.25 kg (amount consumed by the person)
μ = 4 kg (average amount consumed by a person)
z = -0.75

Rearranging the formula, we get:
σ = (x - μ) / z
σ = (3.25 - 4) / (-0.75)
σ = 0.75

Therefore, the standard deviation is 0.75 kg which is closest to answer choice B (1 kg).

Question 8

The mean weights for medium navel oranges is 274 grams.Suppose that the standard deviation for the oranges is 92 grams.Which would be more likely,an orange weighing 392 grams or an orange weighing 137 grams? Explain.

A)A 137 gram orange is more likely (z = 1.49)compared with an orange weighing 392 grams (z = 4.24).
B)A 137 gram orange is more likely (z = -1.49)compared with an orange weighing 392 grams (z = 1.28).
C)A 137 gram orange is more likely (z = 1.28)compared with an orange weighing 392 grams (z = -1.49).
D)A 392 gram orange is more likely (z = -1.49)compared with an orange weighing 137 grams (z = 1.28).
E)A 392 gram orange is more likely (z = 1.28)compared with an orange weighing 137 grams (z = -1.49).

Accepted Answer

The answer of The mean weights for medium navel oranges...

Question 9

The mean weight of babies born in a hospital last year was 2.84 kg.Suppose the standard deviation of the weights is 0.95 kg.Which would be more unusual,a baby weighing 1.8 kg or a baby weighing 3.83 kg? Explain.

A)A 1.8 kg baby is more unusual (z = -1.09)compared with a 3.83 kg baby (z = 1.04).
B)A 3.83 kg baby is more unusual (z = 1.90)compared with a 1.8 kg baby (z = 4.05).
C)A 3.83 kg baby is more unusual (z = -1.09)compared with a 1.8 kg baby (z = -1.09).
D)A 1.8 kg baby is more unusual (z = -1.05)compared with a 3.83 kg baby (z = -1.04).
E)A 3.83 kg baby is more unusual (z = 1.04)compared with a 1.8 kg baby (z = -1.09).

Accepted Answer

To determine which baby is more unusual, we need to calculate the z-scores for each weight. For a 1.8 kg baby, the z-score is (1.8-2.84)/0.95 = -1.09. For a 3.83 kg baby, the z-score is (3.83-2.84)/0.95 = 1.04. A negative z-score indicates that the weight is below average, while a positive z-score indicates that the weight is above average. Since the z-score for the 1.8 kg baby is further from zero than the z-score for the 3.83 kg baby, the 1.8 kg baby is more unusual. Therefore, the best choice is A.

Question 10

Two different running shoe manufacturers market running shoes to first time marathon runners.Swift claims a mean shoe life of 600 km,while Enduramax claims a shoe life of 650 km.If the standard deviation for Swift shoes is 54 km and 124 km for Enduramax,which shoe would you purchase before starting your marathon training (where you figure to run 500 km)? Explain.

A)Enduramax.Enduramax shoes are -

\frac { 75 } { 62 }

standard deviations from the mean while Swift shoes are -

\frac { 50 } { 27 }

standard deviations from the mean.
B)Enduramax.The Enduramax shoes have a longer mean shoe life.
C)Swift.Swift shoes are -

\frac { 75 } { 62 }

standard deviations from the mean while Enduramax shoes are -

\frac { 50 } { 27 }

standard deviations from the mean.
D)Swift.Swift shoes are -

\frac { 50 } { 27 }

standard deviations from the mean while Enduramax shoes are -

\frac { 75 } { 62 }

standard deviations from the mean.
E)Enduramax.Enduramax shoes are -

\frac { 50 } { 27 }

standard deviations from the mean while Swift shoes are -

\frac { 75 } { 62 }

standard deviations from the mean.

Accepted Answer

The answer of Two different running shoe manufacturers market running...

Question 11

The weights of children age two average 24 pounds with a standard deviation of 2 pounds.How many standard deviations from the mean is a weight of 17 pounds?&#10;A)3.50 standard deviations below the mean&#10;B)1.29 standard deviations above the mean&#10;C)1.29 standard deviations below the mean&#10;D)1.41 standard deviations above the mean&#10;E)3.50 standard deviations above the mean

Accepted Answer

The answer of The weights of children age two average...

