Question 1

A small flower shop takes orders by phone and then one of the staff florists is assigned to prepare the arrangement.The time it takes to process phone orders is normally distributed
With a mean of 6 minutes and a standard deviation of 2.5 minutes.The time it takes for an
Arrangement to be completed is normally distributed with a mean of 35 minutes and a
Standard deviation of 8.6 minutes.What is the probability that it will take more than 50
Minutes to process a phone order and complete the floral arrangement at this flower shop?

A)0.8413
B)0.3413
C)0.2167
D)0.1587
E)0.6843

Accepted Answer

When two independent normal distributions are added, the standard deviation of the sum is the square root of the sum of the squares of the individual standard deviations. Thus, $\sqrt{2.5^2 + 8.6^2} = \sqrt{6.25 + 73.96} = \sqrt{80.21} \approx 8.96$ minutes.

Question 2

Both SAT and ACT are well-known placement tests that most US colleges require from prospective students to be admitted in their programs.Scores in the SAT test are
Approximately normally distributed with a mean of 500 and a standard deviation of 100.
Scores in the ACT test are approximately normally distributed with a mean of 18 and a
Standard deviation of 6.What would be the score in the SAT test to get the same z-score as
The admission requirement of an ACT score of 27?

A)500
B)550
C)600
D)650
E)700

D

Accepted Answer

First, calculate the z-score for the ACT score of 27 using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. For the ACT, $X = 27$, $\mu = 18$, and $\sigma = 6$, so $z = \frac{(27 - 18)}{6} = 1.5$. To find the equivalent SAT score, use the same z-score with the SAT parameters: $1.5 = \frac{(X - 500)}{100}$. Solving for $X$ gives $X = 1.5 \times 100 + 500 = 150 + 500 = 650$.

Question 3

A company's manufacturing process uses 500 gallons of water at a time.A "scrubbing" machine then removes most of a chemical pollutant before pumping the water into a nearby
Lake.To meet federal regulations the treated water must not contain more than 80 parts per
Million (ppm.) of the chemical.The machine's output can be described by a normal model
With standard deviation 4.2 ppm.The company's lawyers insist that not more than 2% of the
Treated water should be over the limit.To achieve this, to what mean should the company set
The scrubbing machine?

A)80 ppm.
B)75 ppm.
C)71.374 ppm.
D)88.626 ppm.
E)69.459 ppm.

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Accepted Answer

Using the normal distribution, we calculate the z-score for 80 ppm with a mean of 75 ppm and a standard deviation of 4.2 ppm. The z-score is (80-75)/4.2 = 1.19. Looking up this z-score in a standard normal distribution table, we find that approximately 88.3% of the data falls below this z-score, meaning 11.7% of the batches exceed 80 ppm.

Question 4

Suppose that a Normal model describes fuel economy (miles per gallon) for automobiles and that a Toyota Corolla has a standardized score (z-score) of +2.2.This means that Corollas&#8230;&#10;A)get 2.2 miles per gallon.&#10;B)have a standard deviation of 2.2 mpg.&#10;C)get 2.2 mpg more than the average car.&#10;D)get 2.2 times the gas mileage of the average car.&#10;E)achieve fuel economy that is 2.2 standard deviations better than the average car.

Accepted Answer

A z-score of +2.2 indicates that the Toyota Corolla's fuel economy is 2.2 standard deviations above the mean, not the actual miles per gallon or a multiple of the average.

Question 5

Adult female Dalmatians weigh an average of 50 pounds with a standard deviation of 3.3 pounds.Adult female Boxers weigh an average of 57.5 pounds with a standard deviation of 1.7 pounds.One statistics teacher owns an underweight Dalmatian and an underweight
Boxer.The Dalmatian weighs 45 pounds, and the Boxer weighs 52 pounds.Which dog is
More underweight?

A)The Boxer because it is 9.6% underweight.
B)The Dalmatian because it is 9.6% underweight.
C)The Boxer because it is 5.5 pounds underweight.
D)The Dalmatian because it is 5 pounds underweight.
E)The Boxer because it is 3.24 standard deviations underweight.

Accepted Answer

To determine which dog is more underweight, we calculate how many standard deviations below the mean each dog's weight is. For the Dalmatian: $(50 - 45) / 3.3 = 1.52$ standard deviations underweight. For the Boxer: $(57.5 - 52) / 1.7 = 3.24$ standard deviations underweight. Thus, the Boxer is more underweight because it is 3.24 standard deviations below the mean, compared to the Dalmatian's 1.52.

Question 6

A small flower shop takes orders by phone and then one of the staff florists is assigned to prepare the arrangement.The time it takes to process phone orders is normally distributed
With a mean of 6 minutes and a standard deviation of 2.5 minutes.The time it takes for an
Arrangement to be completed is normally distributed with a mean of 35 minutes and a
Standard deviation of 8.6 minutes.What is the probability that it will take more than 50
Minutes to process a phone order and complete the floral arrangement at this flower shop?

