Question 1

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 1,000 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally
Distributed. What proportion of bottles have volumes greater than 1,002 mL?

Accepted Answer

Using the Z-score formula, Z = (X - μ) / σ, where X is the value of interest (1,002 mL), μ is the mean (1,000 mL), and σ is the standard deviation (7 mL). The Z-score for 1,002 mL is (1002 - 1000) / 7 = 0.2857. Looking up this Z-score in a standard normal distribution table or using a calculator gives a right-tail probability of approximately 0.3875, indicating the proportion of bottles with volumes greater than 1,002 mL.

Question 2

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.8 and standard deviation 9.2. What proportion of women have blood pressures between 70.4 and
83.4?

Accepted Answer

The proportion of women with blood pressures between 70.4 and 83.4 is the area under the normal distribution curve between those two values. Using technology, we can find that area to be 0.4821.

Question 3

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 1,002 milliliters (mL) and standard deviation 4 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes less than 1,002 mL?

Accepted Answer

Given that the fill volumes are normally distributed with a mean of 1,002 mL, the proportion of bottles with volumes less than the mean (1,002 mL) is 0.5 or 50%. This is a property of the normal distribution, where the data is symmetrically distributed around the mean.

Question 4

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 9.0 pounds and standard deviation 2.5 pound(s). Find the 75th percentile of the weights.

Accepted Answer

The 75th percentile corresponds to a z-score of approximately 0.675. Using the formula for a normal distribution, X = μ + zσ, where μ = 9.0 and σ = 2.5, the calculation is 9.0 + 0.675(2.5) = 10.6875, which rounds to 10.67 lb.

Question 5

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.1 and standard deviation 9.5. What proportion of Women have blood pressures lower than 67.9?

Accepted Answer

The answer of A recent study reported that diastolic blood...

Question 6

A bottler of drinking water fills plastic bottles with a mean volume of 1,007 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes less than 1,015 mL?

Accepted Answer

To solve this problem, we need to standardize the volume using the formula:

z = (x - μ) / σ

where x is the value we want to find the proportion for (in this case, 1,015 mL), μ is the mean (1,007 mL), and σ is the standard deviation (6 mL).

z = (1,015 - 1,007) / 6 = 1.33

We then use a standard normal table (or calculator) to find the proportion of bottles with volumes less than 1,015 mL, which is the area to the left of z = 1.33.

Using a standard normal table or calculator, we find that the proportion is approximately 0.9082.

Therefore, the best choice is B, which gives the correct proportion of bottles with volumes less than 1,015 mL.

Question 7

A bottler of drinking water fills plastic bottles with a mean volume of 998 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes between 988 mL and 991 mL?

Accepted Answer

To solve the problem, we need to standardize the values first:

Z1 = (988 - 998) / 7 = -1.429
Z2 = (991 - 998) / 7 = -1.000

Then, we look up the corresponding probabilities from the standard normal distribution table:

P(Z < -1.429) = 0.0764
P(Z < -1.000) = 0.1587

Finally, we subtract the two probabilities to get the proportion of bottles with volumes between 988 mL and 991 mL:

P(-1.429 < Z < -1.000) = 0.1587 - 0.0764 = 0.0823

Therefore, the answer is C, 0.0823.

Question 8

Use technology to solve the following problem: A normal population has a mean $\mu = 28$ and standard deviation $\sigma = 6$ What proportion of the population is between 18 and 26?

Accepted Answer

To find the proportion of the population between 18 and 26, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$. For X = 18, $Z = \frac{18 - 28}{6} = -1.67$. For X = 26, $Z = \frac{26 - 28}{6} = -0.33$. Using a standard normal distribution table or calculator, the area between Z = -1.67 and Z = -0.33 is approximately 0.3217.

Question 9

A normal population has a mean $\mu = 30$ and standard deviation $\sigma = 9$ . What proportion of the population is between 23 and 32?

Accepted Answer

We need to find the proportion of the population that falls within the range of 23 to 32. First, we need to standardize the range using the formula:

z = (x - mu) / sigma

Plugging in the values, we get:

z1 = (23 - 30) / 9 = -0.7778
z2 = (32 - 30) / 9 = 0.2222

Next, we need to find the area under the standard normal curve between these two z-scores. Using a standard normal table or a calculator, we can find:

Area between -0.7778 and 0.2222 = 0.3694

Therefore, the proportion of the population between 23 and 32 is 0.3694 or approximately 0.37. The closest answer choice is C.

