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Sagot :
Given the marks obtained by pupils in a test:
[tex]\[ \begin{array}{lllll} 8 & 4 & 8 & 2 & 8 \\ 6 & 8 & 8 & 8 & 10 \\ 8 & 9 & 8 & 6 & 10 \\ 2 & 2 & 8 & 6 & 6 \end{array} \][/tex]
Let's address each part of the problem step by step.
### a) Construct a frequency distribution table for the data.
To construct the frequency distribution table, we need to count the occurrences of each mark:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Mark} & \text{Tally} & \text{Frequency} \\ \hline 2 & ||| & 3 \\ \hline 4 & | & 1 \\ \hline 6 & |||| & 4 \\ \hline 8 & ||||||||| & 9 \\ \hline 9 & | & 1 \\ \hline 10 & || & 2 \\ \hline \end{array} \][/tex]
### b) What is the modal mark?
The modal mark is the mark that appears most frequently in the data set. From the frequency distribution table, we see that the mark 8 appears 9 times, which is more frequent than any other mark.
Thus, the modal mark is [tex]\(8\)[/tex].
### c) Calculate the mean mark.
The mean mark can be calculated using the formula:
[tex]\[ \text{Mean} = \frac{\sum (\text{Mark} \times \text{Frequency})}{\sum (\text{Frequency})} \][/tex]
From the frequency distribution, the total of the marks times their respective frequencies is:
[tex]\[ 2 \times 3 + 4 \times 1 + 6 \times 4 + 8 \times 9 + 9 \times 1 + 10 \times 2 = 6 + 4 + 24 + 72 + 9 + 20 = 135 \][/tex]
The total number of students (sum of frequencies) is:
[tex]\[ 3 + 1 + 4 + 9 + 1 + 2 = 20 \][/tex]
Therefore, the mean mark is:
[tex]\[ \text{Mean} = \frac{135}{20} = 6.75 \][/tex]
### d) How many pupils score more than 7 marks?
We need to count the number of pupils who scored more than 7 marks. From the frequency table:
- 8 marks: 9 pupils
- 9 marks: 1 pupil
- 10 marks: 2 pupils
Thus, the total number of pupils who scored more than 7 marks is:
[tex]\[ 9 + 1 + 2 = 12 \][/tex]
### e) What is the probability that a student chosen at random obtained 2 marks?
The probability can be calculated as the ratio of the number of students who obtained 2 marks to the total number of students.
From the frequency table, the number of students who obtained 2 marks is 3, and the total number of students is 20.
Therefore, the probability that a student chosen at random obtained 2 marks is:
[tex]\[ \frac{3}{20} = 0.15 \][/tex]
### Summary Table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Mark} & \text{Tally} & \text{Frequency} & f \times x \\ \hline 2 & ||| & 3 & 6 \\ \hline 4 & | & 1 & 4 \\ \hline 6 & |||| & 4 & 24 \\ \hline 8 & ||||||||| & 9 & 72 \\ \hline 9 & | & 1 & 9 \\ \hline 10 & || & 2 & 20 \\ \hline \text{Total} & & 20 & 135 \\ \hline \end{array} \][/tex]
[tex]\[ \begin{array}{lllll} 8 & 4 & 8 & 2 & 8 \\ 6 & 8 & 8 & 8 & 10 \\ 8 & 9 & 8 & 6 & 10 \\ 2 & 2 & 8 & 6 & 6 \end{array} \][/tex]
Let's address each part of the problem step by step.
### a) Construct a frequency distribution table for the data.
To construct the frequency distribution table, we need to count the occurrences of each mark:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Mark} & \text{Tally} & \text{Frequency} \\ \hline 2 & ||| & 3 \\ \hline 4 & | & 1 \\ \hline 6 & |||| & 4 \\ \hline 8 & ||||||||| & 9 \\ \hline 9 & | & 1 \\ \hline 10 & || & 2 \\ \hline \end{array} \][/tex]
### b) What is the modal mark?
The modal mark is the mark that appears most frequently in the data set. From the frequency distribution table, we see that the mark 8 appears 9 times, which is more frequent than any other mark.
Thus, the modal mark is [tex]\(8\)[/tex].
### c) Calculate the mean mark.
The mean mark can be calculated using the formula:
[tex]\[ \text{Mean} = \frac{\sum (\text{Mark} \times \text{Frequency})}{\sum (\text{Frequency})} \][/tex]
From the frequency distribution, the total of the marks times their respective frequencies is:
[tex]\[ 2 \times 3 + 4 \times 1 + 6 \times 4 + 8 \times 9 + 9 \times 1 + 10 \times 2 = 6 + 4 + 24 + 72 + 9 + 20 = 135 \][/tex]
The total number of students (sum of frequencies) is:
[tex]\[ 3 + 1 + 4 + 9 + 1 + 2 = 20 \][/tex]
Therefore, the mean mark is:
[tex]\[ \text{Mean} = \frac{135}{20} = 6.75 \][/tex]
### d) How many pupils score more than 7 marks?
We need to count the number of pupils who scored more than 7 marks. From the frequency table:
- 8 marks: 9 pupils
- 9 marks: 1 pupil
- 10 marks: 2 pupils
Thus, the total number of pupils who scored more than 7 marks is:
[tex]\[ 9 + 1 + 2 = 12 \][/tex]
### e) What is the probability that a student chosen at random obtained 2 marks?
The probability can be calculated as the ratio of the number of students who obtained 2 marks to the total number of students.
From the frequency table, the number of students who obtained 2 marks is 3, and the total number of students is 20.
Therefore, the probability that a student chosen at random obtained 2 marks is:
[tex]\[ \frac{3}{20} = 0.15 \][/tex]
### Summary Table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \text{Mark} & \text{Tally} & \text{Frequency} & f \times x \\ \hline 2 & ||| & 3 & 6 \\ \hline 4 & | & 1 & 4 \\ \hline 6 & |||| & 4 & 24 \\ \hline 8 & ||||||||| & 9 & 72 \\ \hline 9 & | & 1 & 9 \\ \hline 10 & || & 2 & 20 \\ \hline \text{Total} & & 20 & 135 \\ \hline \end{array} \][/tex]
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