Question 12

Mario's poker winnings average $319 per week with a standard deviation of $56.How many standard deviations from the mean is winning $195?&#10;A)About 1.11 standard deviations below the mean&#10;B)About 2.21 standard deviations above the mean&#10;C)About 1.64 standard deviations below the mean&#10;D)About 2.21 standard deviations below the mean&#10;E)About 1.11 standard deviations above the mean

Accepted Answer

The answer of Mario's poker winnings average $319 per week...

Question 13

The average size of forest fires last year was 968 acres with a standard deviation of 182 acres.How many standard deviations from the mean is a forest fire consuming 245 acres?&#10;A)About 3.97 standard deviations below the mean&#10;B)About 1.35 standard deviations above the mean&#10;C)About 3.97 standard deviations above the mean&#10;D)About 1.99 standard deviations below the mean&#10;E)About 1.99 standard deviations above the mean

Accepted Answer

The answer of The average size of forest fires last...

Question 14

A town's snowfall in December averages 10 cm with a standard deviation of 10 cm while in February,the average snowfall is 42 cm with a standard deviation of 16 cm.In which month is it more likely to snow 32 cm? Explain.

A)February.Snowfall of 32 cm is

\frac { 11 } { 5 }

from the mean while snowfall of 32 cm is -

\frac { 5 } { 8 }

from the mean in December.
B)December.Snowfall of 32 cm is -

\frac { 5 } { 8 }

from the mean while snowfall of 32 cm is

\frac { 11 } { 5 }

from the mean in February.
C)It is equally likely in either month.One can't predict Mother Nature.
D)December.Snowfall of 32 cm is

\frac { 11 } { 5 }

from the mean while snowfall of 32 cm is -

\frac { 5 } { 8 }

from the mean in February.
E)February.Snowfall of 32 cm is -

\frac { 5 } { 8 }

from the mean while snowfall of 32 cm is

\frac { 11 } { 5 }

from the mean in December.

Accepted Answer

The answer of A town's snowfall in December averages 10...

Question 15

Two different running shoe manufacturers market running shoes to first time marathon runners.Swift claims a mean shoe life of 600 km,while Enduramax claims a shoe life of 650 km.If the standard deviation for Swift shoes is 54 km and 124 km for Enduramax,which shoe would you purchase before starting your marathon training (where you figure to run 500 km)? Explain.

A)Enduramax.Enduramax shoes are -

\frac { 75 } { 62 }

standard deviations from the mean while Swift shoes are -

\frac { 50 } { 27 }

standard deviations from the mean.
B)Enduramax.The Enduramax shoes have a longer mean shoe life.
C)Swift.Swift shoes are -

\frac { 75 } { 62 }

standard deviations from the mean while Enduramax shoes are -

\frac { 50 } { 27 }

standard deviations from the mean.
D)Swift.Swift shoes are -

\frac { 50 } { 27 }

standard deviations from the mean while Enduramax shoes are -

\frac { 75 } { 62 }

standard deviations from the mean.
E)Enduramax.Enduramax shoes are -

\frac { 50 } { 27 }

standard deviations from the mean while Swift shoes are -

\frac { 75 } { 62 }

standard deviations from the mean.

Accepted Answer

The answer of Two different running shoe manufacturers market running...

Question 16

You heard that the average number of years of experience among stockbrokers is 14 years.You can't remember the standard deviation somebody with 8 years' experience has a z-score of -2.Find the standard deviation.

A)3 years
B)9 months
C)6 months
D)9 years
E)3 months

Accepted Answer

The answer of You heard that the average number of...

Question 17

The average number of average number of hours per day a university student spends on homework is 5 hours with a standard deviation of 1.25 hours.How many standard deviations from the mean is 2 hours spent on homework?

A)2.40 standard deviations below the mean
B)1.20 standard deviations below the mean
C)2.50 standard deviations above the mean
D)1.20 standard deviations above the mean
E)2.40 standard deviations above the mean

Accepted Answer

The answer of The average number of average number of...

