Question

Gus and his friends had a contest to see who could throw a football the farthest. The length of each friend's longest throw, in yards, is shown below. 22, 25, 30, 18, 27, 25, 19, 26, 33, 20 Find the range, median, the first and third quartiles, and the interquartile range.

Answer 1

Given :

• Gus and his friends had a contest to see who could throw a football the farthest.
• The length of each friend's longest throw, in yards, is shown below. 22, 25, 30, 18, 27, 25, 19, 26, 33, 20.

To Find :

• Range.
• Median.
• The first quartiles.
• The third quartiles.
• The interquartile range.

Solution :

Range :

We know that range is the difference between the highest and lowest value.

So, first we have to arrange the values in ascending order.

In ascending order: 18, 19, 20, 22, 25, 25, 26, 27, 30, 33.

Now,

• Lowest Value = 18
• Highest Value = 33

⇒ Range = Highest Value - Lowest Value

⇒ Range = 33 - 18

⇒ Range = 15

∴ Range = 15.

The range is 15.

Median :

Here we can see that,

• no of observations (n) = 10 (even)

So, formula of median for even observations is,

$\bf Median=\left[\dfrac{\left(\dfrac{n}{2}\right)^{th\ observation}+\left(\dfrac{n}{2}+1\right)^{th\ observation}}{2}\right]$

where,

• n = 10

$\\ :\implies\sf Median=\left[\dfrac{\left(\dfrac{10}{2}\right)^{th\ observation}+\left(\dfrac{10}{2}+1\right)^{th\ observation}}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{\left(\dfrac{\cancel{10}}{\not{2}}\right)^{th\ observation}+\left(\dfrac{\cancel{10}}{\not{2}}+1\right)^{th\ observation}}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{\left(5\right)^{th\ observation}+\left(5+1\right)^{th\ observation}}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{\left(5\right)^{th\ observation}+\left(6\right)^{th\ observation}}{2}\right]$

From ascending order,

$\\ :\implies\sf Median=\left[\dfrac{25+25}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{50}{2}\right]$

$\\ :\implies\sf Median=\left[\cancel{\dfrac{50}{2}}\right]$

$\\ :\implies\sf Median=\left[25\right]$

$\\ \therefore\boxed{\bf Median=25.}$

Median is 25.

First Quartile (Q₁) :

For even observations formula of Q₁ is,

$\bf Q_1=\dfrac{n}{4}$

where,

• n = 10

$\\ :\implies\sf Q_1=\dfrac{10}{4}$

$\\ \therefore\boxed{\bf Q_1=2.5}$

First Quartile (Q₁) is 2.5

Third Quartile (Q₃) :

For even observations formula of Q₃ is,

$\bf Q_3=\dfrac{3n}{4}$

where,

• n = 10

$\\ :\implies\sf Q_3=\dfrac{3(10)}{4}$

$\\ :\implies\sf Q_3=\dfrac{30}{4}$

$\\ \therefore\boxed{\bf Q_3=7.5}$

Third Quartile (Q₃) is 7.5

Inner-Quartile Range (Q₂) :

⇒ Q₂ = Q₃ - Q₁

⇒ Q₂ = 7.5 - 2.5

⇒ Q₂ = 5

∴ Q₂ = 5.

Inner-Quartile Range (Q₂) is 5.

Answer 2

Step-by-step explanation:

Given :

Gus and his friends had a contest to see who could throw a football the farthest.

The length of each friend's longest throw, in yards, is shown below. 22, 25, 30, 18, 27, 25, 19, 26, 33, 20.

To Find :

Range.

Median.

The first quartiles.

The third quartiles.

The interquartile range.

Solution :

Range :

We know that range is the difference between the highest and lowest value.

So, first we have to arrange the values in ascending order.

In ascending order: 18, 19, 20, 22, 25, 25, 26, 27, 30, 33.

Now,

Lowest Value = 18

Highest Value = 33

⇒ Range = Highest Value - Lowest Value

⇒ Range = 33 - 18

⇒ Range = 15

∴ Range = 15.

The range is 15.

Median :

Here we can see that,

no of observations (n) = 10 (even)

So, formula of median for even observations is,

$\bf Median=\left[\dfrac{\left(\dfrac{n}{2}\right)^{th\ observation}+\left(\dfrac{n}{2}+1\right)^{th\ observation}}{2}\right]$

where,

n = 10

$\\ :\implies\sf Median=\left[\dfrac{\left(\dfrac{10}{2}\right)^{th\ observation}+\left(\dfrac{10}{2}+1\right)^{th\ observation}}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{\left(\dfrac{\cancel{10}}{\not{2}}\right)^{th\ observation}+\left(\dfrac{\cancel{10}}{\not{2}}+1\right)^{th\ observation}}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{\left(5\right)^{th\ observation}+\left(5+1\right)^{th\ observation}}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{\left(5\right)^{th\ observation}+\left(6\right)^{th\ observation}}{2}\right]$

From ascending order,

$\\ :\implies\sf Median=\left[\dfrac{25+25}{2}\right]$

$\\ :\implies\sf Median=\left[\dfrac{50}{2}\right]$

$\\ :\implies\sf Median=\left[\cancel{\dfrac{50}{2}}\right]$

$\\ :\implies\sf Median=\left[25\right]$

$\\ \therefore\boxed{\bf Median=25.}$

Median is 25.

First Quartile (Q₁) :

For even observations formula of Q₁ is,

$\bf Q_1=\dfrac{n}{4}$

where,

n = 10

$\\ :\implies\sf Q_1=\dfrac{10}{4}$

$\\ \therefore\boxed{\bf Q_1=2.5}$

First Quartile (Q₁) is 2.5

Third Quartile (Q₃) :

For even observations formula of Q₃ is,

$\bf Q_3=\dfrac{3n}{4}$

where,

n = 10

$\\ :\implies\sf Q_3=\dfrac{3(10)}{4}$

$\\ :\implies\sf Q_3=\dfrac{30}{4}$

$\\ \therefore\boxed{\bf Q_3=7.5}$

Third Quartile (Q₃) is 7.5

Inner-Quartile Range (Q₂) :

⇒ Q₂ = Q₃ - Q₁

⇒ Q₂ = 7.5 - 2.5

⇒ Q₂ = 5

∴ Q₂ = 5.

Inner-Quartile Range (Q₂) is 5.