Question

Find the mean of the first 10 prime no.

Answer 1

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.

Complete step-by-step answer:
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 $=\ \dfrac{129}{10}\ =\ 12.9$
Hence, the answer is 12.9.

Note: Just keep in mind that 1 is not a prime number and we ignore it for verifying ‘1’ as a prime number.
And another thing to keep in mind that 2 is the smallest prime number which is even too. So, the only even number which is prime is 2. So, do not ignore it while counting the first 10 prime numbers; be careful with ‘2’ as well.
We don’t need to check any even number except 2 for proving it to be a prime number, as 2 will be one factor of it always.
Add the numbers very carefully for calculating the mean and don’t leave or miss any prime number. So, be careful with it as well.