Question

Classify the following as either a discrete random variable or a continuous random variable.

(a) The airspeed of an unladen swallow.

(b) The number of rodents of unusual size that Buttercup and Westley battle each time they cross through the swamp.

(c) The time it takes for Bob Ross to complete a painting.

(d) The number of push-ups the Clemson Tiger mascot does during a football game.

(e) The amount of Gatorade (in ounces) NFL players consume during a game.

The U.S. Census Bureau collects demographics concerning the number of people in families per household. Assume that the distribution of the number of people per household is contained in the following table, where X = number of people in families per household.

X P(X)

2 0.27

3 0.25

4 0.28

5 0.13

6 0.04

7 0.03

(a) We denote the expected value as µ and the standard deviation as σ. Compute µ− σ and µ+σ. What proportion of households have between µ − σ and µ + σ?

Answer 1

#Discrete variable  can take on only integer values or whole number.
Example : Number of students in the class, number of defective items in a lot , number of apples in a basket.
We can not say that number of students in the calss room is 22.5. it must be a whole number.

#Continuous Data can take any value (within a range), it could be an integer or decimal value.
For example : Height or weight of an individuals..we can say that height of an individual is 5.7 feet or 6 feet,
time for deliver a courier ,length of a leaf.

a) The airspeed of an unladen swallow : Continuous

Because airspeed could be an integer or decimal value.

b) The number of rodents of unusual size that Buttercup and Westley battle each time they cross through the swamp.

Discrete , because number of rodents must be an integer.

c)The time it takes for Bob Ross to complete a painting : Continuous

d) The number of push-ups the Clemson Tiger mascot does during a football game : Discrete

e) The amount of Gatorade (in ounces) NFL players consume during a game : Continuous.

#2

X = number of people in families per household.

µ =
r P(X)
and σ = $\sqrt{\sum x^2*P(X) - \mu ^2}$

0.27 0.54 1.08 0.25 0.75 2.25 4.48 4 0.28 1.12 0.13 0.65 3.25 0.04 0.24 1.44 7 0.03 0.21 1.47 Total 3.51 13.97

µ = 3.51

σ =  

V13.97- (3.51)2

σ = 1.2845

µ − σ = 3.51 - 1.2845 = 2.2255 ~ 2

µ + σ. = 3.51 + 1.2845 = 4.7945 ~ 5

So proportion of households have between 2 and 5 is

P(2)+P(3)+P(4)+P(5) = 0.27 + 0.25+0.28 + 0.13 = 0.93

So 93% of households have between µ − σ and µ + σ