Question 18

A town's average snowfall is 40 cm per year with a standard deviation of 12 cm.How many standard deviations from the mean is a snowfall of 64 cm?&#10;A)0.44 standard deviations below the mean&#10;B)2.00 standard deviations above the mean&#10;C)2.00 standard deviations below the mean&#10;D)1.60 standard deviations above the mean&#10;E)0.44 standard deviations above the mean

Accepted Answer

The answer of A town's average snowfall is 40 cm...

Question 19

The average number of babies born in a certain town each year is 218 with a standard deviation of 26.How many standard deviations from the mean is a year with 387 babies born?&#10;A)3.25 standard deviations below the mean&#10;B)6.50 standard deviations above the mean&#10;C)3.25 standard deviations above the mean&#10;D)6.50 standard deviations below the mean&#10;E)1.78 standard deviations above the mean

Accepted Answer

The answer of The average number of babies born in...

Question 20

The local basketball team averages 65% from the free throw line.A player who makes 72.5% of his free throws has a z-score of 1.5.Find the standard deviation.&#10;A)5%&#10;B)15%&#10;C)3%&#10;D)1%&#10;E)10%

Accepted Answer

The answer of The local basketball team averages 65% from...

Question 21

The average amount of snowfall in a certain town is 54 cm per year.Last year's snowfall was 67.5 cm which corresponded to a z-score of 1.5.Find the standard deviation.&#10;A)5 cm&#10;B)30 cm&#10;C)9 cm&#10;D)20 cm&#10;E)1 cm

Accepted Answer

The answer of The average amount of snowfall in a...

Question 22

On an exam Bob scored 75% and was 1.5 standard deviations above the mean of 63%.What was the standard deviation?&#10;A)12%&#10;B)18%&#10;C)16%&#10;D)8%&#10;E)1.5%

Accepted Answer

The answer of On an exam Bob scored 75% and...

Question 23

The speed vehicles traveled on a local road was recorded for one month.The speeds ranged from 48 km/h to 63 km/h with a mean speed of 57 km/h and a standard deviation of 6 km/h.The quartiles and median speeds were 51 km/h,60 km/h,and 54 km/h.Suppose increased patrols reduced speeds by 7%.Find the new mean and standard deviation.Express your answer in exact decimals.

A)Mean: 53.01 km/h,SD: 5.58 km/h
B)Mean: 53.01 km/h,SD: 6 km/h
C)Mean: 60.99 km/h,SD: 6.42 km/h
D)Mean: 3.99 km/h,SD: 0.42 km/h
E)Mean: 60.99 km/h,SD: 6 km/h

Accepted Answer

The answer of The speed vehicles traveled on a local...

Question 24

Here are some summary statistics for annual snowfall in a certain town compiled over the last 15 years: lowest $\text { amount } = 19$ cm, $\text { mean } = 47$ cm, $\text { median } = 40$ cm, $\text { range } = 81$ cm, $\mathrm { IQR } = 54$ $\mathrm { Q } 1 = 19 \text {, }$ standard $\text { deviation } = 10$ cm.Suppose snowfall was tracked for 5 additional years and the annual snowfall was found to increase by 20%.Find the new mean and standard deviation.

A)Mean: 9.4 cm,SD: 2 cm
B)Mean: 37.6 cm,SD: 10 cm
C)Mean: 56.4 cm,SD: 10 cm
D)Mean: 37.6 cm,SD: 8 cm
E)Mean: 56.4 cm,SD: 12 cm

Accepted Answer

The answer of Here are some summary statistics for annual...

Question 25

The average weight of a newborn infant is 2.7 kg.An infant that weighs 3.15 kg has a z-score of 1.Find the standard deviation.&#10;A)0.05 kg&#10;B)0.14 kg&#10;C)1.35 kg&#10;D)0.45 kg&#10;E)2.7 kg

Accepted Answer

The answer of The average weight of a newborn infant...

