Question

X        1        2       3         4      5       6

     P(x)   .05      .20   .25   .25    .20    .05

1. What is the mean  of the random variable?
2. What is the standard deviation of the random variable?
3. Calculate Pr( 2 <   X   5)
4. Why would this be considered a valid probability distribution?

Answer 1

We solve part a. to d. using following table

T	PCC	2PC)	2-PC
1	0.05	0.05	0.05
2	0.2	0.4	0.8
3	0.25	0.75	2.25
4	0.25	1	4
5	0.2	1	5
6	0.05	0.3	1.8
	$\sum P(x)=$1	$\sum xP(x)=$3.5	$\sum x^{2}P(x)=$13.9

a. What is the mean of the random variable?

Answer :-  We calculate the mean using following formula

$Mean = \sum xP(x)=3.5$

b. What is the standard deviation of the random variable?

Answer :-  We calculate the standard deviation($\sigma$) using following formula

$\sigma =\sqrt{\sum x^{2}P(x)-(\sum xP(x))^{2}}$

  $=\sqrt{13.9-(3.5)^2}$

$\sigma =1.284523258$

Standard deviation = $\sigma =1.284523258$

c. Calculate Pr( 2 < X < 5)

Answer :- We calculate Pr( 2 <  X < 5) using following way

---------(1)

We convert above X into z using following formula

$z=\frac{X-\mu }{\sigma }$ --------------(2)

Here,

$Mean = \mu = 3.5$

Standard deviation = $\sigma =1.284523258$

Using equation (2) in equation (1) we get

$P(\frac{2-3.5}{1.284523258}<\frac{X-\mu }{\sigma }<\frac{5-3.5}{1.284523258})$

-------(3)

We calculate the above probability using Excel function as below

=NORMSDIST(z)

We use Excel function for equation (3) and then press Enter

$=NORMSDIST(1.167748416)-NORMSDIST(-1.167748416)$

$=0.757091739$

Pr( 2 <  X < 5) = 0.757091739

d. Why would this be considered a valid probability distribution?

Answer :- The above probability distribution is valid because sum of the probabilities of the outcomes equal to 1.

That is $\sum P(x)=1$

A probability function is a function which assigns probabilities to the values of a random variable. All the probabilities must be between 0 and 1 inclusive. The sum of the probabilities of the outcomes must be 1.

So, the above probability distribution is considered as a valid probability distribution.

Note :-

Dear student here in part b. and c. I have give you answers without rounding, if you have need to answers with rounding then please convert it into appropriate decimal places. and If you have any query from part a. to d. please mention it in comment section. I will definitely help you.