Answer :
Alright, let's solve the problem step-by-step.
First, let's denote the given data clearly:
- Number of social networking accounts: 0, 1, 2, 3, 4, 5
- Number of students: 4, 6, 3, [tex]\( x \)[/tex], 3, 2
### Part (a): Finding the greatest possible value of [tex]\( x \)[/tex] so that the median is 2
To determine the median, we need to find the value in the middle of the dataset when it is ordered.
1. Calculate the total number of students:
[tex]\[ \text{Total students} = 4 + 6 + 3 + x + 3 + 2 = 18 + x \][/tex]
2. Determine the position of the median:
Since the total number of students is [tex]\( 18 + x \)[/tex]:
- If [tex]\( 18 + x \)[/tex] is even, the median will be the average of the [tex]\( \left(\frac{18 + x}{2}\right)^{th} \)[/tex] and [tex]\( \left(\frac{18 + x}{2} + 1\right)^{th} \)[/tex] values.
- If [tex]\( 18 + x \)[/tex] is odd, the median is the [tex]\( \left(\frac{18 + x + 1}{2}\right)^{th} \)[/tex] value.
3. Set the condition for the median to be 2:
- We need the middle value(s) of the ordered list to be 2.
4. Find the possible values of [tex]\( x \)[/tex]:
To ensure the median is 2, the total number before and after the cumulative count up to 2 must balance around the center.
Let's check these conditions step by step.
- For [tex]\( x \)[/tex] students with 3 accounts, the total number of students is [tex]\( 18 + x \)[/tex].
- To ensure the median is 2, entries before and within the category "2" should lie exactly in the median range.
Evaluate:
4 (students with 0 accounts) + 6 (students with 1 account) + 3 (students with 2 accounts) = 13 + x students.
To keep the median at 2, more than half the students must own up to 2 accounts.
Therefore, [tex]\( 18 + x \)[/tex] (total number of students)
Median should be 2:
Thus, if:
[tex]\[ 13 \geq \frac{18 + x}{2} \rightarrow 26 \geq 18 + x \rightarrow x \leq 8 \][/tex]
Thus, the highest possible value of [tex]\( x \)[/tex] ensuring median of 2 stays in [tex]\( 2 \)[/tex] category:
[tex]\( x = 8 \)[/tex]
### Part (b): Finding the smallest possible value of [tex]\( x \)[/tex] so that the mode is 3
Mode is the number that appears most frequently in the dataset.
1. Current frequencies:
- 4 students have 0 accounts
- 6 students have 1 account
- 3 students have 2 accounts
- [tex]\( x \)[/tex] students have 3 accounts
- 3 students have 4 accounts
- 2 students have 5 accounts
2. Requirement for mode to be 3:
The mode is 3 if [tex]\( x > \)[/tex] any other frequencies.
Frequencies in table are already laid out:
- Maximum so far is 6 (mode = 1),
Thus,
[tex]\[ x > 6 \][/tex]
So the smallest possible [tex]\( x \)[/tex] to make the mode 3 is:
[tex]\[ x = 7 \][/tex]
To conclude:
(a) The greatest possible value of [tex]\( x \)[/tex] so that the median is 2 is [tex]\( \boxed{8} \)[/tex].
(b) The smallest possible value of [tex]\( x \)[/tex] so that the mode is 3 is [tex]\( \boxed{7} \)[/tex].
First, let's denote the given data clearly:
- Number of social networking accounts: 0, 1, 2, 3, 4, 5
- Number of students: 4, 6, 3, [tex]\( x \)[/tex], 3, 2
### Part (a): Finding the greatest possible value of [tex]\( x \)[/tex] so that the median is 2
To determine the median, we need to find the value in the middle of the dataset when it is ordered.
1. Calculate the total number of students:
[tex]\[ \text{Total students} = 4 + 6 + 3 + x + 3 + 2 = 18 + x \][/tex]
2. Determine the position of the median:
Since the total number of students is [tex]\( 18 + x \)[/tex]:
- If [tex]\( 18 + x \)[/tex] is even, the median will be the average of the [tex]\( \left(\frac{18 + x}{2}\right)^{th} \)[/tex] and [tex]\( \left(\frac{18 + x}{2} + 1\right)^{th} \)[/tex] values.
- If [tex]\( 18 + x \)[/tex] is odd, the median is the [tex]\( \left(\frac{18 + x + 1}{2}\right)^{th} \)[/tex] value.
3. Set the condition for the median to be 2:
- We need the middle value(s) of the ordered list to be 2.
4. Find the possible values of [tex]\( x \)[/tex]:
To ensure the median is 2, the total number before and after the cumulative count up to 2 must balance around the center.
Let's check these conditions step by step.
- For [tex]\( x \)[/tex] students with 3 accounts, the total number of students is [tex]\( 18 + x \)[/tex].
- To ensure the median is 2, entries before and within the category "2" should lie exactly in the median range.
Evaluate:
4 (students with 0 accounts) + 6 (students with 1 account) + 3 (students with 2 accounts) = 13 + x students.
To keep the median at 2, more than half the students must own up to 2 accounts.
Therefore, [tex]\( 18 + x \)[/tex] (total number of students)
Median should be 2:
Thus, if:
[tex]\[ 13 \geq \frac{18 + x}{2} \rightarrow 26 \geq 18 + x \rightarrow x \leq 8 \][/tex]
Thus, the highest possible value of [tex]\( x \)[/tex] ensuring median of 2 stays in [tex]\( 2 \)[/tex] category:
[tex]\( x = 8 \)[/tex]
### Part (b): Finding the smallest possible value of [tex]\( x \)[/tex] so that the mode is 3
Mode is the number that appears most frequently in the dataset.
1. Current frequencies:
- 4 students have 0 accounts
- 6 students have 1 account
- 3 students have 2 accounts
- [tex]\( x \)[/tex] students have 3 accounts
- 3 students have 4 accounts
- 2 students have 5 accounts
2. Requirement for mode to be 3:
The mode is 3 if [tex]\( x > \)[/tex] any other frequencies.
Frequencies in table are already laid out:
- Maximum so far is 6 (mode = 1),
Thus,
[tex]\[ x > 6 \][/tex]
So the smallest possible [tex]\( x \)[/tex] to make the mode 3 is:
[tex]\[ x = 7 \][/tex]
To conclude:
(a) The greatest possible value of [tex]\( x \)[/tex] so that the median is 2 is [tex]\( \boxed{8} \)[/tex].
(b) The smallest possible value of [tex]\( x \)[/tex] so that the mode is 3 is [tex]\( \boxed{7} \)[/tex].