Question 10

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.9 and standard deviation 8.8. What proportion of women have blood pressures lower than 65.8?

Accepted Answer

We can use the normal distribution function on a calculator or statistical software to find the area to the left of the given value of 65.8. Using a standard normal distribution with a mean of 0 and standard deviation of 1 (by standardizing the original distribution), we find that the z-score corresponding to a diastolic blood pressure of 65.8 is z = (65.8 - 80.9) / 8.8 = -1.71. Looking up the area to the left of this value using a normal distribution table or a calculator, we find a probability of approximately 0.0431 or 4.31%. Therefore, about 4.31% of adult women in the United States have diastolic blood pressures lower than 65.8 mmHg.

Question 11

A bottler of drinking water fills plastic bottles with a mean volume of 1,006 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have
Volumes greater than 1,009 mL?

Accepted Answer

To find the proportion of bottles with volumes greater than 1,009 mL, we use the Z-score formula: Z = (X - μ) / σ, where X is the value in question (1,009 mL), μ is the mean (1,006 mL), and σ is the standard deviation (6 mL). Plugging in the values gives Z = (1009 - 1006) / 6 = 0.5. Looking up 0.5 on the standard normal distribution table gives a left-tail value of approximately 0.6915, meaning the right tail (greater than 1,009 mL) is 1 - 0.6915 = 0.3085.

Question 12

Use technology to solve the following problem: A normal population has a mean $\mu = 31$ and standard deviation $\sigma = 7$ What is the probability that a randomly chosen value will be greater than 18?

Accepted Answer

To find the probability that a randomly chosen value from a normal distribution is greater than 18, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X = 18$, $\mu = 31$, and $\sigma = 7$. Calculating the Z-score gives us $Z = \frac{18 - 31}{7} = -1.857$. Looking up this Z-score in a standard normal distribution table or using a calculator, we find the probability of being less than 18 is approximately 0.0316. Therefore, the probability of being greater than 18 is $1 - 0.0316 = 0.9684$.

Question 13

Use technology to solve the following problem: A normal population has a mean $\mu = 28$ and standard deviation $\sigma = 8$ What is the 58th percentile of the population?

Accepted Answer

To find the 58th percentile of a normal distribution with a mean ($\mu$) of 28 and a standard deviation ($\sigma$) of 8, you can use a Z-table or a statistical calculator. The 58th percentile corresponds to a Z-score that can be found by looking up the percentile in a Z-table or using an inverse normal function in a calculator. Using the formula for converting Z-scores to raw scores, $X = \mu + Z\sigma$, where $X$ is the raw score, $\mu$ is the mean, $Z$ is the Z-score, and $\sigma$ is the standard deviation. For the 58th percentile, the Z-score is approximately 0.20 (since 0.58 corresponds to a Z-score of about 0.20 in the standard normal distribution). Plugging in the values: $X = 28 + 0.20(8) = 29.6$.

Question 14

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.6 and standard deviation 9.9. What proportion of women have blood pressures higher than 85.8?

Accepted Answer

To find the proportion of women with blood pressures higher than 85.8, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest (85.8), $\mu$ is the mean (80.6), and $\sigma$ is the standard deviation (9.9). Calculating the Z-score gives us $Z = \frac{85.8 - 80.6}{9.9} = 0.5253$. Looking up this Z-score in a standard normal distribution table or using a calculator's normal distribution function, we find that the area to the left of this Z-score is approximately 0.7003, meaning the area to the right (the proportion of women with blood pressures higher than 85.8) is 1 - 0.7003 = 0.2997.

Question 15

A normal population has a mean $\mu = 14$ and standard deviation $\sigma = 2$ . What is the 43rd percentile of the population?

Accepted Answer

To find the 43rd percentile of a normal distribution, we use the Z-score formula and look up the corresponding Z-score for the 43rd percentile in a standard normal distribution table. The 43rd percentile corresponds to a Z-score of approximately -0.176. Using the formula $X = \mu + Z\sigma$, where $X$ is the value at the 43rd percentile, $\mu = 14$, $\sigma = 2$, and $Z = -0.176$, we calculate $X = 14 + (-0.176)(2) = 14 - 0.352 = 13.648$. Rounding to two decimal places, the closest answer is 13.64.

