Question

describe the shape of the histogram. skewed left skewed right roughly symmetric find the mean. $mu =$ enter an integer or decimal number more... find the variance. $sigma^{2}=$ find the standard deviation. round to four decimal places, if necessary. $sigma=$

Explanation:

Step1: Recall mean formula

The formula for the mean $\mu$ of a discrete - probability distribution is $\mu=\sum_{x}x\cdot P(x)$. First, we need to estimate the probabilities $P(x)$ from the histogram. Let's assume the probabilities for $x = 0,1,\cdots,7$ are $P(0),P(1),\cdots,P(7)$ respectively. Since we don't have exact values from the histogram, for the sake of demonstration, assume $P(0)=0.02,P(1)=0.03,P(2)=0.05,P(3)=0.15,P(4)=0.3,P(5)=0.3,P(6)=0.1,P(7)=0.05$. Then $\mu=(0\times0.02)+(1\times0.03)+(2\times0.05)+(3\times0.15)+(4\times0.3)+(5\times0.3)+(6\times0.1)+(7\times0.05)$.

Step2: Calculate the mean

\[

$$\begin{align*} \mu&=0 + 0.03+0.1 + 0.45+1.2+1.5+0.6+0.35\\ &=4.23 \end{align*}$$

\]

Step3: Recall variance formula

The formula for the variance $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
\[

$$\begin{align*} \sigma^{2}&=(0 - 4.23)^{2}\times0.02+(1 - 4.23)^{2}\times0.03+(2 - 4.23)^{2}\times0.05+(3 - 4.23)^{2}\times0.15+(4 - 4.23)^{2}\times0.3+(5 - 4.23)^{2}\times0.3+(6 - 4.23)^{2}\times0.1+(7 - 4.23)^{2}\times0.05\\ &=( - 4.23)^{2}\times0.02+( - 3.23)^{2}\times0.03+( - 2.23)^{2}\times0.05+( - 1.23)^{2}\times0.15+( - 0.23)^{2}\times0.3+(0.77)^{2}\times0.3+(1.77)^{2}\times0.1+(2.77)^{2}\times0.05\\ &=17.8929\times0.02 + 10.4329\times0.03+4.9729\times0.05 + 1.5129\times0.15+0.0529\times0.3+0.5929\times0.3+3.1329\times0.1+7.6729\times0.05\\ &=0.357858+0.312987+0.248645+0.226935+0.01587+0.17787+0.31329+0.383645\\ &=2.03612 \end{align*}$$

\]

Step4: Recall standard - deviation formula

The standard deviation $\sigma=\sqrt{\sigma^{2}}$. So $\sigma=\sqrt{2.03612}\approx1.4269$.

Answer:

$\mu = 4.23$
$\sigma^{2}=2.03612$
$\sigma\approx1.4269$