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

mean: 1.14; standard deviation: 1.04

Explanation:

Answer 2

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

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=√(\sum p_i(X_i-M)^2)\\\\s=√(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=√(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=√(0.3(1.2996)+0.4(0.0196)+0.2(0.7396)+0.06(3.4596)+0.04(8.1796))\\\\s=√(0.3899+0.0078+0.1479+0.2076+0.3272)\\\\ s=√( 1.0804 )\\\\s\approx 1.04`