Question 26

The free throw percentages for the participants of a basketball tournament were compiled.The percents ranged from 40% to 96% with a mean of 53% and a standard deviation of 8%.The quartiles and median percentages were 50%,84%,and 53%.This year's tournament percentages are 3% higher than last year's tournament percentages.Find last year's tournament mean,median,and standard deviation of the free throw percentages.

A)Mean: 56%,median: 56%,SD: 11%
B)Mean: 53%,median: 56%,SD: 11%
C)Mean: 51.5%,median: 53%,SD: 7.8%
D)Mean: 51.5%,median: 54.4%,SD: 7.8%
E)Mean: 50%,median: 50%,SD: 5%

Accepted Answer

The answer of The free throw percentages for the participants...

Question 27

Here are some summary statistics for last year's basketball team scoring output: lowest $\text { score } = 18$ points, $\text { mean } = 58$ points, $\text { median } = 52$ points, $\text { range } = 97$ points, $\mathrm { IQR } = 46$ $\mathrm { Q } 1 = 23 \text {, }$ standard $\text { deviation } = 9$ points.Suppose the opponents' scoring output was 10% lower.Find the opponents' mean and standard deviation.

A)Mean: 52.2 points,SD: 8.1 points
B)Mean: 52.2 points,SD: 9 points
C)Mean: 63.8 points,SD: 9.9 points
D)Mean: 63.8 points,SD: 9 points
E)Mean: 5.8 points,SD: 0.9 points

Accepted Answer

The answer of Here are some summary statistics for last...

Question 28

Here are the summary statistics for the monthly payroll for an accounting firm: $\text { lowest salary } = \$ 60,000 \text {, }$ $\text { mean salary } = \$ 140,000 \text {, }$ median = $100,000,range = $240,000, IQR = $120,000, $\text { first quartile } = \$ 70,000$ standard deviation = $80,000.
Suppose that business has been good and the company gives every employee a $\$ 10,000$ raise.Give the new value of each of the summary statistics.

A)Minimum: 60,000; Mean: 140,000; Median: 100,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
B)Minimum: 70,000; Mean: 150,000; Median: 100,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
C)Minimum: 70,000; Mean: 140,000; Median: 110,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
D)Minimum: 70,000; Mean: 150,000; Median: 110,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000
E)Minimum: 60,000; Mean: 150,000; Median: 110,000; Range: 240,000; IQR: 120,000; Q1: 80,000; SD: 80,000

Accepted Answer

The answer of Here are the summary statistics for the...

Question 29

A family of four averages 85 loads of laundry per month.One such family does 62 loads per month and has a z-score of -2.3.Find the standard deviation.&#10;A)20 loads&#10;B)10 loads&#10;C)1 load&#10;D)2 loads&#10;E)3 loads

Accepted Answer

The answer of A family of four averages 85 loads...

Question 30

The average attendance for your local hockey team is 450,000 per game.The attendance during the last game was only 320,000 which corresponded to a z-score of -16.25.Find the standard deviation.&#10;A)1000 people&#10;B)17,000 people&#10;C)3000 people&#10;D)20,000 people&#10;E)8000 people

Accepted Answer

The answer of The average attendance for your local hockey...

Question 31

The average speed cars travel on a road is 77 km/h.A car travelling 60 km/h has a z-score of -1.7.Find the standard deviation.&#10;A)40 km/h&#10;B)10 km/h&#10;C)30 km/h&#10;D)0.5 km/h&#10;E)1 km/h

Accepted Answer

The answer of The average speed cars travel on a...

Question 32

Here are some statistics for the annual Wildcat golf tournament: lowest $\text { score } = 58 \text {, }$ mean $\text { score } = 95 \text {, }$ median = 103,range = 92,IQR = 110, $\mathrm { Q } 1 = 38 \text {, }$ standard $\text { deviation } = 15 \text {. }$ Suppose it was very windy and all the golfers' scores went up by 7 strokes.Tell the new value for each of the summary statistics.

