# Probability Distribution Mean (Expectation), Variance :: Problems

 Problem Back to Problems Page
A bakery has the following schedule of daily demand for cakes.
 No. of cakes demanded (in hundreds) Probability 0 1 2 3 4 5 6 7 8 9 0.02 0.07 0.09 0.12 0.2 0.2 0.18 0.1 0.01 0.01
Find the expected number of cakes demanded per day

 [Expectation: 436; Variance: 3.5304 ; Standard Deviation : +1.879]

 Solution

Let "x" represent the number of cakes demanded (in hundreds) per day.

The values carried by "x" are 0, 1, 2, ... ,9 ==> "x" is finite.

Sum of the probabilities of the various values of "x"

= 0.02 + 0.07 + 0.09 + 0.12 + 0.20 + 0.20 + 0.18 + 0.10 + 0.01 + 0.01
= 1

The values of "x" represent the values in the range of a discrete random variable
⇒ "X" is a discrete random variable with range = {0, 1, 2, ... .9}

The given distribution therefore, represents a discrete probability distribution of "X"

Calculations for Mean and Standard Deviations

x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 0 0.02 0 0 0 1 0.07 0.07 1 0.07 2 0.09 0.18 4 0.36 3 0.12 0.36 9 1.08 4 0.20 0.80 16 3.20 5 0.20 1.00 25 5.00 6 0.18 1.08 36 6.48 7 0.10 0.70 49 4.90 8 0.01 0.08 64 0.64 9 0.01 0.09 81 0.81 1.00 4.36 22.54

Since the given distribution is the probability distribution of a discrete random variable "X", Σ p = 1.

From the probability distribution:

Expected (Mean) Number of cakes demanded per day
⇒ Expectation/Mean of the distribution
 ⇒ E (x) (Or) x = Σ px = 4.36 (in hundreds) = 436
Variance of the number of caked demanded per day
⇒ Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 22.54 − (4.36)2 = 22.54 − 19.0096 = 3.5304
Standard Deviation of the number of cakes demanded per day
⇒ standard deviation of the distribution
 ⇒ SD (x) = + √ Var (x) ⇒ SD (x) = + √ 3.5304 ⇒ SD (x) = + 1.879

 Credit : Vijayalakshmi Desu