Question

use the histogram below that depicts population data to complete each of the following:
a. compute the mean and standard deviation.
b. classify the score of x = 10 as being an extreme or typical score in the distribution. justify your response.
c. classify the score of x = 2 as being an extreme or typical score in the distribution. justify your response.

Explanation:

Step1: Determine frequencies

Let's assume the frequencies from the histogram for \(x = 1,2,\cdots,10\) are \(f_1,f_2,\cdots,f_{10}\). From the histogram, if we assume \(f_1 = 1,f_2=1,f_3 = 1,f_6=3,f_7 = 1,f_8=1,f_9 = 1,f_{10}=1\) and \(f_4=f_5 = 0\). The total number of data - points \(N=\sum_{i = 1}^{10}f_i=1 + 1+1+0 + 0+3+1+1+1+1=10\).

Step2: Calculate the mean \(\bar{x}\)

The formula for the mean of a discrete - data set is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_if_i}{N}\).
\[

$$\begin{align*} \sum_{i = 1}^{10}x_if_i&=1\times1 + 2\times1+3\times1+4\times0+5\times0+6\times3+7\times1+8\times1+9\times1+10\times1\\ &=1+2 + 3+0+0+18+7+8+9+10\\ &=58 \end{align*}$$

\]
So, \(\bar{x}=\frac{58}{10}=5.8\).

Step3: Calculate the variance \(\sigma^{2}\)

The formula for the variance is \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2f_i}{N}\).
\[

$$\begin{align*} \sum_{i = 1}^{10}(x_i - 5.8)^2f_i&=(1 - 5.8)^2\times1+(2 - 5.8)^2\times1+(3 - 5.8)^2\times1+(4 - 5.8)^2\times0+(5 - 5.8)^2\times0+(6 - 5.8)^2\times3+(7 - 5.8)^2\times1+(8 - 5.8)^2\times1+(9 - 5.8)^2\times1+(10 - 5.8)^2\times1\\ &=(-4.8)^2\times1+(-3.8)^2\times1+(-2.8)^2\times1+(-1.8)^2\times0+(-0.8)^2\times0+(0.2)^2\times3+(1.2)^2\times1+(2.2)^2\times1+(3.2)^2\times1+(4.2)^2\times1\\ &=23.04+14.44 + 7.84+0+0 + 0.12+1.44+4.84+10.24+17.64\\ &=79.6 \end{align*}$$

\]
\(\sigma^{2}=\frac{79.6}{10}=7.96\), and the standard deviation \(\sigma=\sqrt{7.96}\approx2.82\).

Step4: Classify \(X = 10\)

We calculate the z - score for \(x = 10\) using the formula \(z=\frac{x-\bar{x}}{\sigma}\). \(z=\frac{10 - 5.8}{2.82}=\frac{4.2}{2.82}\approx1.49\). Since \(|z|\lt2\), \(X = 10\) is a typical score.

Step5: Classify \(X = 2\)

We calculate the z - score for \(x = 2\) using the formula \(z=\frac{x-\bar{x}}{\sigma}\). \(z=\frac{2 - 5.8}{2.82}=\frac{-3.8}{2.82}\approx - 1.35\). Since \(|z|\lt2\), \(X = 2\) is a typical score.

Answer:

a. Mean \(\bar{x}=5.8\), Standard deviation \(\sigma\approx2.82\)
b. The mean is the central - tendency measure and the standard deviation is a measure of the spread of the data.
c. \(X = 10\) is a typical score (\(z\approx1.49\) and \(|z|\lt2\)). \(X = 2\) is a typical score (\(z\approx - 1.35\) and \(|z|\lt2\))