Question 1

the highest 9%

Accepted Answer

A) -1.34 
B) 1.39 
C) 1.26 
D) 1.34 
E) 1.45 
A) -1.34 
B) 1.39 
C) 1.26 
D) 1.34 
E) 1.45

Question 2

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.

Accepted Answer

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. 
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.

Question 3

the lowest 4%

Accepted Answer

A) -1.48 
B) -1.89 
C) -1.75 
D) -1.63 
E) 1.75 
A) -1.48 
B) -1.89 
C) -1.75 
D) -1.63 
E) 1.75

Question 4

0 < z < 3.01

Accepted Answer

A) 99.87% 
B) 43.67% 
C) 49.87% 
D) 12.17% 
E) 50.13% 
A) 99.87% 
B) 43.67% 
C) 49.87% 
D) 12.17% 
E) 50.13%

Question 5

After increased patrol,33% of vehicles on a road travel below 45 km/h with a standard deviation of 5.1 km/h.Assuming a Normal model is appropriate,find the mean speed.

Accepted Answer

A) 1.68 km/h 
B) 47.24 km/h 
C) -42.76 km/h 
D) 50.1 km/h 
E) 14..85 km/h 
A) 1.68 km/h 
B) 47.24 km/h 
C) -42.76 km/h 
D) 50.1 km/h 
E) 14..85 km/h

Question 6

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.

Accepted Answer

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 
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

Question 7

σ = 10,68% below 200; μ = ?

Accepted Answer

A) 19.53 
B) 136 
C) 0.68 
D) 195.3 
E) 0.47 
A) 19.53 
B) 136 
C) 0.68 
D) 195.3 
E) 0.47

Question 8

-2.41 < z < 0

Accepted Answer

A) 49.10% 
B) 9.48% 
C) 52.16% 
D) 50.80% 
E) 49.20% 
A) 49.10% 
B) 9.48% 
C) 52.16% 
D) 50.80% 
E) 49.20%

Question 9

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.

Accepted Answer

A) 20 loads 
B) 10 loads 
C) 1 load 
D) 2 loads 
E) 3 loads 
A) 20 loads 
B) 10 loads 
C) 1 load 
D) 2 loads 
E) 3 loads

Question 10

For a recent English exam,use the Normal model N(73,9.2)to find the score that represents the 90th percentile.

Accepted Answer

A) 63.8 
B) 61.2 
C) 81.3 
D) 84.8 
E) 82.2 
A) 63.8 
B) 61.2 
C) 81.3 
D) 84.8 
E) 82.2

Question 11

For a recent English exam,use the Normal model N(73,9.2)to find the score that represents the 60th percentile.

Accepted Answer

A) 75.3 
B) 70.7 
C) 82.2 
D) 63.8 
E) 43.8 
A) 75.3 
B) 70.7 
C) 82.2 
D) 63.8 
E) 43.8

Question 12

μ = 60.0,4.01% below 53; σ = ?

Accepted Answer

A) 4.0 
B) 3.87 
C) 1.58 
D) 2.0 
E) 8.0 
A) 4.0 
B) 3.87 
C) 1.58 
D) 2.0 
E) 8.0

Question 13

After increased patrol,cars on a road travel at speeds averaging 55 km/h.If 93% of vehicles travel below 68 km/h,what approximate standard deviation could represent this model (assuming a Normal model is appropriate)?

Accepted Answer

A) 30.41 
B) 8.78 
C) 63.24 
D) 51.15 
E) -8.78 
A) 30.41 
B) 8.78 
C) 63.24 
D) 51.15 
E) -8.78

Question 14

On a recent English exam,if 94% of scores fell above 65 points and the standard deviation is 7.5,find the mean score (assuming the Normal model is appropriate).

Accepted Answer

A) 7.05 
B) -53.34 
C) 76.66 
D) 61.75 
E) 72.5 
A) 7.05 
B) -53.34 
C) 76.66 
D) 61.75 
E) 72.5

Question 15

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.

Accepted Answer

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). 
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).

Question 16

Mario's poker winnings average $319 per week with a standard deviation of $56.How many standard deviations from the mean is winning $195?

Accepted Answer

A) About 1.11 standard deviations below the mean 
B) About 2.21 standard deviations above the mean 
C) About 1.64 standard deviations below the mean 
D) About 2.21 standard deviations below the mean 
E) About 1.11 standard deviations above the mean 
A) About 1.11 standard deviations below the mean 
B) About 2.21 standard deviations above the mean 
C) About 1.64 standard deviations below the mean 
D) About 2.21 standard deviations below the mean 
E) About 1.11 standard deviations above the mean

Question 17

the lowest 9%

Accepted Answer

A) -1.39 
B) -1.34 
C) -1.45 
D) 1.34 
E) -1.26 
A) -1.39 
B) -1.34 
C) -1.45 
D) 1.34 
E) -1.26

Question 18

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?

Accepted Answer

A)    ; 81.5% 
B)    ; 41% 
C)    ; 95% 
D)    ; 68% 
E)    ; 54.5% 
A)    ; 81.5% 
B)    ; 41% 
C)    ; 95% 
D)    ; 68% 
E)    ; 54.5%

Question 19

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.

Accepted Answer

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 
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

Question 20

the lowest 40%

Accepted Answer

A) 0.50 
B) 0.57 
C) -0.57 
D) 0.25 
E) -0.25 
A) 0.50 
B) 0.57 
C) -0.57 
D) 0.25 
E) -0.25

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

the highest 9%

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.

the lowest 4%

0 < z < 3.01

After increased patrol,33% of vehicles on a road travel below 45 km/h with a standard deviation of 5.1 km/h.Assuming a Normal model is appropriate,find the mean speed.

σ = 10,68% below 200; μ = ?

-2.41 < z < 0

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.

For a recent English exam,use the Normal model N(73,9.2)to find the score that represents the 90th percentile.

For a recent English exam,use the Normal model N(73,9.2)to find the score that represents the 60th percentile.

μ = 60.0,4.01% below 53; σ = ?

After increased patrol,cars on a road travel at speeds averaging 55 km/h.If 93% of vehicles travel below 68 km/h,what approximate standard deviation could represent this model (assuming a Normal model is appropriate)?

On a recent English exam,if 94% of scores fell above 65 points and the standard deviation is 7.5,find the mean score (assuming the Normal model is appropriate).

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.

Mario's poker winnings average $319 per week with a standard deviation of $56.How many standard deviations from the mean is winning $195?

the lowest 9%

Assuming a Normal model applies,a town's average annual snowfall (in cm)is modeled by N(43,2)\mathrm { N } ( 43,2 )N(43,2) Draw and label the Normal model.What percent of snowfall is between 39 cm and 45 cm?

the lowest 40%

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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?