Question

60 randomly selected students were asked how many siblings were in their family. Let X = the number of pairs of siblings in the student's family. The results are as follows:

Siblings	Frequency
1	13
2	22
3	15
4	6
5	3
6	0
7	1

Round your answers to two decimal places.

The mean is:

The median is:

The sample standard deviation is:

The first quartile is:

The third quartile is:

What percent of the respondents have had less than 3 siblings? %

20% of all respondents have had at most how many siblings?

Answer 1

Let X be the Number of siblings, and f the frequency

Siblings (X)	Frequency (f)
1	13
2	22
3	15
4	6
5	3
6	0
7	1

The mean is:

ans: The mean is 2.47

The median is:

The number of observations n =60. When n is even, the median is the average of (n/2)th and (n/2+1)th observations.

In this case, the median is the average of 60/2=30th and 60/2+1=31st observations.

To identify the 30th and 31st observations, we get the cumulative frequencies

Siblings (X)	Frequency (f)	Cumulative Frequency
1	13	13
2	22	35
3	15	50
4	6	56
5	3	59
6	0	59
7	1	60

We can see that 30th and 31st observations both have the value of X=2. Hence the median is 2

ans: Median is 2

The sample standard deviation is:

ans: The sample standard deviation is 1.24

The first quartile is:

Using the Tukey's method, the first quartile is the median of the sample data below the median. The median is the average of 30th and 31st observations.

Hence the first quartile is the median of first 30 observations. Since n=30 is even, the median of this set is the average of 30/2=15th and 16th observations. From our earlier table of cumulative frequencies, we can see that the 15th and 16th observations both are X=2. Hence the first quartile is 2

ans: The first quartile is 2

The third quartile is:

Using the Tukey's method, the third quartile is the median of the sample data above the median. The median is the average of 30th and 31st observations.

Hence the third quartile is the median of last 30 observations. Since n=30 is even, the median of this set is the average of 30/2=15th and 16th observations, which are 45th and 46th observations in the full dataset. From our earlier table of cumulative frequencies, we can see that the 45th and 46th observations both are X=3. Hence the third quartile is 3

What percent of the respondents have had less than 3 siblings? %

The total number of observations, which are less than 3 is

13+22=35

Siblings (X)	Frequency (f)	Cumulative Frequency
1	13	13
2	22	35
3	15	50
4	6	56
5	3	59
6	0	59
7	1	60

Using this, the percent of the respondents have had less than 3 siblings is

ans: percent of the respondents have had less than 3 siblings is 58.33%

20% of all respondents have had at most how many siblings?

20% of the respondents correspond to 60*0.20=12

That is we want to know the value of X (number of siblings) less than equal to which there are 12 respondents.

We can see that for X<=1 there are 13 observations.

Hence we can say that X<=1 has 20% of the respondents.

ans: 20% of all respondents have had at most 1 sibling.