Question

In a pizza takeout restaurant, the following probability distribution was obtained for the number of toppings ordered on a large pizza. Find the mean and standard deviation for the random variable.

Answer 1

Answer:

The random variable (number of toppings ordered on a large pizza) has a mean of 1.14 and a standard deviation of 1.04.

Step-by-step explanation:

The question is incomplete:

The probability distribution is:

x   P(x)

0   0.30

1    0.40

2   0.20

3   0.06

4   0.04

The mean can be calculated as:

$M=\sum p_iX_i=0.3\cdot 0+0.4\cdot 1+0.2\cdot 2+0.06\cdot 3+0.04\cdot 4\\\\M=0+0.4+0.4+0.18+0.16\\\\M=1.14$

(pi is the probability of each class, Xi is the number of topping in each class)

The standard deviation is calculated as:

$s=\sqrt{\sum p_i(X_i-M)^2}\\\\s=\sqrt{0.3(0-1.14)^2+0.4(1-1.14)^2+0.2(2-1.14)^2+0.06(3-1.14)^2+0.04(4-1.14)^2}\\\\s=\sqrt{0.3(-1.14)^2+0.4(-0.14)^2+0.2(0.86)^2+0.06(1.86)^2+0.04(2.86)^2}\\\\ s=\sqrt{0.3(1.2996)+0.4(0.0196)+0.2(0.7396)+0.06(3.4596)+0.04(8.1796)}\\\\s=\sqrt{0.3899+0.0078+0.1479+0.2076+0.3272}\\\\ s=\sqrt{ 1.0804 }\\\\s\approx 1.04$