What is the mean, median, and mode of 31, 28, 30, 31, 30?
Answer
544.2k+ views
Hint: We first explain the concepts of mean, median and mode. The formulae to calculate mean and the median term are then determined. We shall use the formulae on the given set of data and obtain the answer to the above problem.
Complete step-by-step solution:
In order to solve this question, let us first define what mean, median and mode are. Mean for a set of data represents the average value for the given set of data. This can be obtained by taking the sum of all the terms divided by the total number of terms.
$\Rightarrow Mean=\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+\ldots +{{x}_{n}}}{n}$
Here, n represents the number of elements in the set of data.
Median is the middle number in the set of data. It is considered only for a set of data arranged in ascending or descending order. The formula for the term can be calculated as:
$\Rightarrow Median=\text{Value of }{{\left( \dfrac{n+1}{2} \right)}^{th}}\text{ term}$
The above is for data containing an odd number of elements, that is if n is odd. For n being even, the formula is,
$\Rightarrow Median=\text{Value of }{{\left( \dfrac{n}{2} \right)}^{th}}\text{ and }{{\left( \dfrac{n+2}{2} \right)}^{th}}\text{ term}$
Mode is defined as the value whose frequency is the most in a given set of data. It is the value that occurs the greatest number of times in the given data set.
Using the above concepts, we calculate mean, median, and mode of 31, 28, 30, 31, 30. The mean for the given data can be written as
$\Rightarrow \overline{x}=\dfrac{31+28+30+31+30}{5}$
Adding all the terms in the numerator,
$\Rightarrow \overline{x}=\dfrac{150}{5}$
Dividing the numerator by the denominator,
$\Rightarrow \overline{x}=30$
Hence, the mean for the given data is 30.
Now to calculate the mean, we first need to arrange the data in ascending order. Arranging it in ascending order gives us: 28, 30, 30, 31, 31. Now since there are an odd number of terms, n is 5, we use the first formula.
$\Rightarrow Median=\text{Value of }{{\left( \dfrac{5+1}{2} \right)}^{th}}\text{ term}$
Simplifying the value in brackets,
$\Rightarrow Median=\text{Value of }{{\text{3}}^{rd}}\text{ term}$
Value of the third term in the given order is 30.
Hence, the median is 30.
Mode can be calculated as the number that occurs most frequently and the term that occurs the greatest number of times is 30 which is twice. Even 31 occurs twice, therefore, the mode is 30 and 31.
Hence, the mean is 30, median is 30 and mode contains two terms 30 and 31.
Note: It is important to know the basic concepts of mean, median and mode. These form the basics of mathematical statistics and we need to know them to solve such questions easily. We need to note that to calculate the median, we need not arrange the data only in an ascending order. We can also arrange it in descending order and calculate the value of the median.
Complete step-by-step solution:
In order to solve this question, let us first define what mean, median and mode are. Mean for a set of data represents the average value for the given set of data. This can be obtained by taking the sum of all the terms divided by the total number of terms.
$\Rightarrow Mean=\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+\ldots +{{x}_{n}}}{n}$
Here, n represents the number of elements in the set of data.
Median is the middle number in the set of data. It is considered only for a set of data arranged in ascending or descending order. The formula for the term can be calculated as:
$\Rightarrow Median=\text{Value of }{{\left( \dfrac{n+1}{2} \right)}^{th}}\text{ term}$
The above is for data containing an odd number of elements, that is if n is odd. For n being even, the formula is,
$\Rightarrow Median=\text{Value of }{{\left( \dfrac{n}{2} \right)}^{th}}\text{ and }{{\left( \dfrac{n+2}{2} \right)}^{th}}\text{ term}$
Mode is defined as the value whose frequency is the most in a given set of data. It is the value that occurs the greatest number of times in the given data set.
Using the above concepts, we calculate mean, median, and mode of 31, 28, 30, 31, 30. The mean for the given data can be written as
$\Rightarrow \overline{x}=\dfrac{31+28+30+31+30}{5}$
Adding all the terms in the numerator,
$\Rightarrow \overline{x}=\dfrac{150}{5}$
Dividing the numerator by the denominator,
$\Rightarrow \overline{x}=30$
Hence, the mean for the given data is 30.
Now to calculate the mean, we first need to arrange the data in ascending order. Arranging it in ascending order gives us: 28, 30, 30, 31, 31. Now since there are an odd number of terms, n is 5, we use the first formula.
$\Rightarrow Median=\text{Value of }{{\left( \dfrac{5+1}{2} \right)}^{th}}\text{ term}$
Simplifying the value in brackets,
$\Rightarrow Median=\text{Value of }{{\text{3}}^{rd}}\text{ term}$
Value of the third term in the given order is 30.
Hence, the median is 30.
Mode can be calculated as the number that occurs most frequently and the term that occurs the greatest number of times is 30 which is twice. Even 31 occurs twice, therefore, the mode is 30 and 31.
Hence, the mean is 30, median is 30 and mode contains two terms 30 and 31.
Note: It is important to know the basic concepts of mean, median and mode. These form the basics of mathematical statistics and we need to know them to solve such questions easily. We need to note that to calculate the median, we need not arrange the data only in an ascending order. We can also arrange it in descending order and calculate the value of the median.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

