Answer
Verified
393.3k+ views
Hint: First, we will find the value of \[k\] using the formula \[\sum {{P_i}} = 1\]. Then we will now find the value of \[{{\text{X}}^2}{\text{P}}\] and \[{\text{XP}}\] from the above table for mean, using the formula, \[\sum {{\text{XP}}} \] and variance, using the formula, \[{\text{Variance}} = \sum {{{\text{X}}^2}{\text{P}}} - {\left( {\sum {{\text{XP}}} } \right)^2}\]
Complete step by step answer:
We will now form a table to find the value of the product \[\sum {{P_i}} \] for the mean.
We know that the sum of \[P\left( {X = {x_i}} \right)\]is always equals to 1, that is, \[\sum {{P_i}} = 1\].
Substituting the values of \[\sum {{P_i}} \] to find the value of \[k\] from the above formula, we get
\[ \Rightarrow 15k = 1\]
Dividing the above equation by 15 on both sides, we get
\[
\Rightarrow \dfrac{{15k}}{{15}} = \dfrac{1}{{15}} \\
\Rightarrow k = \dfrac{1}{{15}} \\
\]
Hence, the value of \[k\] is \[\dfrac{1}{{15}}\].
We will now find the value of \[{{\text{X}}^2}{\text{P}}\] \[{\text{XP}}\]from the above table for median and mode.
Substituting the value of \[k\] in \[\sum {{\text{XP}}} \] and \[\sum {{{\text{X}}^2}{\text{P}}} \], we get
\[
\Rightarrow \sum {{\text{XP}}} = 55 \times \dfrac{1}{{15}} \\
\Rightarrow \sum {{\text{XP}}} = 3.66 \\
\]
\[
\Rightarrow \sum {{{\text{X}}^2}{\text{P}}} = 225 \times \dfrac{1}{{15}} \\
\Rightarrow \sum {{{\text{X}}^2}{\text{P}}} = 15 \\
\]
Therefore, the mean of the given data is \[3.66\].
We know that the formula to calculate variance is \[{\text{Variance}} = \sum {{{\text{X}}^2}{\text{P}}} - {\left( {\sum {{\text{XP}}} } \right)^2}\].
Substituting these values in the above formula of variance, we get
\[
\Rightarrow {\text{Variance}} = 15 - {\left( {3.66} \right)^2} \\
\Rightarrow {\text{Variance}} = 15 - 13.4 \\
\Rightarrow {\text{Variance}} = 1.6 \\
\]
Hence, the variance of the given data is \[1.6\].
Note: In solving these types of questions, students should know the formulae of mean, median, and mode. The question is really simple, students should note down the values from the problem really carefully, else the answer can be wrong. One should take care while finding the mean, variance, and avoid calculation mistakes.
Complete step by step answer:
We will now form a table to find the value of the product \[\sum {{P_i}} \] for the mean.
\[X = {x_i}\] | \[P\left( {X = {x_i}} \right)\] |
1 | \[k\] |
2 | \[2k\] |
3 | \[3k\] |
4 | \[4k\] |
5 | \[5k\] |
\[\sum P = 15k\] |
We know that the sum of \[P\left( {X = {x_i}} \right)\]is always equals to 1, that is, \[\sum {{P_i}} = 1\].
Substituting the values of \[\sum {{P_i}} \] to find the value of \[k\] from the above formula, we get
\[ \Rightarrow 15k = 1\]
Dividing the above equation by 15 on both sides, we get
\[
\Rightarrow \dfrac{{15k}}{{15}} = \dfrac{1}{{15}} \\
\Rightarrow k = \dfrac{1}{{15}} \\
\]
Hence, the value of \[k\] is \[\dfrac{1}{{15}}\].
We will now find the value of \[{{\text{X}}^2}{\text{P}}\] \[{\text{XP}}\]from the above table for median and mode.
\[{\text{X}}\] | \[{\text{P}}\] | \[{\text{XP}}\] | \[{{\text{X}}^2}{\text{P}}\] |
1 | \[k\] | \[k\] | \[k\] |
2 | \[2k\] | \[4k\] | \[8k\] |
3 | \[3k\] | \[9k\] | \[27k\] |
4 | \[4k\] | \[16k\] | \[64k\] |
5 | \[5k\] | \[25k\] | \[125k\] |
\[\sum {\text{P}} = 15k\] | \[\sum {{\text{XP}}} = 55k\] | \[\sum {{{\text{X}}^2}{\text{P}}} = 225k\] |
Substituting the value of \[k\] in \[\sum {{\text{XP}}} \] and \[\sum {{{\text{X}}^2}{\text{P}}} \], we get
\[
\Rightarrow \sum {{\text{XP}}} = 55 \times \dfrac{1}{{15}} \\
\Rightarrow \sum {{\text{XP}}} = 3.66 \\
\]
\[
\Rightarrow \sum {{{\text{X}}^2}{\text{P}}} = 225 \times \dfrac{1}{{15}} \\
\Rightarrow \sum {{{\text{X}}^2}{\text{P}}} = 15 \\
\]
Therefore, the mean of the given data is \[3.66\].
We know that the formula to calculate variance is \[{\text{Variance}} = \sum {{{\text{X}}^2}{\text{P}}} - {\left( {\sum {{\text{XP}}} } \right)^2}\].
Substituting these values in the above formula of variance, we get
\[
\Rightarrow {\text{Variance}} = 15 - {\left( {3.66} \right)^2} \\
\Rightarrow {\text{Variance}} = 15 - 13.4 \\
\Rightarrow {\text{Variance}} = 1.6 \\
\]
Hence, the variance of the given data is \[1.6\].
Note: In solving these types of questions, students should know the formulae of mean, median, and mode. The question is really simple, students should note down the values from the problem really carefully, else the answer can be wrong. One should take care while finding the mean, variance, and avoid calculation mistakes.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE