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Sagot :
To approach this problem, let's break it down into the given parts:
#### (a) The mean score of the class is 6. Find the:
##### (i) Value of [tex]\( k \)[/tex]
To find the value of [tex]\( k \)[/tex], let's first understand the distribution of scores given in the table.
| Score ([tex]\(x\)[/tex]) | 11 | 3 | 4 | 5 | 7 | 8 | 9 | 17 |
|---|---|---|---|---|---|---|---|---|
| Number of Pupils ([tex]\(f\)[/tex]) | 13 | 5 | 3 | [tex]\( k^2 + 1 \)[/tex] | 6 | 2 | 3 | 4 |
The mean score, [tex]\(\bar{x}\)[/tex], is given by:
[tex]\[ \bar{x} = \frac{\sum (x \times f)}{\sum f} \][/tex]
Given that the mean score is 6, we can write the equation as:
[tex]\[ 6 = \frac{11 \cdot 13 + 3 \cdot 5 + 4 \cdot 3 + 5 \cdot (k^2 + 1) + 7 \cdot 6 + 8 \cdot 2 + 9 \cdot 3 + 17 \cdot 4}{13 + 5 + 3 + (k^2 + 1) + 6 + 2 + 3 + 4} \][/tex]
Firstly, calculate the numerator (the sum of the product of scores and their frequencies):
[tex]\[ 11 \cdot 13 + 3 \cdot 5 + 4 \cdot 3 + 5 \cdot (k^2 + 1) + 7 \cdot 6 + 8 \cdot 2 + 9 \cdot 3 + 17 \cdot 4 \][/tex]
Let's compute specific terms:
[tex]\[ 11 \cdot 13 = 143,\quad 3 \cdot 5 = 15,\quad 4 \cdot 3 = 12,\quad 5 \cdot (k^2 + 1) = 5k^2 + 5,\quad 7 \cdot 6 = 42,\quad 8 \cdot 2 = 16,\quad 9 \cdot 3 = 27,\quad 17 \cdot 4 = 68 \][/tex]
Sum these values:
[tex]\[ 143 + 15 + 12 + 5k^2 + 5 + 42 + 16 + 27 + 68 = 328 + 5k^2 \][/tex]
Next, calculate the denominator (the sum of the frequencies):
[tex]\[ 13 + 5 + 3 + (k^2 + 1) + 6 + 2 + 3 + 4 \][/tex]
Sum these values:
[tex]\[ 13 + 5 + 3 + k^2 + 1 + 6 + 2 + 3 + 4 = 37 + k^2 \][/tex]
Thus, we have the equation:
[tex]\[ 6 = \frac{328 + 5k^2}{37 + k^2} \][/tex]
Cross-multiply to solve for [tex]\( k \)[/tex]:
[tex]\[ 6 (37 + k^2) = 328 + 5k^2 \][/tex]
[tex]\[ 222 + 6k^2 = 328 + 5k^2 \][/tex]
[tex]\[ 6k^2 - 5k^2 = 328 - 222 \][/tex]
[tex]\[ k^2 = 106 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ k = \sqrt{106} \][/tex]
Thus, the value of [tex]\( k \)[/tex] is [tex]\(\sqrt{106}\)[/tex].
##### (ii) Median
To find the median score, arrange the scores in ascending order based on their frequencies and identify the middle value.
#### (b) Draw a bar chart for the distribution
Create a bar chart by plotting scores on the x-axis and the frequency of students on the y-axis. Each score will be represented by its respective frequency from the table.
#### (c) Probability that a randomly picked pupil scored less than 6
Calculate this probability by summing the frequencies of the scores less than 6 and divide by the total number of pupils.
Combine the frequencies for scores 3, 4, and 5:
[tex]\[ \text{Frequency for score 3} = 5 \][/tex]
[tex]\[ \text{Frequency for score 4} = 3 \][/tex]
[tex]\[ \text{Frequency for score 5} = k^2 + 1 = \sqrt{106}^2 + 1 = 106 + 1 = 107 \][/tex]
Sum these frequencies:
[tex]\[ 5 + 3 + 107 = 115 \][/tex]
Probability:
[tex]\[ P(\text{score} < 6) = \frac{115}{\text{Total frequency}} = \frac{115}{150} = \frac{23}{30} \][/tex]
So, the probability that a randomly picked pupil scored less than 6 is [tex]\(\frac{23}{30}\)[/tex].