A)Lowest score: 65,mean: 102,median: 110,range: 92,IQR: 110,Q1: 45,SD: 15
B)Lowest score: 65,mean: 102,median: 110,range: 85,IQR: 110,Q1: 45,SD: 15
C)Lowest score: 65,mean: 102,median: 110,range: 85,IQR: 117,Q1: 45,SD: 15
D)Lowest score: 65,mean: 95,median: 103,range: 85,IQR: 110,Q1: 45,SD: 15
E)Lowest score: 65,mean: 102,median: 110,range: 85,IQR: 110,Q1: 45,SD: 22

Accepted Answer

The answer of Here are some statistics for the annual...

Question 33

An English instructor gave a final exam and found a mean score of 69 points and a standard deviation of 5.8 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Then find the interval for the central 68% of the scores.

A)

<strong>An English instructor gave a final exam and found a mean score of 69 points and a standard deviation of 5.8 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Then find the interval for the central 68% of the scores.</strong> A) ; 57.4 to 80.6 points B) ; 63.2 to 74.8 points C) ; 63.2 to 74.8 points D) ; 57.4 to 69 points E) ; 69 to 80.6 points <div style=padding-top: 35px>

; 57.4 to 80.6 points
B)

; 63.2 to 74.8 points
C)

; 63.2 to 74.8 points
D)

; 57.4 to 69 points
E)

; 69 to 80.6 points

Accepted Answer

The answer of An English instructor gave a final exam...

Question 34

Here are some summary statistics for the size of forest fires last year: smallest $\text { fire } = 92$ acres, $\text { mean } = 462$ acres, $\text { median } = 462$ acres, $\text { range } = 7908$ acres, $\mathrm { IQR } = 375 \text {, }$ $Q 1 = 185 \text {, }$ standard $\text { deviation } = 57$ acres.If it costs $1200 per acre to fight the fires,find the minimum,median,standard deviation,and IQR of the cost.

A)Minimum: 1292; median: 1662; SD: 1257; IQR: 1575
B)Minimum: 92; median: 462; SD: 57; IQR: 375
C)Minimum: 110,400; median: 554,400; SD: 68,400; IQR: 450,000
D)Minimum: 110,400; median: 462; SD: 68,400; IQR: 375
E)Minimum: 110,400; median: 554,400; SD: 57; IQR: 375

Accepted Answer

The answer of Here are some summary statistics for the...

Question 35

The systolic blood pressure of $\text { 18-year-old }$ women is normally distributed with a mean of $120 \mathrm {~mm } \mathrm {~Hg }$ and a standard deviation of $12 \mathrm {~mm} \mathrm {~Hg } .$ Draw and label the Normal model for systolic blood pressure.What percentage of $\text { 18-year-old }$ women have a systolic blood pressure between $96 \mathrm {~mm} \mathrm {~Hg }$ and $144 \mathrm {~mm} \mathrm {~Hg? }$&#10;A)   ; 95%&#10;B)   ; 99.7%&#10;C)   ; 68%&#10;D)   ; 34%&#10;E)   ; 84%

Accepted Answer

The answer of The systolic blood pressure of \(\text {...

Question 36

Here are some summary statistics for the recent English exam: lowest $\text { score } = 31 \text {, }$ mean $\text { score } = 66 \text {, }$ $\text { median } = 80.2 \text {, }$ $\text { range } = 79 \text {, }$ $\mathrm { IQR } = 58$ $\mathrm { Q } 1 = 26$ standard $\text { deviation } = 8.3 \text {. }$ Suppose the students did not study for the exam and each score went down 15%.Tell the new value for each of the summary statistics.Express your answer in exact decimals.

A)Lowest score: 35.65,mean: 75.9,median: 92.23,range: 90.85,IQR: 66.7,Q1: 22.1, SD: 9.545
B)Lowest score: 26.35,mean: 56.1,median: 68.17,range: 79,IQR: 49.3,Q1: 22.1,SD: 7.055
C)Lowest score: 26.35,mean: 56.1,median: 68.17,range: 67.15,IQR: 49.3,Q1: 22.1, SD: 7.055
D)Lowest score: 4.65,mean: 9.9,median: 12.03,range: 11.85,IQR: 8.7,Q1: 22.1,SD: 1.245
E)Lowest score: 26.35,mean: 56.1,median: 68.17,range: 79,IQR: 58,Q1: 22.1,SD: 7.055

Accepted Answer

The answer of Here are some summary statistics for the...

