Question 1

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

Accepted Answer

The z-score formula is Z = (X - μ) / σ, where Z is the z-score, X is the value, μ is the mean, and σ is the standard deviation. Plugging in the given values: 1.5 = (67.5 - 54) / σ. Solving for σ gives σ = 9 cm.

Question 2

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

Accepted Answer

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

Question 3

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

To find the new mean speed, we need to subtract 7% from the original mean speed.

New Mean Speed = 57 km/h - 0.07(57 km/h) = 53.01 km/h

To find the new standard deviation, we can use the formula:

New SD = Old SD * (1 - 0.07) = 6 km/h * 0.93 = 5.58 km/h

Therefore, the correct choice is A.

Question 4

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

When all data points increase by a percentage, the mean increases by the same percentage, but the standard deviation also increases by that percentage. The new mean is $47 \times 1.2 = 56.4$ cm, and the new standard deviation is $10 \times 1.2 = 12$ cm.

Question 5

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

Accepted Answer

The z-score formula is Z = (X - μ) / σ, where Z is the z-score, X is the value, μ is the mean, and σ is the standard deviation. Plugging in the given values: 1 = (3.15 - 2.7) / σ. Solving for σ gives σ = 0.45 kg.

Question 6

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

Since this year's percentages are 3% higher, we need to add 3% to the mean and the quartiles of last year's percentages to get this year's mean and quartiles. 
Therefore, the median of last year's percentages would still be 53%. 
Next, we can use the formula to find the new standard deviation of the percentages: 
New SD = Old SD * (1 + Percent Increase as a Decimal) 
New SD = 8 * (1 + 0.03) 
New SD = 8.24 
Therefore, last year's tournament mean, median, and standard deviation of the percentages are: 
Mean: 53% + 3% = 56% 
Median: 53% 
Standard Deviation: 7.8% (rounded to the nearest tenth) 
Hence, the best choice is D.

Question 7

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 opponents' mean score is 10% lower than the team's mean score, so $58 - 0.10 \times 58 = 52.2$ points. The standard deviation, being a measure of dispersion that does not depend on the mean, remains unchanged when all values are scaled by the same factor, so it remains $9$ points. However, because we're applying a percentage decrease to the mean, not the standard deviation, the standard deviation for the opponents' scores actually decreases by the same percentage, making it $9 - 0.10 \times 9 = 8.1$ points.

Question 8

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 minimum value cannot be greater than the original minimum, so options A and E are eliminated. 
- The mean will increase by the amount of the raise, which is $50,000 for every employee. Since there are no other changes in the data, the mean will increase from $100,000 to $150,000. Option B has the correct mean, but the minimum value is too high. 
- The median is not affected by the raise since it is simply the middle value of the data set. Therefore, option C is incorrect since the median should remain at $100,000. 
- The range and IQR will remain the same since they are based on the minimum and maximum values and the quartiles, which are not affected by the raise. Therefore, the only correct option is D.

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

Accepted Answer

We can use the formula for z-score:
z = (x - μ) / σ

where
z = -2.3
x = 62 loads
μ = 85 loads (mean)
σ = standard deviation

Solving for σ, we get:

σ = (x - μ) / z
σ = (62 - 85) / (-2.3)
σ ≈ 10

Therefore, the standard deviation is approximately 10 loads per month. So the best choice is B.

Question 10

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

Accepted Answer

The z-score formula is $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value from the dataset, $\mu$ is the mean, and $\sigma$ is the standard deviation. Plugging in the given values: $-16.25 = \frac{(320,000 - 450,000)}{\sigma}$. Solving for $\sigma$ gives $\sigma = 8,000$ people.

Question 11

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.
A)40 km/h
B)10 km/h
C)30 km/h
D)0.5 km/h
E)1 km/h

Accepted Answer

We know that the average speed of cars on the road is 77 km/h. We also know that a car traveling 60 km/h has a z-score of -1.7.

Using the formula for z-score, we can calculate the difference between the car's speed and the average speed of cars on the road in terms of standard deviations:

z = (x - μ) / σ

where x is the car's speed, μ is the average speed of cars on the road, and σ is the standard deviation.

Substituting the given values, we get:

-1.7 = (60 - 77) / σ

Solving for σ, we get:

σ = (60 - 77) / -1.7

σ ≈ 10

Therefore, the standard deviation of the speeds of cars on the road is approximately 10 km/h. The best choice is B) 10 km/h.

Question 12

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

Adding a constant to every score in a data set increases measures of central tendency (mean, median, Q1) and location (lowest score) by that constant but does not change measures of spread (range, IQR, standard deviation).

Question 13

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 14

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

To find the cost, we need to multiply the size of each fire (11 acres in this case) with the cost per acre ($1200). Thus, the smallest fire would cost $13,200 to fight, and all the other fires would cost the same. Therefore, we can simply find the minimum, median, standard deviation, and IQR of a data set consisting of six values, each equal to $13,200. The minimum would be $13,200, the median would be $79,200, the standard deviation would be $0, and the IQR would be $0. Since choice C is the only one that has these values, it is the correct answer.

Question 15

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 $144 \mathrm {~mm} \mathrm {~Hg? }$
A)  ; 95%
B)  ; 99.7%
C)  ; 68%
D)  ; 34%
E)  ; 84%

Accepted Answer

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

Question 16

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 17

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)  ; $31 to $79
B)  ; $39 to $71
C)  ; $47 to $71
D)  ; $47 to $63
E)  ; $39 to $63

Accepted Answer

To draw the Normal model, we use the mean of $55 as the center and the standard deviation of $8 to determine how spread out the distribution is. We can label the model as follows:

Mean = $55
Standard deviation = $8

Using the Empirical Rule, we know that approximately 68% of bills will be within 1 standard deviation of the mean. Therefore, we can estimate the interval for the central 68% as follows:

Lower bound = $55 - $8 = $47
Upper bound = $55 + $8 = $63

So, the best choice is D, which corresponds to $47 to $63 as the interval for the central 68% of bills.

Question 18

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

Since the slowest time this year was 11 minutes, we know that the slowest time last year would have been 8% better, or 10.168 minutes (11 - 0.88). We can use this to find the mean time for last year's race:

mean = (10.168 + 8.68 + 8.68 + 7.28 + 7.28 + 6.32)/6 = 7.69 minutes

Next, we need to find the standard deviation for last year's race. Since the race results were 8% better last year, we can use the formula for scaling standard deviation:

new SD = old SD / scaling factor

scaling factor = 1 - percent increase as a decimal = 1 - 0.08 = 0.92

So, the new SD is:

SD = 0.88 / 0.92 = 0.9565 minutes

Rounding to two decimal places, we get:

Mean: 7.69 minutes
SD: 0.96 minutes

Therefore, the best choice is (D) with a mean of 72.68 minutes and standard deviation of 10.12 minutes.

Question 19

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.
A)$10,000
B)$500
C)$15,000
D)$1000
E)$4000

Accepted Answer

The z-score formula is (X - μ) / σ = z. Here, X = $18,900, μ = $23,700, and z = -1.2. Solving for σ gives σ = ($23,700 - $18,900) / -1.2 = $4,000.

Question 20

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

Accepted Answer

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

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

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.

On an exam Bob scored 75% and was 1.5 standard deviations above the mean of 63%.What was the standard deviation?

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.

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.

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 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.

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.

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 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.

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