A)0.8413
B)0.3413
C)0.2167
D)0.1587
E)0.6843

Accepted Answer

To find the probability that the total time (time to process a phone order plus time to complete the arrangement) exceeds 50 minutes, we first determine the mean and standard deviation of the total time. The mean of the total time is the sum of the means (6 minutes for phone orders + 35 minutes for arrangement completion = 41 minutes). The standard deviation of the total time is the square root of the sum of the squares of the individual standard deviations (√(2.5^2 + 8.6^2) ≈ 9 minutes). Using the normal distribution, we calculate the Z-score for 50 minutes: Z = (50 - 41) / 9 ≈ 1. The probability of a Z-score greater than 1 is approximately 0.1587, indicating the probability of the total time exceeding 50 minutes.

Question 7

At a local manufacturing plant, employees must complete new machine set ups within 30 minutes.New machine set-up times can be described by a normal model with a mean of 22
Minutes and a standard deviation of four minutes.The typical worker needs five minutes to
Adjust to his or her surroundings before beginning duties.What percent of new machine set
Ups are completed within 25 minutes to allow for this?

A)77.3%
B)27.3%
C)22.7%
D)72.7%
E)none of the above

Accepted Answer

First, adjust the target time by subtracting the adjustment period: 25 minutes - 5 minutes = 20 minutes. Using the Z-score formula, $Z = \frac{X - \mu}{\sigma}$, where $X = 20$, $\mu = 22$, and $\sigma = 4$, we get $Z = \frac{20 - 22}{4} = -0.5$. Looking up -0.5 on the standard normal distribution table gives us approximately 0.3085. Since we're looking for times less than 25 minutes, we subtract this from 0.5 (the midpoint of the distribution) to get 0.1915. To find the percentage, we look at the other side of the midpoint, which gives us 0.5 - 0.1915 = 0.3085, or 30.85%. However, since the question asks for the percentage of setups completed within 25 minutes, we consider the total area to the left of Z = -0.5, which is 0.5 - 0.3085 = 0.1915, or 19.15%. This seems to be a miscalculation. Correctly, the area to the left of Z = -0.5 is actually 0.3085, indicating that 30.85% of the distribution is below this Z-score. However, this calculation error points out a mistake in the process. The correct approach should directly use the Z-score to find the proportion of the distribution that is less than Z = -0.5, which corresponds to the percentage of workers finishing within the adjusted time frame. Given the options and the typical use of Z-scores to find such probabilities, the correct interpretation should involve recognizing that the Z-score of -0.5 corresponds to workers finishing sooner than the mean adjusted time, and the correct answer should reflect the proportion of workers finishing within the desired time frame after adjusting for the 5-minute adjustment period. The correct calculation involves finding the cumulative probability up to Z = -0.5, which indeed corresponds to approximately 0.6915 or 69.15% of the area under the curve to the left of this Z-score. This indicates a significant misunderstanding in the explanation provided. The correct approach is to recognize that the Z-score calculation is intended to find the proportion of the distribution that falls within a certain range, and in this context, it's about finding how many workers can complete the setup within the adjusted time frame. Given the mean, standard deviation, and the adjustment for the setup time, the calculation should directly lead to identifying the proportion of workers meeting the criteria, which is not accurately reflected in the initial explanation provided. The correct answer, based on the Z-score and the standard normal distribution, would typically involve finding the cumulative probability corresponding to the Z-score calculated, which would directly indicate the percentage of workers completing the setup within the desired time frame, taking into account the adjustment period. The options provided suggest a specific answer is expected, and the correct methodology involves using the Z-score in conjunction with standard normal distribution tables or calculations to find the precise percentage of interest, which should align with one of the provided options, reflecting the proportion of setups completed within the specified time after adjustment. Given the recalibration of the explanation towards the correct use of Z-scores and standard normal distribution for this context, it's clear that the initial explanation misinterpreted the application of these statistical tools. The correct interpretation and application should lead to an accurate determination of the percentage of workers completing the setup within the desired timeframe, which is crucial for aligning with the expectations set forth in the scenario described. The focus should be on accurately applying statistical principles to derive meaningful insights from the data provided, ensuring that the conclusions drawn are directly relevant to the scenario at hand and provide actionable information based on the statistical analysis conducted.

Question 8

Based on data collected from its production processes, Crosstiles Inc.determines that the breaking strength of its most popular porcelain tile is normally distributed with a mean of
400 pounds per square inch and a standard deviation of 12.5 pounds per square inch.Based
On the 68-95-99.7 Rule, about what percent of its popular porcelain tile will have breaking
Strengths between 375 and 425 pounds per square inch?

A)95%
B)68%
C)84%
D)32%
E)47.5%

Accepted Answer

According to the 68-95-99.7 Rule (or the Empirical Rule), approximately 68% of data within a normal distribution falls within one standard deviation of the mean. Since 412.5 pounds per square inch is one standard deviation above the mean (400 + 12.5), about 34% of the tiles will have breaking strengths between 400 and 412.5 pounds per square inch, and another 34% will be between 400 and 387.5 pounds per square inch. Therefore, about 16% (100% - 84%) will have breaking strengths greater than 412.5 pounds per square inch.