Question 37

The amount of Jen's monthly phone bill is normally distributed with a mean of $55 and a standard deviation of $8.Draw and label the Normal model for Jen's monthly phone bill.In what interval would you expect the central 68% of bills to be found?

A)

<strong>The amount of Jen's monthly phone bill is normally distributed with a mean of $55 and a standard deviation of $8.Draw and label the Normal model for Jen's monthly phone bill.In what interval would you expect the central 68% of bills to be found?</strong> A) ; $31 to $79 B) ; $39 to $71 C) ; $47 to $71 D) ; $47 to $63 E) ; $39 to $63 <div style=padding-top: 35px>

; $31 to $79
B)

; $39 to $71
C)

; $47 to $71
D)

; $47 to $63
E)

; $39 to $63

Accepted Answer

The answer of The amount of Jen's monthly phone bill...

Question 38

Here are some summary statistics for all of the runners in a local 12 kilometre race: slowest $\text { time } = 129$ minutes, $\text { mean } = 79$ minutes, $\text { median } = 79$ minutes, $\text { range } = 99$ minutes, $\mathrm { IQR } = 64 \text {, }$ $\mathrm { Q } 1 = 32$ standard $\text { deviation } = 11$ minutes.Suppose last year's race results were better by 8%.Find last year's mean and standard deviation.Express your answer in exact decimals.

A)Mean: 6.32 minutes,SD: 0.88 minutes
B)Mean: 85.32 minutes,SD: 11 minutes
C)Mean: 85.32 minutes,SD: 11.88 minutes
D)Mean: 72.68 minutes,SD: 10.12 minutes
E)Mean: 72.68 minutes,SD: 11 minutes

Accepted Answer

The answer of Here are some summary statistics for all...

Question 39

A salesman's commission averages $23,700 per year.Last year his commission was $18,900 which corresponded to a z-score of -1.2.Find the standard deviation.&#10;A)$10,000&#10;B)$500&#10;C)$15,000&#10;D)$1000&#10;E)$4000

Accepted Answer

The answer of A salesman's commission averages $23,700 per year.Last...

Question 40

The average score on a mathematics test was 64%.A student who scored 80% had a z-score of 2.Find the standard deviation.&#10;A)16%&#10;B)8%&#10;C)3%&#10;D)20%&#10;E)1%

Accepted Answer

The answer of The average score on a mathematics test...

Question 41

-2.41 < z < 0&#10;A)49.10%&#10;B)9.48%&#10;C)52.16%&#10;D)50.80%&#10;E)49.20%

Accepted Answer

The answer of -2.41 < z < 0&#10;A)49.10%&#10;B)9.48%&#10;C)52.16%&#10;D)50.80%&#10;E)49.20%...

Question 42

An English instructor gave a final exam and found a mean score of 69 points and a standard deviation of 5.8 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Then find the interval for the central 68% of the scores.

A)

; 57.4 to 80.6 points
B)

; 63.2 to 74.8 points
C)

; 63.2 to 74.8 points
D)

; 57.4 to 69 points
E)

; 69 to 80.6 points

Accepted Answer

The answer of An English instructor gave a final exam...

Question 43

The lengths of human pregnancies can be described by a Normal model with a mean of 268 days and a standard deviation of 15 days.What percentage can we expect for a pregnancy that will last at least 300 days?

A)98.34%
B)1.66%
C)1.79%
D)48.34%
E)1.99%

Accepted Answer

The answer of The lengths of human pregnancies can be...

Question 44

For a recent English exam,use the Normal model N(73,9.2)to find the percent of scores between 56 and 87.Round to the nearest tenth of a percent.&#10;A)90.4%&#10;B)3.2%&#10;C)3.37%&#10;D)9.7%&#10;E)96.6%

Accepted Answer

The answer of For a recent English exam,use the Normal...

