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4 votes
Age, f

15, 1
14, 1
13, 1
12, 1
11, 1
10, 1
9, 0
8, 2
7, 2

1. Which measure of central tendency would be the most appropriate to describe the variable Gender? Age?
2. Calculate Mode, Median and Mean for Age. 3. Calculate the range, variance and standard deviation (use the formula for population variance and SD) for Age.
4. What are the ages of participants that reflect 68% of your sample?​

2 Answers

5 votes

Answer:

Explanation:

Since the variable Gender is not given in the data, we cannot compute any measure of central tendency for it. For the variable Age, the most appropriate measure of central tendency would be the mode, since all the observations have the same value for this variable.

To calculate the mode, median and mean for Age, we can first arrange the data in ascending order of age:

7, 2

8, 2

9, 0

10, 1

11, 1

12, 1

13, 1

14, 1

15, 1

The mode is the most frequently occurring value, which is 1 for all ages. The median is the middle value when the data is arranged in ascending order. Since there are 9 observations, the median is the average of the 5th and 6th values, which are 12 and 13. So, the median is (12+13)/2 = 12.5. The mean is the sum of all the values divided by the number of observations, which is:

(72 + 82 + 90 + 101 + 111 + 121 + 131 + 141 + 15*1) / 9 = 11.44 (rounded to two decimal places)

Therefore, the mode is 1, the median is 12.5 and the mean is 11.44.

To calculate the range, variance and standard deviation for Age, we can use the following formulas (assuming the data is a population):

Range = maximum value - minimum value = 15 - 7 = 8

Mean = 11.44 (computed in step 2)

Variance = Σ(x - μ)^2 / N, where μ is the mean, x is each value, and N is the number of observations. Plugging in the values, we get:

Variance = [(7-11.44)^2 + (8-11.44)^2 + (9-11.44)^2 + (10-11.44)^2 + (11-11.44)^2 + (12-11.44)^2 + (13-11.44)^2 + (14-11.44)^2 + (15-11.44)^2] / 9 = 5.72 (rounded to two decimal places)

Standard deviation = square root of variance = √5.72 = 2.39 (rounded to two decimal places)

Therefore, the range is 8, the variance is 5.72, and the standard deviation is 2.39.

Since the data is approximately normally distributed, we can use the empirical rule to estimate the ages that reflect 68% of the sample. According to the empirical rule, 68% of the data falls within one standard deviation of the mean. So, we can compute the lower and upper limits of this range by subtracting and adding one standard deviation from the mean, respectively:

Lower limit = mean - standard deviation = 11.44 - 2.39 = 9.05 (rounded to two decimal places)

Upper limit = mean + standard deviation = 11.44 + 2.39 = 13.83 (rounded to two decimal places)

Therefore, the ages of participants that reflect 68% of the sample are between 9.05 and 13.83 years old.

User Jim Clay
by
7.2k points
1 vote

Answer:

It seems like there is a mistake in the variable provided. The variable mentioned is "Gender" but the data provided is about "Age" and "f" (frequency). Assuming that "f" is the frequency of participants at each age, here are the answers to your questions:

For the variable "f", the most appropriate measure of central tendency would be the mode, which represents the most frequently occurring value. For the variable "Age", the most appropriate measure of central tendency would be the median, as it is less affected by outliers than the mean.

Mode = 8; Median = 11; Mean = 10.11 (rounded to two decimal places)

Range = 15 - 7 = 8; Variance = 8.99 (rounded to two decimal places); Standard deviation = 2.99 (rounded to two decimal places)

To calculate the age range that reflects 68% of the sample, we need to find the mean age and the standard deviation. The mean age is 10.11, and the standard deviation is 2.99. 68% of the data falls within one standard deviation of the mean. Using the empirical rule, we can say that the ages of participants that reflect 68% of the sample are between (10.11 - 2.99) and (10.11 + 2.99), which is between 7.12 and 13.10.

Explanation:

User Ashish Bista
by
6.3k points