Answer:
b. E(X)= 44.61
c. V(X)= 2.9979
d. √V(X)= 1.73
Explanation:
Hello!
Given the variable X: Number of orders per week.
The raw data is in the attachment.
a.
The conditions that distribution should be met to be considered a distribution of probability are two:
-All probabilities should be between 0 and 1
-The sum of all probabilities should be 1, ∑p(x)=1
Looking at the given data, the first condition checks.
There are no observed probabilities below 0 or above one, so the first condition checks.
∑p(x)=0.03+ 0.10+ 0.15+ 0.17+ 0.25+ 0.15+ 0.10+ 0.05 = 1
The second condition checks.
This can be considered a probability distribution.
b.
To calclate the mean of this data set you have do the following calculation:
E(X)= ∑Xp(x)= 41*0.03+ 42*0.10+ 43*0.15+ 44*0.17+ 45*0.25+ 46*0.15+ 47*0.10+ 48*0.05 = 44.61
c.
The variance is a measurement of dispersion and you can calculate it as:
V(X)= ∑X²p(x)-(∑Xp(x))²
∑X²p(x)= 41²*0.03+ 42²*0.10+ 43²*0.15+ 44²*0.17+ 45²*0.25+ 46²*0.15+ 47²*0.10+ 48²*0.05 = 1993.05
V(X)= ∑X²p(x)-(∑Xp(x))²= 1993.05-(44.61)²= 2.9979
d.
The standard deviation is the square root of the variance:
√V(X)= √2.9979= 1.73
I hope this helps!