#### (a) The mean score of the class is 6. Find the:
##### (i) Value of [tex]\( k \)[/tex]
To find the value of [tex]\( k \)[/tex], let's first understand the distribution of scores given in the table.
| Score ([tex]\(x\)[/tex]) | 11 | 3 | 4 | 5 | 7 | 8 | 9 | 17 |
|---|---|---|---|---|---|---|---|---|
| Number of Pupils ([tex]\(f\)[/tex]) | 13 | 5 | 3 | [tex]\( k^2 + 1 \)[/tex] | 6 | 2 | 3 | 4 |
The mean score, [tex]\(\bar{x}\)[/tex], is given by:
[tex]\[ \bar{x} = \frac{\sum (x \times f)}{\sum f} \][/tex]
Given that the mean score is 6, we can write the equation as:
[tex]\[ 6 = \frac{11 \cdot 13 + 3 \cdot 5 + 4 \cdot 3 + 5 \cdot (k^2 + 1) + 7 \cdot 6 + 8 \cdot 2 + 9 \cdot 3 + 17 \cdot 4}{13 + 5 + 3 + (k^2 + 1) + 6 + 2 + 3 + 4} \][/tex]
Firstly, calculate the numerator (the sum of the product of scores and their frequencies):
[tex]\[ 11 \cdot 13 + 3 \cdot 5 + 4 \cdot 3 + 5 \cdot (k^2 + 1) + 7 \cdot 6 + 8 \cdot 2 + 9 \cdot 3 + 17 \cdot 4 \][/tex]
Let's compute specific terms:
[tex]\[ 11 \cdot 13 = 143,\quad 3 \cdot 5 = 15,\quad 4 \cdot 3 = 12,\quad 5 \cdot (k^2 + 1) = 5k^2 + 5,\quad 7 \cdot 6 = 42,\quad 8 \cdot 2 = 16,\quad 9 \cdot 3 = 27,\quad 17 \cdot 4 = 68 \][/tex]
Sum these values:
[tex]\[ 143 + 15 + 12 + 5k^2 + 5 + 42 + 16 + 27 + 68 = 328 + 5k^2 \][/tex]
Next, calculate the denominator (the sum of the frequencies):
[tex]\[ 13 + 5 + 3 + (k^2 + 1) + 6 + 2 + 3 + 4 \][/tex]
Sum these values:
[tex]\[ 13 + 5 + 3 + k^2 + 1 + 6 + 2 + 3 + 4 = 37 + k^2 \][/tex]
Thus, we have the equation:
[tex]\[ 6 = \frac{328 + 5k^2}{37 + k^2} \][/tex]
Cross-multiply to solve for [tex]\( k \)[/tex]:
[tex]\[ 6 (37 + k^2) = 328 + 5k^2 \][/tex]
[tex]\[ 222 + 6k^2 = 328 + 5k^2 \][/tex]
[tex]\[ 6k^2 - 5k^2 = 328 - 222 \][/tex]
[tex]\[ k^2 = 106 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ k = \sqrt{106} \][/tex]
Thus, the value of [tex]\( k \)[/tex] is [tex]\(\sqrt{106}\)[/tex].
##### (ii) Median
To find the median score, arrange the scores in ascending order based on their frequencies and identify the middle value.
#### (b) Draw a bar chart for the distribution
Create a bar chart by plotting scores on the x-axis and the frequency of students on the y-axis. Each score will be represented by its respective frequency from the table.
#### (c) Probability that a randomly picked pupil scored less than 6
Calculate this probability by summing the frequencies of the scores less than 6 and divide by the total number of pupils.
Combine the frequencies for scores 3, 4, and 5:
[tex]\[ \text{Frequency for score 3} = 5 \][/tex]
[tex]\[ \text{Frequency for score 4} = 3 \][/tex]
[tex]\[ \text{Frequency for score 5} = k^2 + 1 = \sqrt{106}^2 + 1 = 106 + 1 = 107 \][/tex]
Sum these frequencies:
[tex]\[ 5 + 3 + 107 = 115 \][/tex]
Probability:
[tex]\[ P(\text{score} < 6) = \frac{115}{\text{Total frequency}} = \frac{115}{150} = \frac{23}{30} \][/tex]
So, the probability that a randomly picked pupil scored less than 6 is [tex]\(\frac{23}{30}\)[/tex].
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