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?
A)1.80 standard deviations above the mean
B)1.80 standard deviations below the mean
C)0.69 standard deviations above the mean
D)0.80 standard deviations above the mean
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?
A)About 1.33 standard deviations above the mean
B)About 1.33 standard deviations below the mean
C)About 4.67 standard deviations above the mean
D)About 2.67 standard deviations above the mean
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?
A)1.75 standard deviations below the mean
B)1.36 standard deviations below the mean
C)3.50 standard deviations below the mean
D)3.50 standard deviations above the mean
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?
A)0.83 standard deviations above the mean
B)1.83 standard deviations below the mean
C)2.00 standard deviations above the mean
D)2.00 standard deviations below the mean
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.
A)4 kg
B)1 kg
C)0.1 kg
D)5 kg
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 both shoes is 51 km,which shoe would you purchase before starting your marathon training (where you figure to run 500 km)? Explain.
A)Swift.Swift shoes are - $\frac { 50 } { 17 }$ standard deviations from the mean while Enduramax shoes are - $\frac { 100 } { 51 }$ standard deviations from the mean.
B)Enduramax.Enduramax shoes are - $\frac { 50 } { 17 }$ standard deviations from the mean while Swift shoes are - $\frac { 100 } { 51 }$ standard deviations from the mean.
C)Swift.I only need to run 500 km and the Swift will last 100 km beyond that.
D)Enduramax.Enduramax shoes are - $\frac { 100 } { 51 }$ standard deviations from the mean while Swift shoes are - $\frac { 50 } { 17 }$ standard deviations from the mean.
E)Swift.Swift shoes are - $\frac { 100 } { 51 }$ standard deviations from the mean while Enduramax shoes are - $\frac { 50 } { 17 }$ 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?
A)3.50 standard deviations below the mean
B)1.29 standard deviations above the mean
C)1.29 standard deviations below the mean
D)1.41 standard deviations above the mean
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?
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

Accepted Answer

To find how many standard deviations from the mean $195 is, we use the formula: (X - μ) / σ, where X is the value ($195), μ is the mean ($319), and σ is the standard deviation ($56). So, (195 - 319) / 56 = -124 / 56 = -2.21. This means $195 is about 2.21 standard deviations below the mean.

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?
A)About 3.97 standard deviations below the mean
B)About 1.35 standard deviations above the mean
C)About 3.97 standard deviations above the mean
D)About 1.99 standard deviations below the mean
E)About 1.99 standard deviations above the mean

Accepted Answer

To find how many standard deviations from the mean a value is, you subtract the mean from the value and divide by the standard deviation: (245 - 968) / 182 = -723 / 182 ≈ -3.97. This means the fire consuming 245 acres is about 3.97 standard deviations below the mean.

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 choice is based on calculating how many standard deviations 500 km (the expected running distance) is from the mean life of each shoe. For Swift, $ \frac{600 - 500}{54} = \frac{100}{54} = \frac{50}{27} $ standard deviations from the mean. For Enduramax, $ \frac{650 - 500}{124} = \frac{150}{124} = \frac{75}{62} $ standard deviations from the mean. Since Swift shoes are closer to their mean life when running 500 km, they are a better choice for this specific training distance.

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 z-score formula is $Z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Plugging in the values, $-2 = \frac{(8 - 14)}{\sigma}$, solving for $\sigma$ gives $\sigma = 3$ years.

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

To find how many standard deviations from the mean 2 hours is, use the formula: (X - μ) / σ, where X is 2 hours, μ (mean) is 5 hours, and σ (standard deviation) is 1.25 hours. This calculation gives (2 - 5) / 1.25 = -3 / 1.25 = -2.4, indicating 2.40 standard deviations below the mean.

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?
A)0.44 standard deviations below the mean
B)2.00 standard deviations above the mean
C)2.00 standard deviations below the mean
D)1.60 standard deviations above the mean
E)0.44 standard deviations above the mean

Accepted Answer

To find how many standard deviations from the mean a value is, you subtract the mean from the value and divide by the standard deviation: (64 cm - 40 cm) / 12 cm = 24 cm / 12 cm = 2.00 standard deviations above the mean.

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?
A)3.25 standard deviations below the mean
B)6.50 standard deviations above the mean
C)3.25 standard deviations above the mean
D)6.50 standard deviations below the mean
E)1.78 standard deviations above the mean

Accepted Answer

The z-score for a year with 387 babies born is (387 - 218) / 26 = 6.50. This means that a year with 387 babies born is 6.50 standard deviations above the mean.

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.
A)5%
B)15%
C)3%
D)1%
E)10%

Accepted Answer

The z-score is calculated as (x - mu) / sigma, where x is the value of the data point, mu is the mean, and sigma is the standard deviation. We know that the player's z-score is 1.5, and that he makes 72.5% of his free throws. The mean is 65%, so we can plug these values into the formula to get:1.5 = (72.5 - 65) / sigma1.5 = 7.5 / sigmasigma = 7.5 / 1.5sigma = 5%

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

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?

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?

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?

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?

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.

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.

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.

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?

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 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?

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.

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.

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 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?

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?

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.