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# Find the mean of the first 10 prime no.

Last updated date: 13th Jun 2024
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Hint: Prime number is a positive integer which has two factors only, first factor would be ‘1’ and other one is the number itself. Hence, find the first 10 prime numbers. ‘1’is not considered for prime numbers.

Mean of n numbers ${{a}_{1}},{{a}_{2}},{{a}_{3}}......{{a}_{n}}$ can be given by relation.
Mean $=\ \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+..........{{a}_{n}}}{n}$
We know that the prime number is a positive integer which has only two factors, first one is ‘1’ and other is the number itself.
So, here we need to find the sum of the first ten prime numbers, hence, we have to find the first 10 prime numbers and hence the summation.
So, let us observe the numbers starting from 2 as 1 is not included for the prime number because 1 has only factor i.e. 1. So, it is not fulfilling the criteria for prime numbers that prime numbers have two factors.
So, let’s start to find the prime numbers: -
So, factors of 2 $=\ 1,\ 2$
Here, 2 is a prime number as 2 has only two factors i.e. 1 and 2.
$\to$ factors of 3 $=\ 1,\ 3$
so, 3 is a prime number as well.
$\to$ factors of 4 $=\ 1,\ 2,\ 4$
Now, we can observe that ‘4’ has three factors so 4 is not a prime number.
Now, we need to observe that each even number has ‘2’ as a factor. So, we do not need to check the even numbers (except 2) for prime numbers. So, let’s check only for odd numbers (which are not divisible by 2).
$\to$ so, factors of 5 $=\ 1,\ 5$
Hence, 5 is ${{3}^{rd}}$prime number.
$\to$ factors of 7 $=\ 1,\ 7$
so, 7 is${{4}^{th}}$prime number.
$\to$ factors of 9 $=\ 1,\ 3,\ 9$
so, 9 is not a prime number.
$\to$ factors of 11 $=\ 1,\ 11$
so, 11 is the ${{5}^{th}}$ prime number.
$\to$ factors of 13 $=\ 1,\ 13$
so, 13 is the ${{6}^{th}}$ prime number.
$\to$ factors of 15 $=\ 1,\ 3,\ 5$
so, 15 is not a prime number.
$\to$ factors of 17 $=\ 1,\ 17$
so, 17 is the ${{7}^{th}}$ prime number.
$\to$ factors of 19 $=\ 1,\ 19$
so, 19 is the ${{8}^{th}}$ prime number.
$\to$ factors of 21 $=\ 1,\ 3,\ 7,\ 21$
so, 21 is not a prime number.
$\to$ factors of 23 $=\ 1,\ 23$
so, 23 is the ${{9}^{th}}$ prime number.
$\to$ factors of 25 $=\ 1,\ 5,\ 25$
so, 25 is not a prime number.
$\to$ factors of 27 $=\ 1,\ 3,\ 9,\ 27$
so, 27 is not a prime number.
$\to$ factors of 29 $=\ 1,\ 29$
so, 29 is the ${{10}^{th}}$ prime number.
Hence, first 10 prime numbers are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29
So, sum of all 10 numbers can be given as
Now, we know that mean of n numbers $\left( {{a}_{1}},{{a}_{2}},{{a}_{3}}..........{{a}_{n}} \right)$ can be given by relation
Mean $=\ \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+..........{{a}_{n}}}{n}$………………………………..(i)
Where n is the count of numbers of $\left( {{a}_{1}},{{a}_{2}},{{a}_{3}}..........{{a}_{n}} \right)$.
Hence, mean of first 10 prime numbers can be given as
Mean $=\ \dfrac{2+3+5+7+11+13+17+19+23+29}{10}$
Mean $=\ \dfrac{129}{10}\ =\ 12.9$