In Gallup's Annual Consumption Habits Poll, telephone interviews were conducted for a random sample of 1,014 adults aged 18 and

Question

13 days ago

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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.

Mathematics

1 answer:

babunello [8.4K] · Answer 1 · 2026-03-14T19:18:16+03:00

Answer:

(b)E(x)=1.3087

(c)Variance of x =1.7119

(d)E(y)=2.0447

Step-by-step explanation:

The random variable x refers to the average number of cups of coffee consumed daily.

The total number of respondents equals 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 for 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 for 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)

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

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

E(y)=2.0447

The expected value for y surpasses that of x.