Question 1

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.

Accepted Answer

A) 0.05 kg 
B) 0.14 kg 
C) 1.35 kg 
D) 0.45 kg 
E) 2.7 kg 
A) 0.05 kg 
B) 0.14 kg 
C) 1.35 kg 
D) 0.45 kg 
E) 2.7 kg

Question 2

The annual snowfall in a town (in cm)can be represented by the Normal model N(41,5.3).What amount of snowfall would you expect to be unusually low for this town?

Accepted Answer

A) 25.1 cm 
B) 56.9 cm 
C) 38.35 cm 
D) 35.7 cm 
E) 46.3 cm 
A) 25.1 cm 
B) 56.9 cm 
C) 38.35 cm 
D) 35.7 cm 
E) 46.3 cm

Question 3

μ = 15.2,15.87% above 16.1; σ = ?

Accepted Answer

A) 1.35 
B) 0.83 
C) 1.25 
D) 0.90 
E) 1.80 
A) 1.35 
B) 0.83 
C) 1.25 
D) 0.90 
E) 1.80

Question 4

The average score on a mathematics test was 64%.A student who scored 80% had a z-score of 2.Find the standard deviation.

Accepted Answer

A) 16% 
B) 8% 
C) 3% 
D) 20% 
E) 1% 
A) 16% 
B) 8% 
C) 3% 
D) 20% 
E) 1%

Question 5

After increased patrol,cars on a road travel at speeds averaging 54 km/h.If 10% of vehicles travel above 70 km/h,what approximate standard deviation could represent this model (assuming a Normal model is appropriate)?

Accepted Answer

A) 35.94 
B) 12.50 
C) 7 
D) 5.4 
E) -12.50 
A) 35.94 
B) 12.50 
C) 7 
D) 5.4 
E) -12.50

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.

Accepted Answer

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. 
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.

Question 7

the middle 96%

Accepted Answer

A) 0 to 2.05 
B) -2.33 to 2.33 
C) -3.00 to 3.00 
D) -1.75 to 1.75 
E) -2.05 to 2.05 
A) 0 to 2.05 
B) -2.33 to 2.33 
C) -3.00 to 3.00 
D) -1.75 to 1.75 
E) -2.05 to 2.05

Question 8

the highest 86%

Accepted Answer

A) -1.08 
B) 0.8051 
C) 1.08 
D) 0.5557 
E) -1.02 
A) -1.08 
B) 0.8051 
C) 1.08 
D) 0.5557 
E) -1.02

Question 9

A Normal Probability plot is a

Accepted Answer

A) Histogram. 
B) Scatterplot. 
C) Boxplot. 
D) Stem-and-leaf plot. 
A) Histogram. 
B) Scatterplot. 
C) Boxplot. 
D) Stem-and-leaf plot.

Question 10

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.

Accepted Answer

A) 5 cm 
B) 30 cm 
C) 9 cm 
D) 20 cm 
E) 1 cm 
A) 5 cm 
B) 30 cm 
C) 9 cm 
D) 20 cm 
E) 1 cm

Question 11

z > 0.59

Accepted Answer

A) 25.47% 
B) 72.24% 
C) 22.24% 
D) 27.76% 
E) 21.90% 
A) 25.47% 
B) 72.24% 
C) 22.24% 
D) 27.76% 
E) 21.90%

Question 12

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.

Accepted Answer

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 
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

Question 13

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.

Accepted Answer

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 
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

Question 14

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%.

Accepted Answer

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 
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

Question 15

μ = 0.38,20% above 0.50; σ = ?

Accepted Answer

A) 1.43 
B) 0.20 
C) 0.143 
D) 0.1 
E) 0.84 
A) 1.43 
B) 0.20 
C) 0.143 
D) 0.1 
E) 0.84

Question 16

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.

Accepted Answer

A) 1000 people 
B) 17,000 people 
C) 3000 people 
D) 20,000 people 
E) 8000 people 
A) 1000 people 
B) 17,000 people 
C) 3000 people 
D) 20,000 people 
E) 8000 people

Question 17

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?

Accepted Answer

A) 40.13% 
B) 9.87% 
C) 47.15% 
D) 38.21% 
E) 59.87% 
A) 40.13% 
B) 9.87% 
C) 47.15% 
D) 38.21% 
E) 59.87%

Question 18

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.

Accepted Answer

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. 
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.

Question 19

z > -1.82

Accepted Answer

A) 3.44% 
B) -3.44% 
C) 96.56% 
D) 46.56% 
E) 92.57% 
A) 3.44% 
B) -3.44% 
C) 96.56% 
D) 46.56% 
E) 92.57%

Question 20

Based on the Normal model for snowfall in a certain town N(57,8),how many cm of snow would represent the 85th percentile?

Accepted Answer

A) 49 cm 
B) 48.5 cm 
C) 65 cm 
D) 65.3 cm 
E) 48.7 cm 
A) 49 cm 
B) 48.5 cm 
C) 65 cm 
D) 65.3 cm 
E) 48.7 cm

Exam 5: The Standard Deviation As a Ruler and the Normal Model

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.

The annual snowfall in a town (in cm)can be represented by the Normal model N(41,5.3).What amount of snowfall would you expect to be unusually low for this town?

μ = 15.2,15.87% above 16.1; σ = ?

The average score on a mathematics test was 64%.A student who scored 80% had a z-score of 2.Find the standard deviation.

After increased patrol,cars on a road travel at speeds averaging 54 km/h.If 10% of vehicles travel above 70 km/h,what approximate standard deviation could represent this model (assuming a Normal model is appropriate)?

the middle 96%

the highest 86%

A Normal Probability plot is a

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.

z > 0.59

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%.

μ = 0.38,20% above 0.50; σ = ?

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.

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?

z > -1.82

Based on the Normal model for snowfall in a certain town N(57,8),how many cm of snow would represent the 85th percentile?

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