Question 1

If an urn contains 2 red balls, 2 blue balls, and 1 green ball, the probability of drawing a green ball is .10.

Accepted Answer

The probability of drawing a green ball is 1 out of 5, or 0.20, since there is 1 green ball and a total of 5 balls.

Question 2

Empirical distributions are not based on mathematics or logic.

Accepted Answer

Empirical distributions are based on observed data, not on mathematics or logic.

Question 3

The inflection points on the normal curve are at the mean and the median.

Accepted Answer

The inflection points on the normal curve are one standard deviation away from the mean on either side, not at the mean or the median.

Question 4

A point on the normal curve with .40 of the curve beyond it has .10 between it and the mean.

Accepted Answer

The area between the mean and a point on the normal curve is equal to the area beyond that point. Therefore, if 0.40 of the curve is beyond a point, then 0.10 of the curve must be between that point and the mean.

Question 5

The normal distribution is symmetrical about the median.

Accepted Answer

The normal distribution is symmetrical about its mean, and since the mean and median are equal in a perfectly normal distribution, it is also symmetrical about the median.

Question 6

Extreme scores on the normal curve are those between the mean and the median.

Accepted Answer

Extreme scores on the normal curve are those that lie far from the mean, not between the mean and the median. In a perfectly normal distribution, the mean and median are the same.

Question 7

The numerator of the z score in Chapter 7 is the difference between a distribution's mean and its median.

Accepted Answer

The numerator of the z score is the difference between an individual score and the mean of the distribution, not the difference between the mean and the median.

Question 8

The area of a portion of a theoretical distribution is always equal to the probability of the events covered by that area.

Accepted Answer

In a theoretical probability distribution, the area under the curve representing the distribution over a given interval directly corresponds to the probability of events within that interval.

Question 9

Your textbook used positive and negative z scores as measures along the X axis.

Accepted Answer

Z-scores are indeed used to measure the position of a score in terms of standard deviations away from the mean along the X-axis in a normal distribution. Positive z-scores indicate values above the mean, while negative z-scores indicate values below the mean.

Question 10

The size of the theoretical distributions described in Chapter 7 does not depend on the size of the population.

Accepted Answer

The size of theoretical distributions, such as the normal distribution, is determined by their parameters (mean, standard deviation) rather than the size of the population from which samples are drawn.

Question 11

Data Set 7-1: Suppose an urn (a kind of jar that seems to be the best natural habitat for marbles) contained 3 red, 6 black, 5 blue, 2 yellow and 4 green marbles.
-Refer to Data Set 7-1. The probability of drawing a blue marble is

Accepted Answer

The total number of marbles is 3 (red) + 6 (black) + 5 (blue) + 2 (yellow) + 4 (green) = 20 marbles. The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is 5/20 = 0.25.

Question 12

Data Set 7-1: Suppose an urn (a kind of jar that seems to be the best natural habitat for marbles) contained 3 red, 6 black, 5 blue, 2 yellow and 4 green marbles.
-Refer to Data Set 7-1. The probability of drawing a black or a green marble is

Accepted Answer

The total number of marbles is 3 + 6 + 5 + 2 + 4 = 20. The probability of drawing a black or green marble is the sum of the probabilities of drawing a black marble and drawing a green marble, which is (6/20) + (4/20) = 10/20 = 0.5. None of the provided alternatives match this calculation.

Question 13

Data Set 7-1: Suppose an urn (a kind of jar that seems to be the best natural habitat for marbles) contained 3 red, 6 black, 5 blue, 2 yellow and 4 green marbles.
-Refer to Data Set 7-1. The probability of drawing a marble that is red, yellow, black or green is

Accepted Answer

The total number of marbles is 3 (red) + 6 (black) + 5 (blue) + 2 (yellow) + 4 (green) = 20 marbles. The probability of drawing a marble that is red, yellow, black, or green is the sum of the probabilities of drawing each of these colors. This is (3 + 6 + 2 + 4) / 20 = 15 / 20 = 0.75.

Question 14

Data Set 7-2: In the Fall of 1902 there were 184 seniors, 179 juniors, 267 sophomores, and 353 freshmen enrolled at a small college.
-In Data Set 7-2 the probability of picking a student at random and getting a freshman is

Accepted Answer

To find the probability of picking a freshman at random, divide the number of freshmen by the total number of students. Freshmen: 353, Total students: 184 (seniors) + 179 (juniors) + 267 (sophomores) + 353 (freshmen) = 983. Probability = 353 / 983 ≈ 0.359.

Question 15

Data Set 7-2: In the Fall of 1902 there were 184 seniors, 179 juniors, 267 sophomores, and 353 freshmen enrolled at a small college.
-In Data Set 7-2 the probability of picking a student at random and getting either a junior or a senior is

Accepted Answer

The answer of Data Set 7-2: In the Fall of...

Question 16

Data Set 7-2: In the Fall of 1902 there were 184 seniors, 179 juniors, 267 sophomores, and 353 freshmen enrolled at a small college.
-Data Set 7-2 is

Accepted Answer

This is an empirical distribution because it is based on observed, real-world data regarding the enrollment numbers of students in different years at a college.