Question 16

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6. What proportion of Women have blood pressures between 88.1 and 89.4?

Accepted Answer

The first step is to standardize the values of interest using the formula:
z = (x - μ) / σ
where x is the value of interest, μ is the mean, and σ is the standard deviation.
For the lower limit of 88.1, z = (88.1 - 80.3) / 8.6 = 0.90698
For the upper limit of 89.4, z = (89.4 - 80.3) / 8.6 = 1.05714
Next, we look up the corresponding probabilities using a standard normal distribution table. The probability of z being between 0.90698 and 1.05714 is 0.0368. Therefore, approximately 3.68% of women have diastolic blood pressures between 88.1 and 89.4.

Question 17

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.9. What proportion of Women have blood pressures higher than 85.8?

Accepted Answer

We need to find the proportion of women with blood pressures higher than 85.8, which can be translated into finding the probability that a randomly selected woman has a diastolic blood pressure higher than 85.8. Using the normal distribution and standardizing:

z = (85.8 - 80.3) / 8.9 = 0.62

Using a standard normal distribution table or calculator, we find that the proportion of values exceeding a z-score of 0.62 is 0.2676. Therefore, about 26.76% of women have blood pressures higher than 85.8.

Question 18

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.5 pounds and standard deviation 1.4 pounds. Find the first quartile of the weight.

Accepted Answer

The first quartile (Q1) corresponds to the 25th percentile of a distribution. In a normal distribution, this can be found using the Z-score formula for the percentile in question. The Z-score for the 25th percentile is approximately -0.675. Using the formula $Q1 = \mu + Z(\sigma)$, where $\mu$ is the mean (8.5 pounds) and $\sigma$ is the standard deviation (1.4 pounds), we calculate $Q1 = 8.5 + (-0.675)(1.4) = 8.5 - 0.945 = 7.555$, which rounds to 7.56 pounds.

Question 19

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 990 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes between 985 mL and 988 mL?

Accepted Answer

The answer of Use technology to solve the following problem:...

Question 20

Use technology to solve the following problem: The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.7 pounds and standard deviation 2.0 pound(s). Find the 84th Percentile of the weights.

Accepted Answer

To find the 84th percentile of a normally distributed set of data, you can use the Z-score formula for percentiles, which is Z = (X - μ) / σ, where Z is the Z-score corresponding to the percentile, X is the value at that percentile, μ is the mean, and σ is the standard deviation. The Z-score for the 84th percentile is approximately 1.0 because the 84th percentile falls 1 standard deviation above the mean in a normal distribution. Using the given mean (μ = 8.7) and standard deviation (σ = 2.0), we solve for X: X = Zσ + μ = 1.0(2.0) + 8.7 = 10.7. Therefore, the closest value to 10.7 is 10.68 lb, which is option A.

Quiz 7: The Normal Distribution

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 1,000 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes greater than 1,002 mL?

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.8 and standard deviation 9.2. What proportion of women have blood pressures between 70.4 and 83.4?

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 1,002 milliliters (mL) and standard deviation 4 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes less than 1,002 mL?

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 9.0 pounds and standard deviation 2.5 pound(s) . Find the 75th percentile of the weights.

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.1 and standard deviation 9.5. What proportion of Women have blood pressures lower than 67.9?

A bottler of drinking water fills plastic bottles with a mean volume of 1,007 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes less than 1,015 mL?

A bottler of drinking water fills plastic bottles with a mean volume of 998 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes between 988 mL and 991 mL?

Use technology to solve the following problem: A normal population has a mean $\mu = 28$ and standard deviation $\sigma = 6$ What proportion of the population is between 18 and 26?

A normal population has a mean $\mu = 30$ and standard deviation $\sigma = 9$ . What proportion of the population is between 23 and 32?

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.9 and standard deviation 8.8. What proportion of women have blood pressures lower than 65.8?

A bottler of drinking water fills plastic bottles with a mean volume of 1,006 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes greater than 1,009 mL?

