Question

five numbers 4;8;10;x;y
have a mean of 10 and a standard deviation of 4

find the values of x and y
use the standard deviation formula to do so

Answer 1

Answer:

12 and 16

Step-by-step explanation:

The formula for the mean of a set of data is:

$\overline{x}=\dfrac{\displaystyle \sum x}{n}$

where:

• $\overline{x}$ is the mean.
• x refers to each individual data point in the dataset.
• ∑x represents the sum of all the data points.
• n is the total number of data points.

In this case, the five numbers are 4, 8, 10, x and y, and the mean is 10.

Substituting these values into the mean formula gives:

$\dfrac{4+8+10+x+y}{5}=10$

Simplify and isolate y:

$\dfrac{22+x+y}{5}=10$

$22+x+y=50$

$x+y=28$

$y=28-x$

$\hrulefill$

The formula for the standard deviation of a set of data is:

$\sigma=\sqrt{\dfrac{\displaystyle \sum x^2}{n}-\overline{x}^2}$

where:

• σ is the standard deviation.
• $\overline{x}$ is the mean.
• x refers to each individual data point in the dataset.
• ∑x² represents the sum of the squares of all the data points.
• n is the total number of data points.

Given that the mean is 10 and the standard deviation is 4, then:

$\sqrt{\dfrac{4^2+8^2+10^2+x^2+y^2}{5}-10^2}=4$

Simplify:

$\dfrac{16+64+100+x^2+y^2}{5}-100=16$

$\dfrac{180+x^2+y^2}{5}=116$

$180+x^2+y^2=580$

$x^2+y^2=400$

$\hrulefill$

Now we have created a system of two equations:

$\begin{cases}y=28-x\\x^2+y^2=400\end{cases}$

Substitute the first equation into the second equation and solve for x:

$\begin{aligned}x^2+(28-x)^2&=400\\x^2+784-56x+x^2&=400\\2x^2-56x+384&=0\\2(x^2-28x+192)&=0\\x^2-28x+192&=0\\(x-12)(x-16)&=0\\\\x-12&=0 \implies x=12\\x-16&=0 \implies x=16\end{aligned}$

If we substitute x = 12 into y = 28 - x, we get y = 16.

Similarly, if we substitute x = 16 into y = 28 - x, we get y = 12.

Therefore, the values of x and y are 12 and 16.