Question 17

Data Set 7-3: A population of scores was normally distributed with a mean of 32 and a standard deviation of 3.
-Look at Data Set 7-3. The proportion that scored between 30 and 27 is

Accepted Answer

To find the proportion of scores between 30 and 27, we calculate the z-scores for these values and then find the area between them using the standard normal distribution. For 30: $z = \frac{30 - 32}{3} = -\frac{2}{3}$. For 27: $z = \frac{27 - 32}{3} = -\frac{5}{3}$. Looking up these z-scores in a standard normal distribution table or using a calculator, the area between them represents the proportion of scores in that range, which is approximately .2039.

Question 18

Data Set 7-3: A population of scores was normally distributed with a mean of 32 and a standard deviation of 3.
-For Data Set 7-3, the scores that separate the middle 60 percent from the extremes are

Accepted Answer

To find the scores that separate the middle 60 percent from the extremes in a normally distributed population, we use the concept of z-scores and the properties of the normal distribution. The middle 60% leaves 20% in each tail of the distribution. Looking up the z-score for 20% in the tail in a standard normal distribution table gives us approximately 0.84. Using the formula for converting z-scores to raw scores: X = μ + Zσ, where μ is the mean and σ is the standard deviation, we calculate the scores as follows: X = 32 + (0.84)(3) = 34.52 for the upper score and X = 32 - (0.84)(3) = 29.48 for the lower score.

Question 19

Data Set 7-3: A population of scores was normally distributed with a mean of 32 and a standard deviation of 3.
-Look at Data Set 7-3. The proportion of the distribution with scores of 30 or more is

Accepted Answer

To find the proportion of the distribution with scores of 30 or more, we calculate the z-score for a score of 30 using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. Substituting the given values, we get $z = \frac{30 - 32}{3} = -\frac{2}{3}$. Looking up this z-score in a standard normal distribution table or using a calculator, we find that the area to the right of this z-score (which represents the proportion of scores 30 or more) is approximately .7486.

Question 20

Data Set 7-4: A set of scores with a mean of 16 and a standard deviation of 3.
-Look at Data Set 7-4. What score separates the top 30 percent of the population from the lower 70 percent?

Accepted Answer

The answer of Data Set 7-4: A set of scores...

Quiz 7: Theoretical Distributions Including the Normal Distribution

If an urn contains 2 red balls, 2 blue balls, and 1 green ball, the probability of drawing a green ball is .10.

Empirical distributions are not based on mathematics or logic.

The inflection points on the normal curve are at the mean and the median.

A point on the normal curve with .40 of the curve beyond it has .10 between it and the mean.

The normal distribution is symmetrical about the median.

Extreme scores on the normal curve are those between the mean and the median.

The numerator of the z score in Chapter 7 is the difference between a distribution's mean and its median.

The area of a portion of a theoretical distribution is always equal to the probability of the events covered by that area.

Your textbook used positive and negative z scores as measures along the X axis.

The size of the theoretical distributions described in Chapter 7 does not depend on the size of the population.

Data Set 7-1: Suppose an urn (a kind of jar that seems to be the best natural habitat for marbles) contained 3 red, 6 black, 5 blue, 2 yellow and 4 green marbles. -Refer to Data Set 7-1. The probability of drawing a blue marble is

Data Set 7-1: Suppose an urn (a kind of jar that seems to be the best natural habitat for marbles) contained 3 red, 6 black, 5 blue, 2 yellow and 4 green marbles. -Refer to Data Set 7-1. The probability of drawing a black or a green marble is

Data Set 7-1: Suppose an urn (a kind of jar that seems to be the best natural habitat for marbles) contained 3 red, 6 black, 5 blue, 2 yellow and 4 green marbles. -Refer to Data Set 7-1. The probability of drawing a marble that is red, yellow, black or green is

Data Set 7-2: In the Fall of 1902 there were 184 seniors, 179 juniors, 267 sophomores, and 353 freshmen enrolled at a small college. -In Data Set 7-2 the probability of picking a student at random and getting a freshman is

Data Set 7-2: In the Fall of 1902 there were 184 seniors, 179 juniors, 267 sophomores, and 353 freshmen enrolled at a small college. -In Data Set 7-2 the probability of picking a student at random and getting either a junior or a senior is

Data Set 7-2: In the Fall of 1902 there were 184 seniors, 179 juniors, 267 sophomores, and 353 freshmen enrolled at a small college. -Data Set 7-2 is

Data Set 7-3: A population of scores was normally distributed with a mean of 32 and a standard deviation of 3. -Look at Data Set 7-3. The proportion that scored between 30 and 27 is

Data Set 7-3: A population of scores was normally distributed with a mean of 32 and a standard deviation of 3. -For Data Set 7-3, the scores that separate the middle 60 percent from the extremes are

Data Set 7-3: A population of scores was normally distributed with a mean of 32 and a standard deviation of 3. -Look at Data Set 7-3. The proportion of the distribution with scores of 30 or more is

Data Set 7-4: A set of scores with a mean of 16 and a standard deviation of 3. -Look at Data Set 7-4. What score separates the top 30 percent of the population from the lower 70 percent?