In a pizza takeout restaurant, the following probability distribution was obtained. The random variable x represents the number of toppings for a large pizza. Find the mean and standard deviation. $$\begin{array} { c | c }
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.30 \
1 & 0.40 \
2 & 0.20 \
3 & 0.06 \
4 & 0.04
\end{array}$$
A)mean: 1.14; standard deviation: 1.04
B)mean: 1.54; standard deviation: 1.30
C)mean: 1.30; standard deviation: 2.38
D)mean: 1.30; standard deviation: 1.54

Question

In a pizza takeout restaurant, the following probability distribution was obtained. The random variable x represents the number of toppings for a large pizza. Find the mean and standard deviation. $$\begin{array} { c | c } 
\mathrm { x } & \mathrm { P } ( \mathrm { x } ) \
\hline 0 & 0.30 \
1  & 0.40 \
2 & 0.20 \
3 & 0.06 \
4 & 0.04
\end{array}$$
A)mean: 1.14; standard deviation: 1.04
B)mean: 1.54; standard deviation: 1.30
C)mean: 1.30; standard deviation: 2.38
D)mean: 1.30; standard deviation: 1.54

Quizplus · Accepted Answer

The mean (μ) is calculated as Σ[x*P(x)] = (0*0.30) + (1*0.40) + (2*0.20) + (3*0.06) + (4*0.04) = 1.14. The variance (σ^2) is Σ[(x-μ)^2*P(x)], and the standard deviation (σ) is the square root of the variance, which calculates to approximately 1.04.

In a Pizza Takeout Restaurant, the Following Probability Distribution Was