Use technology to solve the following problem: A normal population has a mean $\mu = 31$ and standard deviation $\sigma = 7$ What is the probability that a randomly chosen value will be greater than 18?

Use technology to solve the following problem: A normal population has a mean $\mu = 28$ and standard deviation $\sigma = 8$ What is the 58th percentile of the population?

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.6 and standard deviation 9.9. What proportion of women have blood pressures higher than 85.8?

A normal population has a mean $\mu = 14$ and standard deviation $\sigma = 2$ . What is the 43rd percentile of the population?

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6. What proportion of Women have blood pressures between 88.1 and 89.4?

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.9. What proportion of Women have blood pressures higher than 85.8?

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.5 pounds and standard deviation 1.4 pounds. Find the first quartile of the weight.

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 990 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes between 985 mL and 988 mL?

Use technology to solve the following problem: The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.7 pounds and standard deviation 2.0 pound(s) . Find the 84th Percentile of the weights.

Quiz 7: The Normal Distribution

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 1,000 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes greater than 1,002 mL?

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.8 and standard deviation 9.2. What proportion of women have blood pressures between 70.4 and 83.4?

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 1,002 milliliters (mL) and standard deviation 4 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes less than 1,002 mL?

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 9.0 pounds and standard deviation 2.5 pound(s) . Find the 75th percentile of the weights.

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.1 and standard deviation 9.5. What proportion of Women have blood pressures lower than 67.9?

A bottler of drinking water fills plastic bottles with a mean volume of 1,007 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes less than 1,015 mL?

A bottler of drinking water fills plastic bottles with a mean volume of 998 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes between 988 mL and 991 mL?

Use technology to solve the following problem: A normal population has a mean μ=28\mu = 28μ=28 and standard deviation σ=6\sigma = 6σ=6 What proportion of the population is between 18 and 26?

A normal population has a mean μ=30\mu = 30μ=30 and standard deviation σ=9\sigma = 9σ=9 . What proportion of the population is between 23 and 32?

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.9 and standard deviation 8.8. What proportion of women have blood pressures lower than 65.8?

A bottler of drinking water fills plastic bottles with a mean volume of 1,006 milliliters (mL) and standard deviation 6 mL. The fill volumes are normally distributed. What proportion of bottles have Volumes greater than 1,009 mL?

Use technology to solve the following problem: A normal population has a mean μ=31\mu = 31μ=31 and standard deviation σ=7\sigma = 7σ=7 What is the probability that a randomly chosen value will be greater than 18?

Use technology to solve the following problem: A normal population has a mean μ=28\mu = 28μ=28 and standard deviation σ=8\sigma = 8σ=8 What is the 58th percentile of the population?

Use technology to solve the following problem: A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.6 and standard deviation 9.9. What proportion of women have blood pressures higher than 85.8?

A normal population has a mean μ=14\mu = 14μ=14 and standard deviation σ=2\sigma = 2σ=2 . What is the 43rd percentile of the population?

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.6. What proportion of Women have blood pressures between 88.1 and 89.4?

A recent study reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 8.9. What proportion of Women have blood pressures higher than 85.8?

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.5 pounds and standard deviation 1.4 pounds. Find the first quartile of the weight.

Use technology to solve the following problem: A bottler of drinking water fills plastic bottles with a mean volume of 990 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally Distributed. What proportion of bottles have volumes between 985 mL and 988 mL?

Use technology to solve the following problem: The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.7 pounds and standard deviation 2.0 pound(s) . Find the 84th Percentile of the weights.

Use technology to solve the following problem: A normal population has a mean $\mu = 28$ and standard deviation $\sigma = 6$ What proportion of the population is between 18 and 26?

A normal population has a mean $\mu = 30$ and standard deviation $\sigma = 9$ . What proportion of the population is between 23 and 32?

Use technology to solve the following problem: A normal population has a mean $\mu = 31$ and standard deviation $\sigma = 7$ What is the probability that a randomly chosen value will be greater than 18?

Use technology to solve the following problem: A normal population has a mean $\mu = 28$ and standard deviation $\sigma = 8$ What is the 58th percentile of the population?

A normal population has a mean $\mu = 14$ and standard deviation $\sigma = 2$ . What is the 43rd percentile of the population?