
An aircraft has 120 passenger seats. The number of seats occupied during 100 fights is given below:
No. of Seats 100-104 104-108 108-112 112-116 116-120 Frequency 15 20 32 18 15
Determine the mean of seats occupied over the flights.
| No. of Seats | 100-104 | 104-108 | 108-112 | 112-116 | 116-120 |
| Frequency | 15 | 20 | 32 | 18 | 15 |
Answer
505.2k+ views
Hint: The question is related to the statistics topic. Here we have to determine the mean of seats occupied over the flights using the given table of grouped data. To find the mean we have a formula i.e., \[\overline X = \dfrac{{\sum {{f_i}{x_i}} }}{N}\] , where \[{f_i}\] is frequency, \[{x_i}\] is the midpoint of class-interval and \[N = \sum f \] on substituting all values in formula we get the required solution.
Complete step by step answer:
On observing the question is in the form of grouped data where Grouped data is data that has been organized into a frequency distribution. The direct method to find the mean of grouped data is: We have to take the midpoint of every class interval as \[{x_i}\] by using a formula \[{x_i} = \dfrac{{Upper\,limit - lower\,limit}}{2}\]. Next, multiply these values of \[{x_i}\] with their respective frequencies \[f\]. Take total and apply the formula \[\overline X = \dfrac{{\sum {{f_i}{x_i}} }}{N}\].
Where, \[\overline X \] is the mean of grouped data
\[\sum {{f_i}{x_i}} \]- sum of the product of mid term of class intervals and respective frequencies,
\[N = \sum f \]- The total sum of frequencies.
Now consider the table of given observations.
Now, consider the formula of mean \[\overline X \] is
\[ \Rightarrow \,\,\,\overline X = \dfrac{{\sum {{f_i}{x_i}} }}{N}\]
On substituting the values, we have
\[ \Rightarrow \,\,\,\overline X = \dfrac{{10992}}{{10}}\]
On division, we get
\[ \Rightarrow \,\,\,\overline X = 109.92\]
But seats cannot be in decimal, so the number of seats is 109.
Therefore, the mean number of seats occupied over the flights is 109.
Note: As we know, the Mean is the average of the numbers i.e., a calculated "central" value of a set of numbers. The mean of grouped data can also be find by another method called step deviation method by using a formula \[\overline X = A + \dfrac{{\sum {{f_i}{d_i}} }}{N}\] , where A is assumed mean and \[{d_i}\] is the deviations are taken from assumed mean.
Complete step by step answer:
On observing the question is in the form of grouped data where Grouped data is data that has been organized into a frequency distribution. The direct method to find the mean of grouped data is: We have to take the midpoint of every class interval as \[{x_i}\] by using a formula \[{x_i} = \dfrac{{Upper\,limit - lower\,limit}}{2}\]. Next, multiply these values of \[{x_i}\] with their respective frequencies \[f\]. Take total and apply the formula \[\overline X = \dfrac{{\sum {{f_i}{x_i}} }}{N}\].
Where, \[\overline X \] is the mean of grouped data
\[\sum {{f_i}{x_i}} \]- sum of the product of mid term of class intervals and respective frequencies,
\[N = \sum f \]- The total sum of frequencies.
Now consider the table of given observations.
| No. of seats | Frequency \[\left( f \right)\] | \[{x_i} = \dfrac{{Upper\,limit - lower\,limit}}{2}\] | \[{f_i}{x_i}\] |
| 100-104 | 15 | 102 | 1530 |
| 104-108 | 20 | 106 | 2120 |
| 108-112 | 32 | 110 | 3520 |
| 112-116 | 18 | 114 | 2052 |
| 116-120 | 15 | 118 | 1770 |
| \[N = \sum f = 100\] | \[\sum {{f_i}{x_i}} = 10992\] |
Now, consider the formula of mean \[\overline X \] is
\[ \Rightarrow \,\,\,\overline X = \dfrac{{\sum {{f_i}{x_i}} }}{N}\]
On substituting the values, we have
\[ \Rightarrow \,\,\,\overline X = \dfrac{{10992}}{{10}}\]
On division, we get
\[ \Rightarrow \,\,\,\overline X = 109.92\]
But seats cannot be in decimal, so the number of seats is 109.
Therefore, the mean number of seats occupied over the flights is 109.
Note: As we know, the Mean is the average of the numbers i.e., a calculated "central" value of a set of numbers. The mean of grouped data can also be find by another method called step deviation method by using a formula \[\overline X = A + \dfrac{{\sum {{f_i}{d_i}} }}{N}\] , where A is assumed mean and \[{d_i}\] is the deviations are taken from assumed mean.
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