Question

Kelsie works at a bicycle shop as a salesperson. She records the number of bicycles she sells daily. Here is the probability distribution of B = the number of bicycles Kelsie sells on a randomly selected day, and T the time she spends filling out daily sales reports. 2 B = # of bicycles sold T = time (minutes) Probability 40 20 10 0.05 0.15 0.50 0.30
a) Calculate E(B)

b) Calculate the mean time she spends filling out daily reports.

c) Calculate the standard deviation for both random variables.

Answer 1

Answer:

a) $E(B)= \sum_{i=1}^n B_i P(B_i) =0*0.3+1*0.5+2*0.15+ 3*0.05=0.95$

b) $E(T)= \sum_{i=1}^n T_i P(T_i) =10*0.3+20*0.5+30*0.15+ 40*0.05=19.5$

c) $E(B^2)= \sum_{i=1}^n B^2_i P(B_i) =0^2 *0.3+1^2 *0.5+2^2 *0.15+ 3^2 *0.05=1.55$

And the variance is given by:

$Var(B) = E(B^2) -[E(B)]^2 = 1.55- [0.95]^2 =0.6475$

And the deviation would be $Sd(B) = \sqrt{0.6475}=0.8047$

$E(T^2)= \sum_{i=1}^n T^2_i P(T_i) =10^2 *0.3+20^2 *0.5+30^2 *0.15+ 40^2 *0.05=445$

And the variance is given by:

$Var(T) = E(T^2) -[E(T)]^2 = 445- [19.5]^2 =64.75$

And the deviation would be $Sd(T) = \sqrt{64.75}=8.047$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".  

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).  

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem  

Part a

For this case we have the following info:

B            0            1             2              3

____________________________________

T           10           20          30            40

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P           0.3          0.5        0.15          0.05

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And we can calculate the expected value for the random variable B like this:

$E(B)= \sum_{i=1}^n B_i P(B_i) =0*0.3+1*0.5+2*0.15+ 3*0.05=0.95$

Part b

Similar to part a we can find the expected value for the random variable T like this:

$E(T)= \sum_{i=1}^n T_i P(T_i) =10*0.3+20*0.5+30*0.15+ 40*0.05=19.5$

Part c

In order to find the variance for B we need to calculate the second moment given by:

$E(B^2)= \sum_{i=1}^n B^2_i P(B_i) =0^2 *0.3+1^2 *0.5+2^2 *0.15+ 3^2 *0.05=1.55$

And the variance is given by:

$Var(B) = E(B^2) -[E(B)]^2 = 1.55- [0.95]^2 =0.6475$

And the deviation would be $Sd(B) = \sqrt{0.6475}=0.8047$

Similar for the random variable T we have:

$E(T^2)= \sum_{i=1}^n T^2_i P(T_i) =10^2 *0.3+20^2 *0.5+30^2 *0.15+ 40^2 *0.05=445$

And the variance is given by:

$Var(T) = E(T^2) -[E(T)]^2 = 445- [19.5]^2 =64.75$

And the deviation would be $Sd(B) = \sqrt{64.75}=8.047$