To find the median number of siblings using the frequency table, we need to first arrange the responses in ascending order:
1, 2.5, 2.5, 3, 3, 6, 6
The middle value is the median. Since there are 7 responses, the middle value is at the (7+1)/2 = 4th position.
The 4th value in the arranged list is 3.
Therefore, the median number of siblings is 3.
A group of students were asked how many siblings they have. Determine the median number of siblings using the frequency table.(1 point)
Responses
3
3
2.5
2.5
6
6
1
8 answers
Loren’s friends jump as far as they can and record their results in inches. Their results include the observations 42, 47, 50, 42, 45, 41, 49, 51, and 44. Which number, if added to the dataset, would represent an outlier in the data?(1 point)
Responses
40
40
44
44
72
72
52
Responses
40
40
44
44
72
72
52
To determine which number would represent an outlier in the given data set, we can use the interquartile range (IQR) technique.
First, we need to find the quartiles of the data set. To do this, we arrange the observations in ascending order:
41, 42, 42, 44, 45, 47, 49, 50, 51
Next, we find the median of the data set. Since the data set has an odd number of observations, the median is the middle value, which is 45.
Now, we divide the data set into two halves. The lower half (Q1) consists of the values 41, 42, 42, 44. The upper half (Q3) consists of the values 47, 49, 50, 51.
To calculate the IQR, we subtract Q1 from Q3:
IQR = Q3 - Q1
= 51 - 44
= 7
According to the IQR technique, any value below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR can be considered an outlier.
Lower limit: Q1 - 1.5 * IQR
= 44 - (1.5 * 7)
= 35.5
Upper limit: Q3 + 1.5 * IQR
= 51 + (1.5 * 7)
= 61.5
Any value below 35.5 or above 61.5 would be considered an outlier.
Out of the given options, only 72 (72 inches) is above the upper limit of 61.5.
Therefore, adding 72 to the data set would represent an outlier.
First, we need to find the quartiles of the data set. To do this, we arrange the observations in ascending order:
41, 42, 42, 44, 45, 47, 49, 50, 51
Next, we find the median of the data set. Since the data set has an odd number of observations, the median is the middle value, which is 45.
Now, we divide the data set into two halves. The lower half (Q1) consists of the values 41, 42, 42, 44. The upper half (Q3) consists of the values 47, 49, 50, 51.
To calculate the IQR, we subtract Q1 from Q3:
IQR = Q3 - Q1
= 51 - 44
= 7
According to the IQR technique, any value below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR can be considered an outlier.
Lower limit: Q1 - 1.5 * IQR
= 44 - (1.5 * 7)
= 35.5
Upper limit: Q3 + 1.5 * IQR
= 51 + (1.5 * 7)
= 61.5
Any value below 35.5 or above 61.5 would be considered an outlier.
Out of the given options, only 72 (72 inches) is above the upper limit of 61.5.
Therefore, adding 72 to the data set would represent an outlier.
Which measures of center does not change when a value of 0 is inserted into these datasets?
Dataset 1:
(12,19,18,12,19,14,12,28,22,10)
Dataset 2:
(12,19,18,12,19,14,12,28,22,10,0)
(1 point)
Responses
The mean stays the same
The mean stays the same
The median does not change
The median does not change
The mode remains the same
The mode remains the same
No measures of center change
No measures of center change
Dataset 1:
(12,19,18,12,19,14,12,28,22,10)
Dataset 2:
(12,19,18,12,19,14,12,28,22,10,0)
(1 point)
Responses
The mean stays the same
The mean stays the same
The median does not change
The median does not change
The mode remains the same
The mode remains the same
No measures of center change
No measures of center change
The median does not change when a value of 0 is inserted into the datasets.
In Dataset 1, the median is 14.5.
In Dataset 2, the median is also 14.5.
Therefore, the correct answer is:
The median does not change.
In Dataset 1, the median is 14.5.
In Dataset 2, the median is also 14.5.
Therefore, the correct answer is:
The median does not change.
The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. What are the first and third quartiles?(1 point)
First Quartile:
Third Quartile:
First Quartile:
Third Quartile:
To find the first and third quartiles from the list of prime numbers, we need to arrange them in ascending order:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Since there are 10 prime numbers, the median (second quartile) will be the 5th value:
Median: 11
Now, we can divide the list into two halves. For the first quartile, we consider the values to the left of the median:
2, 3, 5, 7
The first quartile is the median of these values. Since there are 4 values, the median will be the average of the 2nd and 3rd values:
First Quartile: (3 + 5) / 2 = 8 / 2 = 4
The third quartile will be considered the values to the right of the median:
13, 17, 19, 23, 29
The third quartile is the median of these values. Since there are 5 values, the median will be the 3rd value:
Third Quartile: 19
Therefore, the first quartile is 4 and the third quartile is 19.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Since there are 10 prime numbers, the median (second quartile) will be the 5th value:
Median: 11
Now, we can divide the list into two halves. For the first quartile, we consider the values to the left of the median:
2, 3, 5, 7
The first quartile is the median of these values. Since there are 4 values, the median will be the average of the 2nd and 3rd values:
First Quartile: (3 + 5) / 2 = 8 / 2 = 4
The third quartile will be considered the values to the right of the median:
13, 17, 19, 23, 29
The third quartile is the median of these values. Since there are 5 values, the median will be the 3rd value:
Third Quartile: 19
Therefore, the first quartile is 4 and the third quartile is 19.
Which question helps you understand a cause and effect relationship between words?(1 point)
Responses
How important is it?
How important is it?
Whose is it?
Whose is it?
What happened?
What happened?
Where did it happen?
Responses
How important is it?
How important is it?
Whose is it?
Whose is it?
What happened?
What happened?
Where did it happen?
1 answer