To calculate the mean and standard deviation for the numbers $4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9$:

1. Mean:

The formula for the mean ($\mu$) is:

$\mu = \frac{1}{n} \sum_{i=1}^{n} x_i$

Where $x_i$ are the data points and $n$ is the number of data points.

Given: $n = 9, \quad \sum x_i = 4.1 + 4.2 + 4.3 + 4.4 + 4.5 + 4.6 + 4.7 + 4.8 + 4.9$

$\mu = \frac{4.1 + 4.2 + 4.3 + 4.4 + 4.5 + 4.6 + 4.7 + 4.8 + 4.9}{9}$

$\mu = \frac{40.5}{9} = 4.5$

2. Standard Deviation:

The formula for the standard deviation ($\sigma$) is:

$\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2}$

Using the mean $\mu = 4.5$:

Let's compute the variance first:

$\sigma^2 = \frac{1}{9} \left( (4.1 - 4.5)^2 + (4.2 - 4.5)^2 + (4.3 - 4.5)^2 + (4.4 - 4.5)^2 + (4.5 - 4.5)^2 + (4.6 - 4.5)^2 + (4.7 - 4.5)^2 + (4.8 - 4.5)^2 + (4.9 - 4.5)^2 \right)$

$\sigma^2 = \frac{1}{9} \left( (-0.4)^2 + (-0.3)^2 + (-0.2)^2 + (-0.1)^2 + 0^2 + (0.1)^2 + (0.2)^2 + (0.3)^2 + (0.4)^2 \right)$

$\sigma^2 = \frac{1}{9} \left( 0.16 + 0.09 + 0.04 + 0.01 + 0 + 0.01 + 0.04 + 0.09 + 0.16 \right) = \frac{1}{9} \times 0.6 = 0.0667$

Finally, the standard deviation:

$\sigma = \sqrt{0.0667} \approx 0.258$

Summary:

• Mean: $4.5$
• Standard deviation: $0.258$

Would you like any more details or have questions on this?

Here are 5 related questions:

1. How does standard deviation relate to data spread?
2. What is the difference between population and sample standard deviation?
3. How would you interpret a small versus large standard deviation?
4. How can outliers affect the standard deviation?
5. How does standard deviation help in comparing datasets?

Tip: When dealing with larger data sets, you can use statistical software or a calculator to compute mean and standard deviation efficiently.