Question 45

The test scores from a recent Mathematics test are as follows: 95.5,65.9,93.2,88.6,56.8,50,86.4,54.5,40.9,77.3,79.5,65.9,70.5,77.3,81.8,50,79.5,and 68.2.The mean score was 71.2 with a standard deviation of 15.5.If the Normal model is appropriate,what percent of the scores will be greater than 86.7?

A)15.87%
B)0.15%
C)10%
D)34%
E)2.28%

Accepted Answer

The answer of The test scores from a recent Mathematics...

Question 46

A town's average snowfall is 45 cm per year with a standard deviation of 9 cm.Using a Normal model,what values should border the middle 68% of the model?&#10;A)49.5 cm and 40.5 cm&#10;B)45 cm and 41.94 cm&#10;C)54 cm and 36 cm&#10;D)63 cm and 27 cm&#10;E)46.8 cm and 43.2 cm

Accepted Answer

The answer of A town's average snowfall is 45 cm...

Question 47

The test scores from a recent Mathematics test are as follows: 95.5,65.9,93.2,88.6,56.8,50,86.4,54.5,40.9,77.3,79.5,65.9,70.5,77.3,81.8,50,79.5,and 68.2.The mean score was 71.2 with a standard deviation of 15.5.If the Normal model is appropriate,what percent of the scores will be greater than 86.7?

A)15.87%
B)0.15%
C)10%
D)34%
E)2.28%

Accepted Answer

The answer of The test scores from a recent Mathematics...

Question 48

An English instructor gave a final exam and found a mean score of 70 points and a standard deviation of 5.1 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Then find the percent of scores above 85.3 points.

A)

<strong>An English instructor gave a final exam and found a mean score of 70 points and a standard deviation of 5.1 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Then find the percent of scores above 85.3 points.</strong> A) ; 0.3% B) ; 2.5% C) ; 16% D) ; 0.15% E) ; 5% <div style=padding-top: 35px>

; 0.3%
B)

; 2.5%
C)

; 16%
D)

; 0.15%
E)

; 5%

Accepted Answer

The answer of An English instructor gave a final exam...

Question 49

For a recent English exam,use the Normal model N(73,9.2)to find the percent of scores under 58.Round to the nearest tenth of a percent.&#10;A)94.8%&#10;B)4.2%&#10;C)95.8%&#10;D)1.63%&#10;E)5.2%

Accepted Answer

The answer of For a recent English exam,use the Normal...

Question 50

Assuming a Normal model applies,a town's average annual snowfall (in cm)is modeled by $$\mathrm { N } ( 43,2 )$$ Draw and label the Normal model.What percent of snowfall is between 39 cm and 45 cm?&#10;A)   ; 81.5%&#10;B)   ; 41%&#10;C)   ; 95%&#10;D)   ; 68%&#10;E)   ; 54.5%

Accepted Answer

The answer of Assuming a Normal model applies,a town's average...

Question 51

Assuming a Normal model applies,a town's average annual snowfall (in cm)is modeled by $$\mathrm { N } ( 45,7 )$$ Draw and label the Normal model.What percent of snowfall is between 52 cm and 59 cm?&#10;A)   ; 11%&#10;B)   ; 13.5%&#10;C)   ; 32%&#10;D)   ; 16%&#10;E)   ; 27%

Accepted Answer

The answer of Assuming a Normal model applies,a town's average...

Question 52

0 < z < 3.01&#10;A)99.87%&#10;B)43.67%&#10;C)49.87%&#10;D)12.17%&#10;E)50.13%

Accepted Answer

The answer of 0 < z < 3.01&#10;A)99.87%&#10;B)43.67%&#10;C)49.87%&#10;D)12.17%&#10;E)50.13%...

Question 53

Assuming a Normal model applies,a town's average annual snowfall (in cm)is modeled by $$\mathrm { N } ( 42,8 )$$ Draw and label the Normal model.Then find the interval for the middle 95% of snowfall.&#10;A)   ; 18 to 50 cm&#10;B)   ; 34 to 50 cm&#10;C)   ; 26 to 58 cm&#10;D)   ; 34 to 66 cm&#10;E)   ; 18 to 66 cm

Accepted Answer

The answer of Assuming a Normal model applies,a town's average...