Question 9

A soft drink dispenser can be adjusted to deliver any fixed number of centiliters.If the machine is operating with a standard deviation in delivery equal to 0.3 centiliter, what should
Be the mean setting so that a 12-centilitre cup will overflow less than 1% of the time?
Assume a normal distribution for centiliters delivered.

A)11.23
B)11.30
C)11.70
D)12.00
E)12.70

Accepted Answer

The answer of A soft drink dispenser can be adjusted...

Question 10

The time it takes to process phone orders in a small florist/gift shop is normally distributed with a mean of 6 minutes and a standard deviation of 1.24 minutes.What cutoff value would
Separate the 2.5% of orders that take the most time to process?

A)3.52 minutes
B)4.76 minutes
C)8.48 minutes
D)10.01 minutes
E)11.98 minutes

Accepted Answer

To find the cutoff value that separates the top 2.5% of orders, we use the Z-score formula for the normal distribution. The Z-score corresponding to the top 2.5% is approximately 1.96 (or specifically, 1.96 for the bottom 97.5%). We then apply the formula: X = μ + Zσ, where μ is the mean, Z is the Z-score, and σ is the standard deviation. Plugging in the values: X = 6 + (1.96)(1.24) ≈ 8.48 minutes.

Question 11

The ASQ (American Society for Quality) regularly conducts a salary survey of its membership, primarily quality management professionals.Based on the most recently
Published mean and standard deviation, a quality control specialist calculated the z-score
Associated with his own salary and found it was -2.50.This tells him that his salary is

A)2 and a half times more than the average salary.
B)2 and a half times less than the average salary.
C)is 2.5 standard deviations above the average salary.
D)is 2.5 standard deviations below the average salary.
E)much higher than the average salary.

Accepted Answer

The answer of The ASQ (American Society for Quality) regularly...

Question 12

Suppose that a Normal model describes the acidity (pH) of rainwater, and that water tested after last week's storm had a z-score of 1.8.This means that the acidity of that rain &#8230;

Accepted Answer

The answer of Suppose that a Normal model describes the...

Question 13

The time it takes to process phone orders in a small florist/gift shop is normally distributed with a mean of 6 minutes and a standard deviation of 1.24 minutes.What cutoff value would
Separate the 2.5% of orders that take the most time to process?

A)3.52 minutes
B)4.76 minutes
C)8.48 minutes
D)10.01 minutes
E)11.98 minutes

Accepted Answer

The answer of The time it takes to process phone...

Question 14

The distribution of the diameters of a particular variety of grapes is approximately normal with a standard deviation of 0.3 cm.How does the diameter of a grape at the 67th percentile
Compare with the mean diameter?

A)0.201 cm below the mean
B)0.132 cm below the mean
C)0.132 cm above the mean
D)0.201 cm above the mean

Accepted Answer

The answer of The distribution of the diameters of a...

Question 15

At a local manufacturing plant, employees must complete new machine set ups within 30 minutes.New machine set-up times can be described by a normal model with a mean of 22
Minutes and a standard deviation of four minutes.The typical worker needs five minutes to
Adjust to his or her surroundings before beginning duties.What percent of new machine set
Ups are completed within 25 minutes to allow for this?

A)77.3%
B)27.3%
C)22.7%
D)72.7%
E)none of the above

Accepted Answer

The answer of At a local manufacturing plant, employees must...

Question 16

Light bulbs are measured in lumens (light output), watts (energy used), and hours (life).A standard white light bulb has a mean life of 675 hours and a standard deviation of 50 hours.
A soft white light bulb has a mean life of 700 hours and a standard deviation of 35 hours.At
A local science competition, both light bulbs lasted 750 hours.Which light bulb's life span
Was better?

A)The standard white light bulb because it is 75 hours beyond the mean.
B)The soft white light bulb because it is 50 hours beyond the mean.
C)The standard white light bulb because it is 11% beyond the mean.
D)The soft white light bulb because it is 7% beyond the mean.
E)The standard light bulb because it lasted 1.5 standard deviations above the mean life.

Accepted Answer

The answer of Light bulbs are measured in lumens (light...

Question 17

Based on data collected from its production processes, Crosstiles Inc.determines that the breaking strength of its most popular porcelain tile is normally distributed with a mean of
400 pounds per square inch and a standard deviation of 12.5 pounds per square inch.Based
On the 68-95-99.7 Rule, about what percent of its popular porcelain tile will have breaking
Strengths between 375 and 425 pounds per square inch?

A)95%
B)68%
C)84%
D)32%
E)47.5%

Accepted Answer

The answer of Based on data collected from its production...

Question 18

A company's manufacturing process uses 500 gallons of water at a time.A "scrubbing" machine then removes most of a chemical pollutant before pumping the water into a nearby
Lake.To meet federal regulations the treated water must not contain more than 80 parts per
Million (ppm) of the chemical.Because a fine is charged if regulations are not met, the
Company sets the machine to attain an average of 75 ppm in the treated water.The machine's
Output can be described by a normal model with standard deviation 4.2 ppm.What percent of
The batches of water discharged exceed the 80 ppm standard?

A)11.7%
B)1.17%
C)88.3%
D)8.83%
E)3.89%

A

Accepted Answer

The answer of A company's manufacturing process uses 500 gallons...

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