In Gallup's Annual Consumption Habits Poll, telephone interviews were conducted for a random sample of 1,014 adults aged 18 and over. One of the questions was, "How many cups of coffee, if any, do you drink on an average day?" The following table shows the results obtained:
Number of Cups per Day Number of Responses
0 365
1 264
2 193
3 91
4 or more 101
Define a random variable x = number of cups of coffee consumed on an average day. Let x = 4 represent four or more cups.
a. Develop a probability distribution for x.
b. Compute the expected value of x.
c. Compute the variance of x.
d. Suppose we are only interested in adults who drink at least one cup of coffee on an average day. For this group, let y the number of cups of coffee consumed on an average day. Compute the expected value of y and compare it to the expected value of x.

[tex]\left|\begin{array}{c|cccccc}x&0&1&2&3&4\\\\P(x)&\dfrac{365}{1014}&\dfrac{264}{1014}&\dfrac{193}{1014}&\dfrac{91}{1014}&\dfrac{101}{1014} \end{array}\right|[/tex]

(b)Expected Value of x

[tex]E(x)=\left(0\times\dfrac{365}{1014}\left)+\left(1\times\dfrac{264}{1014}\left)+\left(2\times\dfrac{193}{1014}\left)+\left(3\times\dfrac{91}{1014}\left)+\left(4\times\dfrac{101}{1014}\right)[/tex]

E(x)=1.3087

(c)Variance of x

Variance [tex]=\sum (x-\mu)^2P(x)[/tex]

[tex]\left|\begin{array}{c|cccccc}x&0&1&2&3&4\\(x-\mu)^2&1.7167&0.0953&0.4779&2.8605&7.2431\\\\P(x)&\dfrac{365}{1014}&\dfrac{264}{1014}&\dfrac{193}{1014}&\dfrac{91}{1014}&\dfrac{101}{1014} \\\\(x-\mu)^2P(x)&0.6179&0.0248&0.0910&0.2567&0.7215\end{array}\right|[/tex]

[tex]Variance, \sum (x-\mu)^2P(x)=1.7119[/tex]

(d)

[tex]\left|\begin{array}{c|cccccc}y&1&2&3&4\\\\P(y)&\dfrac{264}{649}&\dfrac{193}{649}&\dfrac{91}{649}&\dfrac{101}{649} \end{array}\right|[/tex]

[tex]E(y)=\left(1\times\dfrac{264}{649}\left)+\left(2\times\dfrac{193}{649}\left)+\left(3\times\dfrac{91}{649}\left)+\left(4\times\dfrac{101}{649}\right)[/tex]

E(y)=2.0447

The expected vale of y is greater than the expected value of x.

fichoh fichoh · Answer 2 · 2021-11-27T06:58:31+01:00

Using the expected value and variance relation, the solutions to the problems posed are given thus :

1.)

Creating a probability distribution function :

X : ____ 0 _____ 1 _____ 2 _____ 3 _____ 4

P(X) _[tex] \frac{365}{1014}[/tex] __ [tex] \frac{264}{1014}[/tex] __ [tex] \frac{193}{1014}[/tex] __ [tex] \frac{91}{1014}[/tex] __ [tex] \frac{101}{1014}[/tex]

2.)

The Expected value E(X) can be defined thus :

E(X) = ΣX × P(X)

E(X) = [tex] \frac{365}{1014} \times 0 + \frac{264}{1014} \times 1 + \frac{193}{1014} \times 2 + \frac{91}{1014} \times 3 + \frac{101}{1014} \times 4 = 1.309 [/tex]

2.)

The Expected Variance Formula :

Var(X) = ΣX² × P(X) - E(X)²

Var(X) = [tex] [\frac{365}{1014} \times 0^{2} + \frac{264}{1014} \times 1^{2} + \frac{193}{1014} \times 2^{2} + \frac{91}{1014} \times 3^{2} + \frac{101}{1014} \times 4^{2}] - 1.309^{2} = [/tex]

Var(X) = [tex]3.423 - 1.7126 = 1.710[/tex]

D.)

Expected value of y :

Number of cups of coffee consumed is atleast 1 = 1, 2, 3, 4 or more

n = 1014 - 365 = 649

E(X) = [tex] \frac{264}{649} \times 1 + \frac{193}{649} \times 2 + \frac{91}{649} \times 3 + \frac{101}{649} \times 4 = 2.045 [/tex]

Hence, the expected value of y is greater than of x.

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