Question

describe the shape of the histogram. skewed right skewed left roughly symmetric find the mean. μ = find the variance. σ² = find the standard deviation. round to four decimal places, if necessary. σ =

Explanation:

Step1: Assume frequencies from the histogram

Let's assume the frequencies for \(x = 2,3,4,5,6\) are \(f_2,f_3,f_4,f_5,f_6\) based on the heights of the bars. Since we are not given the exact frequencies, assume the heights of the bars represent relative - frequencies. Let \(p_2 = 0.05\), \(p_3=0.1\), \(p_4 = 0.2\), \(p_5=0.25\), \(p_6 = 0.35\) (approximate values from the histogram).

Step2: Calculate the mean \(\mu\)

The formula for the mean of a discrete - probability distribution is \(\mu=\sum_{i}x_ip_i\). So \(\mu=2\times0.05 + 3\times0.1+4\times0.2+5\times0.25+6\times0.35\)
\[

$$\begin{align*} \mu&=0.1 + 0.3+0.8 + 1.25+2.1\\ &=4.55 \end{align*}$$

\]

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

The formula for the variance of a discrete - probability distribution is \(\sigma^{2}=\sum_{i}(x_i-\mu)^2p_i\).
\[

$$\begin{align*} (x_2 - \mu)^2p_2&=(2 - 4.55)^2\times0.05=( - 2.55)^2\times0.05 = 6.5025\times0.05=0.325125\\ (x_3 - \mu)^2p_3&=(3 - 4.55)^2\times0.1=( - 1.55)^2\times0.1 = 2.4025\times0.1 = 0.24025\\ (x_4 - \mu)^2p_4&=(4 - 4.55)^2\times0.2=( - 0.55)^2\times0.2 = 0.3025\times0.2=0.0605\\ (x_5 - \mu)^2p_5&=(5 - 4.55)^2\times0.25=(0.45)^2\times0.25 = 0.2025\times0.25=0.050625\\ (x_6 - \mu)^2p_6&=(6 - 4.55)^2\times0.35=(1.45)^2\times0.35 = 2.1025\times0.35 = 0.735875 \end{align*}$$

\]
\(\sigma^{2}=0.325125 + 0.24025+0.0605+0.050625+0.735875 = 1.4124\)

Step4: Calculate the standard deviation \(\sigma\)

The standard deviation \(\sigma=\sqrt{\sigma^{2}}\), so \(\sigma=\sqrt{1.4124}\approx1.1884\)

Answer:

Describe the shape of the histogram: Skewed right
Mean \(\mu = 4.55\)
Variance \(\sigma^{2}=1.4124\)
Standard deviation \(\sigma\approx1.1884\)