Question 54

The volumes of soda in 1 litre cola bottle can be described by a Normal model with a mean of 0.95 L and a standard deviation of 0.04 L.What percentage of bottles can we expect to have a volume less than 0.94 L?&#10;A)40.13%&#10;B)9.87%&#10;C)47.15%&#10;D)38.21%&#10;E)59.87%

Accepted Answer

The answer of The volumes of soda in 1 litre...

Question 55

Assuming a Normal model applies,a town's average annual snowfall (in cm)is modeled by $$\mathrm { N } ( 50,6 )$$ Draw and label the Normal model.About what percent represents snowfall of less than 62 cm?&#10;A)   ; 0.15%&#10;B)   ; 2.5%&#10;C)   ; 84%&#10;D)   ; 99.85%&#10;E)   ; 97.5%

Accepted Answer

The answer of Assuming a Normal model applies,a town's average...

Question 56

For a recent English exam,use the Normal model N(73,9.2)to find the percent of scores over 85.Round to the nearest tenth of a percent.&#10;A)88.5%&#10;B)90.3%&#10;C)9.7%&#10;D)8.1%&#10;E)11.5%

Accepted Answer

The answer of For a recent English exam,use the Normal...

Question 57

A town's average snowfall is 40 cm per year with a standard deviation of 5 cm.According to the Normal model,what percent of snowfall is less than 3 standard deviations from the mean?&#10;A)15.87%&#10;B)99.74%&#10;C)5%&#10;D)2.28%&#10;E)0.26%

Accepted Answer

The answer of A town's average snowfall is 40 cm...

Question 58

An English instructor gave a final exam and found a mean score of 70 points and a standard deviation of 6.8 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Describe the scores of the top 2.5%.

A)

<strong>An English instructor gave a final exam and found a mean score of 70 points and a standard deviation of 6.8 points.Assume that a Normal model can be applied.Draw and label the Normal model for the exam scores.Describe the scores of the top 2.5%.</strong> A) ; Higher than 90.4 points B) ; Higher than 76.8 points C) ; 83.6 points D) ; Higher than 83.6 points E) ; Higher than 76.8 points <div style=padding-top: 35px>

; Higher than 90.4 points
B)

; Higher than 76.8 points
C)

; 83.6 points
D)

; Higher than 83.6 points
E)

; Higher than 76.8 points

Accepted Answer

The answer of An English instructor gave a final exam...

Question 59

z < 1.13&#10;A)89.07%&#10;B)87.08%&#10;C)12.92%&#10;D)84.85%&#10;E)88.09%

Accepted Answer

The answer of z < 1.13&#10;A)89.07%&#10;B)87.08%&#10;C)12.92%&#10;D)84.85%&#10;E)88.09%...

Question 60

A bank's loan officer rates applicants for credit.The ratings can be described by a Normal model with a mean of 200 and a standard deviation of 50.If an applicant is randomly selected,what percentage can be expected to be between 200 and 275?&#10;A)6.68%&#10;B)43.32%&#10;C)93.32%&#10;D)42.37%&#10;E)5.00%

Accepted Answer

The answer of A bank's loan officer rates applicants for...

Question 61

For a recent English exam,use the Normal model N(73,9.2)to find the score that represents the 60th percentile.&#10;A)75.3&#10;B)70.7&#10;C)82.2&#10;D)63.8&#10;E)43.8

Accepted Answer

The answer of For a recent English exam,use the Normal...

Question 62

the lowest 4%&#10;A)-1.48&#10;B)-1.89&#10;C)-1.75&#10;D)-1.63&#10;E)1.75

Accepted Answer

The answer of the lowest 4%&#10;A)-1.48&#10;B)-1.89&#10;C)-1.75&#10;D)-1.63&#10;E)1.75...

Question 63

For a recent English exam,use the Normal model N(73,9.2)to find the score that represents the 90th percentile.&#10;A)63.8&#10;B)61.2&#10;C)81.3&#10;D)84.8&#10;E)82.2