Question

describe the shape of the histogram.
skewed right
skewed left
roughly symmetric
find the mean.
$mu =$
find the variance.
$sigma^{2}=$
find the standard deviation. round to four decimal places.
$sigma=$

Explanation:

Step1: Determine shape of histogram

The tail of the histogram extends to the right (higher - value side), so it is skewed right.

Step2: Assume probabilities for each value

Let's assume the probabilities for \(x = 0,1,2,3,4,5,6\) are \(p_0,p_1,p_2,p_3,p_4,p_5,p_6\) based on the heights of the bars in the histogram. Since the heights are not given numerically, for the sake of calculating mean \(\mu=\sum_{i = 0}^{6}x_ip_i\), if we assume the following (from visual estimate): \(p_0 = 0,p_1=0,p_2 = 0.02,p_3=0.08,p_4 = 0.22,p_5=0.28,p_6=0.4\). Then \(\mu=2\times0.02 + 3\times0.08+4\times0.22 + 5\times0.28+6\times0.4= 0.04+0.24 + 0.88+1.4 + 2.4=5\).

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

The formula for variance is \(\sigma^{2}=\sum_{i = 0}^{6}(x_i-\mu)^2p_i\). \((2 - 5)^2\times0.02+(3 - 5)^2\times0.08+(4 - 5)^2\times0.22+(5 - 5)^2\times0.28+(6 - 5)^2\times0.4=9\times0.02 + 4\times0.08+1\times0.22+0\times0.28 + 1\times0.4=0.18+0.32+0.22+0 + 0.4 = 1.12\).

Step4: Calculate standard deviation \(\sigma\)

The standard deviation \(\sigma=\sqrt{\sigma^{2}}\). So \(\sigma=\sqrt{1.12}\approx1.0583\).

Answer:

Describe the shape of the histogram: Skewed right
\(\mu = 5\)
\(\sigma^{2}=1.12\)
\(\sigma\approx1.0583\)