We are given a dataset of body masses of 6 students: 35, 40, 45, 45, 50, $y^2$. We are given that the sample mean is 41. We need to find the value of $y^2$ (question 21) and then find the sample standard deviation (question 22).

1. Problem Description

We are given a dataset of body masses of 6 students: 35, 40, 45, 45, 50, $y^2$. We are given that the sample mean is

4

1. We need to find the value of $y^2$ (question 21) and then find the sample standard deviation (question 22).

2. Solution Steps

Question 21: Find $y^2$ if the sample mean is

4

1. The formula for the sample mean is:

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

In this case, we have:

$\bar{x} = \frac{35 + 40 + 45 + 45 + 50 + y^2}{6}$

We are given that $\bar{x} = 41$. So,

$41 = \frac{35 + 40 + 45 + 45 + 50 + y^2}{6}$

$41 \times 6 = 35 + 40 + 45 + 45 + 50 + y^2$

$246 = 215 + y^2$

$y^2 = 246 - 215$

$y^2 = 31$

Question 22: Find the sample standard deviation.

The sample standard deviation is given by the formula:

$s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}$

where $n$ is the number of data points, $x_i$ are the individual data points, and $\bar{x}$ is the sample mean.

In this case, $n = 6$, $\bar{x} = 41$, and the data points are 35, 40, 45, 45, 50,

3

1. First, calculate the squared differences from the mean:

$(35 - 41)^2 = (-6)^2 = 36$

$(40 - 41)^2 = (-1)^2 = 1$

$(45 - 41)^2 = (4)^2 = 16$

$(45 - 41)^2 = (4)^2 = 16$

$(50 - 41)^2 = (9)^2 = 81$

$(31 - 41)^2 = (-10)^2 = 100$

Now, sum these squared differences:

$\sum_{i=1}^{6} (x_i - \bar{x})^2 = 36 + 1 + 16 + 16 + 81 + 100 = 250$

Then divide by $n-1 = 6-1 = 5$:

$\frac{250}{5} = 50$

Finally, take the square root:

$s = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \approx 7.071$

3. Final Answer

$y^2 = 31$

Sample standard deviation $s = 5\sqrt{2} \approx 7.071$