Mathematics: 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

In Gallup s Annual Consumption Habits Poll, telephone

Ответ:

alyssamaize

29.01.2022

Mathematics

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) _ $\frac{365}{1014}$ __ $\frac{264}{1014}$ __ $\frac{193}{1014}$ __ $\frac{91}{1014}$ __ $\frac{101}{1014}$

2.)

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

E(X) = ΣX × P(X)

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

2.)

The Expected Variance Formula :

Var(X) = ΣX² × P(X) - E(X)²

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

Var(X) = 3.423 - 1.7126 = 1.710

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) = $\frac{264}{649} \times 1 + \frac{193}{649} \times 2 + \frac{91}{649} \times 3 + \frac{101}{649} \times 4 = 2.045$

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

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Ответ:

joycetleiji1

29.01.2022

Mathematics

(b)E(x)=1.3087

(c)Variance of x =1.7119

(d)E(y)=2.0447

Step-by-step explanation:

Random variable x = number of cups of coffee consumed on an average day.

Total Respondents = 1014

(a)Probability distribution for x.

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

(b)Expected Value of x

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

E(x)=1.3087

(c)Variance of x

Variance $=\sum (x-\mu)^2P(x)$

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

$Variance, \sum (x-\mu)^2P(x)=1.7119$

(d)

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

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