Question

The table displays the probabilities of orders per week at Macy’s. The random variable x displays number of orders per week. (a) Does the table describe a probability distribution? Why or why not? (b) Find the mean. (c) Find the variance. (d) Find the standard deviation. *

Answer 1

Answer:

b. E(X)= 44.61

c. V(X)=  2.9979

d. √V(X)=  1.73

Step-by-step 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!