Basic Level: Mean Median Mode Test! Quiz

1) If the largest value of a data set is doubled, which of the following is false?

The mean increases.

The standard deviation increases.

The interquartile range increases.

The range increases.

When the largest value of a data set is doubled, it only affects the maximum value, which does not impact the interquartile range. The interquartile range is the difference between the upper quartile and the lower quartile, which is determined by the middle 50% of the data. Since doubling the largest value does not affect this middle 50%, the interquartile range remains the same. However, doubling the largest value would affect the mean, standard deviation, and range, causing them to increase.

Explanation

When the largest value of a data set is doubled, it only affects the maximum value, which does not impact the interquartile range. The interquartile range is the difference between the upper quartile and the lower quartile, which is determined by the middle 50% of the data. Since doubling the largest value does not affect this middle 50%, the interquartile range remains the same. However, doubling the largest value would affect the mean, standard deviation, and range, causing them to increase.

2) The nominal level of measurement is represented in which variable below?

Fear of crime

Gender

Temperature

Income

The nominal level of measurement is represented in the variable "gender" because it is a categorical variable that represents distinct categories or groups. In this case, gender can be classified into two categories - male and female, without any inherent order or numerical value attached to them. The nominal level of measurement is the lowest level of measurement, where data is simply categorized into different groups or categories.

Explanation

The nominal level of measurement is represented in the variable "gender" because it is a categorical variable that represents distinct categories or groups. In this case, gender can be classified into two categories - male and female, without any inherent order or numerical value attached to them. The nominal level of measurement is the lowest level of measurement, where data is simply categorized into different groups or categories.

3) A sample of 20 observations has a standard deviation of 3. The sum of the squared deviations of observations from the sample mean is

20

23

29

60

The sum of squared deviations from the mean is a measure of the variability or dispersion of the data. It is calculated by summing the squared differences between each observation and the mean. In this case, since the standard deviation is given as 3, we can infer that the variance is 9. Therefore, the sum of squared deviations can be calculated by multiplying the variance by the sample size (20), giving us a value of 180. However, the question asks for the sum of squared deviations, not the sum of squared deviations divided by the sample size (which would give us the variance). Therefore, we divide 180 by 20 to get 9, which is the correct answer.

Explanation

The sum of squared deviations from the mean is a measure of the variability or dispersion of the data. It is calculated by summing the squared differences between each observation and the mean. In this case, since the standard deviation is given as 3, we can infer that the variance is 9. Therefore, the sum of squared deviations can be calculated by multiplying the variance by the sample size (20), giving us a value of 180. However, the question asks for the sum of squared deviations, not the sum of squared deviations divided by the sample size (which would give us the variance). Therefore, we divide 180 by 20 to get 9, which is the correct answer.

4) The scores of male (M) and female (F) students on a statistics exam are displayed in the following boxplots. The pluses indicate the location of the means. Which of the following is correct?

The mean grade of the females is about 72.

About 75% of the males score above 82.

The median of the male students is about 66.

The scores of the males have a higher variability than the scores of the females

About 25% of the females scored above 72.

The correct answer is that the scores of the males have a higher variability than the scores of the females. This can be inferred from the boxplots, where the box for the males is wider than the box for the females. A wider box indicates a greater spread of scores, indicating higher variability.

Explanation

The correct answer is that the scores of the males have a higher variability than the scores of the females. This can be inferred from the boxplots, where the box for the males is wider than the box for the females. A wider box indicates a greater spread of scores, indicating higher variability.

5) Provided that the data above is normally distributed with a mean of 18 and standard deviation of 6, determine the proportion of students with a 33 or higher.

0.0062

0.0109

0.0124

0.0217

The proportion of students with a 33 or higher can be determined by calculating the z-score for 33 using the formula (x - mean) / standard deviation. The z-score for 33 is (33 - 18) / 6 = 2.5. Looking up the z-score in a standard normal distribution table, we find that the proportion of values above 2.5 is approximately 0.0062. Therefore, the correct answer is 0.0062.

Explanation

The proportion of students with a 33 or higher can be determined by calculating the z-score for 33 using the formula (x - mean) / standard deviation. The z-score for 33 is (33 - 18) / 6 = 2.5. Looking up the z-score in a standard normal distribution table, we find that the proportion of values above 2.5 is approximately 0.0062. Therefore, the correct answer is 0.0062.

6) A professor scaled (curved) the scores on an exam by multiplying the student's raw score by 1.2, then adding 15 points. If the mean and standard deviation of the scores before the curve were 51 and 5, respectively, then the mean and standard deviation of the scaled scores are respectively:

76.2 and 21

76.2 and 6

76.2 and 5

61 and 6

Cannot be determined without knowing if the scores are normally distributed

The professor scaled the scores by multiplying the raw score by 1.2 and adding 15 points. This means that each student's score was increased by 20% and then an additional 15 points were added. Since the mean of the raw scores was 51, the mean of the scaled scores would be 51 multiplied by 1.2 plus 15, which equals 76.2. The standard deviation of the raw scores was 5, and since scaling does not affect the variability, the standard deviation of the scaled scores would remain the same at 5. Therefore, the mean and standard deviation of the scaled scores are 76.2 and 6, respectively.

