Question

exd and statistics problem set

1. using everyday language, briefly explain what the mean, or average, of a population tells us.
2. using everyday language, briefly explain what the standard deviation tells us.
3. use the histograms below to identify which population has a greater standard deviation. justify your choice.

population x
population y

1. a researcher is interested in comparing the height of two populations of plants. her data, in centimeters, are below. calculate the mean, standard deviation, and standard error for the height of the plants in population b to the nearest tenth. record your answer in the chart.

	population a	population b
	7.3	9.3
	9.2	8.8
	8.1	6.4
	8.0	8.9
mean	8.2
standard deviation	0.7	1.3
sem	0.3
2*sem

Explanation:

Step1: Define mean

The mean of a population is the sum of all values in the population divided by the number of values. It gives a central - value representing the 'typical' value in the population. For example, if we have the ages of all people in a town, the mean age tells us the average age of people in that town.

Step2: Define standard deviation

Standard deviation measures how spread out the values in a population are. A low standard deviation means that the values tend to be close to the mean, while a high standard deviation means the values are more spread out from the mean. For instance, if we measure the heights of students in a class, a low standard deviation would mean most students have similar heights, and a high one would mean there is a wide range of heights.

Step3: Compare histograms

For Population X, the data is more clustered around the center (around 100 on the measure axis), while for Population Y, the data is more spread out across the range of measures. Since standard deviation measures spread, Population Y has a greater standard deviation.

Step4: Calculate mean for Population B

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the individual data - points and $n$ is the number of data - points. For Population B, $n = 5$, $x_1=9.6$, $x_2 = 9.3$, $x_3=8.8$, $x_4=6.4$, $x_5=8.9$. $\sum_{i = 1}^{5}x_{i}=9.6 + 9.3+8.8 + 6.4+8.9=43$. So, $\bar{x}=\frac{43}{5}=8.6$.

Step5: Calculate standard deviation for Population B

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n - 1}}$.
$(x_1-\bar{x})=(9.6 - 8.6)=1$, $(x_2-\bar{x})=(9.3 - 8.6)=0.7$, $(x_3-\bar{x})=(8.8 - 8.6)=0.2$, $(x_4-\bar{x})=(6.4 - 8.6)=-2.2$, $(x_5-\bar{x})=(8.9 - 8.6)=0.3$.
$\sum_{i = 1}^{5}(x_{i}-\bar{x})^2=1^2+0.7^2 + 0.2^2+(-2.2)^2+0.3^2=1 + 0.49+0.04 + 4.84+0.09 = 6.46$.
$s=\sqrt{\frac{6.46}{4}}\approx1.3$.

Step6: Calculate standard error for Population B

The formula for the standard error $SEM=\frac{s}{\sqrt{n}}$. Since $s\approx1.3$ and $n = 5$, $SEM=\frac{1.3}{\sqrt{5}}\approx0.6$.

Step7: Calculate $2*SEM$ for Population B

$2*SEM=2\times0.6 = 1.2$.

Answer:

	Population A	Population B
Standard deviation	0.7	1.3
SEM	0.3	0.6
2*SEM	0.6	1.2