
The mean of the following data is 18.75 then the value of p:
Class mark$({x_i})$ 10 15 p 25 30 Frequency $({f_i})$ 5 10 7 8 2
| Class mark$({x_i})$ | 10 | 15 | p | 25 | 30 |
| Frequency $({f_i})$ | 5 | 10 | 7 | 8 | 2 |
Answer
554.7k+ views
Hint: First of all finding the mean and solving the arithmetic mean gives a certain value and which is equated to the given value of the mean. Mean is the arithmetic mean or the average of the given data. The arithmetic mean is the ratio of all the observations to the total number of observations. Here observing how the class marks are repeating with a particular frequency and then solving accordingly.
Complete step-by-step solution:
Here the class marks are given and how many times they are repeating is also given.
The total number of observations is given by: $\sum {{f_i}} $ as it gives the summation of all the observations from which the total number of observations can be obtained.
The sum of the observations is given by : $\sum {{f_i}{x_i}} $, as each observation is repeated for a particular frequency hence multiplying the observation with that particular frequency.
$\therefore $The average here is given by:
$ \Rightarrow {x_n} = \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }}$
Here $\sum\limits_i {{f_i}} = 5 + 10 + 7 + 8 + 2$
And $\sum\limits_i {{f_i}{x_i} = 5(10) + 10(15) + 7(p) + 8(25) + 2(30)} $
Now finding the mean by substituting all the above expressions in the mean expression:
$ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{5(10) + 10(15) + 7(p) + 8(25) + 2(30)}}{{5 + 10 + 7 + 8 + 2}}$
\[ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{50 + 150 + 7p + 200 + 60}}{{32}}\]
\[ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{460 + 7p}}{{32}}\]
Now given the mean of the data, hence equating the above expression to the given mean:
\[ \Rightarrow \dfrac{{460 + 7p}}{{32}} = 18.75\]
\[ \Rightarrow 460 + 7p = 600\]
\[ \Rightarrow 7p = 600 - 460\]
\[ \Rightarrow 7p = 140\]
\[ \Rightarrow p = 20\]
$\therefore p = 20$
The value of p is 20.
Note: Remember that here while solving the mean the total number of observations is considered as the sum of the frequency numbers, as in for how many times a particular observation is repeating, is taken to consideration to summate all the frequency values to get the total number of observations.
Complete step-by-step solution:
Here the class marks are given and how many times they are repeating is also given.
The total number of observations is given by: $\sum {{f_i}} $ as it gives the summation of all the observations from which the total number of observations can be obtained.
The sum of the observations is given by : $\sum {{f_i}{x_i}} $, as each observation is repeated for a particular frequency hence multiplying the observation with that particular frequency.
$\therefore $The average here is given by:
$ \Rightarrow {x_n} = \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }}$
Here $\sum\limits_i {{f_i}} = 5 + 10 + 7 + 8 + 2$
And $\sum\limits_i {{f_i}{x_i} = 5(10) + 10(15) + 7(p) + 8(25) + 2(30)} $
Now finding the mean by substituting all the above expressions in the mean expression:
$ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{5(10) + 10(15) + 7(p) + 8(25) + 2(30)}}{{5 + 10 + 7 + 8 + 2}}$
\[ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{50 + 150 + 7p + 200 + 60}}{{32}}\]
\[ \Rightarrow \dfrac{{\sum\limits_i {{f_i}{x_i}} }}{{\sum\limits_i {{f_i}} }} = \dfrac{{460 + 7p}}{{32}}\]
Now given the mean of the data, hence equating the above expression to the given mean:
\[ \Rightarrow \dfrac{{460 + 7p}}{{32}} = 18.75\]
\[ \Rightarrow 460 + 7p = 600\]
\[ \Rightarrow 7p = 600 - 460\]
\[ \Rightarrow 7p = 140\]
\[ \Rightarrow p = 20\]
$\therefore p = 20$
The value of p is 20.
Note: Remember that here while solving the mean the total number of observations is considered as the sum of the frequency numbers, as in for how many times a particular observation is repeating, is taken to consideration to summate all the frequency values to get the total number of observations.
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