Explanation

The professor scaled the scores by multiplying the raw score by 1.2 and adding 15 points. This means that each student's score was increased by 20% and then an additional 15 points were added. Since the mean of the raw scores was 51, the mean of the scaled scores would be 51 multiplied by 1.2 plus 15, which equals 76.2. The standard deviation of the raw scores was 5, and since scaling does not affect the variability, the standard deviation of the scaled scores would remain the same at 5. Therefore, the mean and standard deviation of the scaled scores are 76.2 and 6, respectively.

7) The following list is a set of data ordered from smallest to largest. All values are integers. 2 12 y y y 15 18 18 19 I. The median and the first quartile cannot be equal. II. The mode is 18. III. 2 is an outlier.

I only

II only

III only

I and III only

I, II, and III

I. The median is the middle value of a set of data when it is arranged in order. In this case, the median is the 5th value, which is one of the "y" values. The first quartile is the value that divides the lower half of the data into two equal parts. Since the median is one of the "y" values and the first quartile is one of the "y" values as well, they cannot be equal. Therefore, statement I is correct.

II. The mode is the value that appears most frequently in a set of data. In this case, the value 18 appears twice, which is more than any other value. Therefore, statement II is correct.

III. An outlier is a value that is significantly different from the other values in a set of data. In this case, the value 2 is significantly smaller than the other values, making it an outlier. Therefore, statement III is correct.

Explanation

I. The median is the middle value of a set of data when it is arranged in order. In this case, the median is the 5th value, which is one of the "y" values. The first quartile is the value that divides the lower half of the data into two equal parts. Since the median is one of the "y" values and the first quartile is one of the "y" values as well, they cannot be equal. Therefore, statement I is correct.

II. The mode is the value that appears most frequently in a set of data. In this case, the value 18 appears twice, which is more than any other value. Therefore, statement II is correct.

III. An outlier is a value that is significantly different from the other values in a set of data. In this case, the value 2 is significantly smaller than the other values, making it an outlier. Therefore, statement III is correct.

8) A scores from 2012 is listed below. Calculate the sample mean and standard deviation 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18

20.50, 5.79

20.50, 5.94

20.85, 5.79

20.85, 5.94

The sample mean is calculated by summing up all the scores and dividing by the total number of scores. In this case, the sum of the scores is 417 and there are 20 scores in total. Therefore, the sample mean is 417/20 = 20.85.

The standard deviation is a measure of the dispersion or spread of the scores. It is calculated by finding the average of the squared differences between each score and the mean, and then taking the square root of that average. In this case, the squared differences from the mean are: 8.22, 196.02, 189.42, 0.42, 0.42, 12.02, 11.62, 0.42, 11.62, 4.82, 8.22, 12.02, 81.62, 81.62, 5.62, 43.22, 8.22, 12.02, 11.62, 5.62. The average of these squared differences is 55.76. Taking the square root of that gives a standard deviation of 5.94.

Explanation

The sample mean is calculated by summing up all the scores and dividing by the total number of scores. In this case, the sum of the scores is 417 and there are 20 scores in total. Therefore, the sample mean is 417/20 = 20.85.

The standard deviation is a measure of the dispersion or spread of the scores. It is calculated by finding the average of the squared differences between each score and the mean, and then taking the square root of that average. In this case, the squared differences from the mean are: 8.22, 196.02, 189.42, 0.42, 0.42, 12.02, 11.62, 0.42, 11.62, 4.82, 8.22, 12.02, 81.62, 81.62, 5.62, 43.22, 8.22, 12.02, 11.62, 5.62. The average of these squared differences is 55.76. Taking the square root of that gives a standard deviation of 5.94.

9) The inter quartile Range for the AGE Attribute is

The interquartile range (IQR) is a measure of statistical dispersion, specifically the range between the first quartile (Q1) and the third quartile (Q3) in a dataset. In this case, the IQR for the AGE attribute is 8, indicating that the range between the 25th and 75th percentiles of the age values is 8. This suggests that the middle 50% of the ages in the dataset fall within a range of 8 years.

Explanation

The interquartile range (IQR) is a measure of statistical dispersion, specifically the range between the first quartile (Q1) and the third quartile (Q3) in a dataset. In this case, the IQR for the AGE attribute is 8, indicating that the range between the 25th and 75th percentiles of the age values is 8. This suggests that the middle 50% of the ages in the dataset fall within a range of 8 years.

Submit

10) A mother has a child and tells all her friends that he has an IQ of 113 on the XYZ scale and is truly exceptionally intelligent. Given that the mean is 100 and the standard deviation 15 in which band on the graph opposite does he fit in? (one correct choice)

BAND -A

BAND -B

BAND -C

BAND -D

BAND -E

The child's IQ score of 113 is above the mean of 100, indicating that he is above average in intelligence. Additionally, the standard deviation of 15 suggests that most scores fall within a range of 15 points above or below the mean. Since the child's score is 13 points above the mean, he falls within the highest band on the graph, BAND -E, which represents scores that are significantly above average.

Explanation

The child's IQ score of 113 is above the mean of 100, indicating that he is above average in intelligence. Additionally, the standard deviation of 15 suggests that most scores fall within a range of 15 points above or below the mean. Since the child's score is 13 points above the mean, he falls within the highest band on the graph, BAND -E, which represents scores that are significantly above average.