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Video Games. A pathological video game user (PVGU) is a video game user that averages 31 or more hours a week of gameplay.

According to the article "Pathological Video Game Use among Youths: A Two-Year Longitudinal Study" (Pediatrics, Vol. 127. No. 2, pp. 319-329) by D. Gentile et al., in 2011, about 9% of children in grades 3-8 were PVGUs. Suppose that, today, seven youths in grades 3-8 are randomly selected.

(a) Assuming that the percentage of PVGUS in grades 3-8 is the same today as it was in 2011, determine the probability distribution for the number, X, who are PVGUs.

(b) Determine and interpret the mean of X.

(c) If, in fact, exactly three of the seven youths selected are PVGUs, would you be inclined to conclude that the percentage of PVGUs in grades 3-8 has increased from the 2011 percentage? Explain your reasoning. Hint: First consider the probability $P(Xâ‰\yen 3)$.

(d) If, in fact, exactly two of the seven youths selected are PVGUs, would you be inclined to conclude the percentage of PVGUs in grades 3-8 has increased from the 2011 percentage? Explain your reasoning.

Step by step solution

01

Part (a) Step 1. Given information.

The given statement is:

PVGUs made up roughly 9% of children in grades 3 through 8.

Seven students in grades 3–8 are chosen at random today.

$p=0.09,n=7$

02

Part (a) Step 2. Find the probability distribution of X.

$P\left(X=x\right)=\left(\begin{array}{c}7\\ x\end{array}\right){\left(0.09\right)}^x\left(1-0.09\right)$

X	$P\left(X=x\right)$
0	$\left(\begin{array}{c}7\\ 0\end{array}\right){\left(0.09\right)}^0{\left(0.91\right)}^7=0.517$
1	$\left(\begin{array}{c}7\\ 1\end{array}\right){\left(0.09\right)}^1{\left(0.91\right)}^6=0.358$
2	$\left(\begin{array}{c}7\\ 2\end{array}\right){\left(0.09\right)}^2{\left(0.91\right)}^5=0.106$
3	$\left(\begin{array}{c}7\\ 3\end{array}\right){\left(0.09\right)}^3{\left(0.91\right)}^4=0.017$
4	$\left(\begin{array}{c}7\\ 4\end{array}\right){\left(0.09\right)}^4{\left(0.91\right)}^3=0.002$
5	$\left(\begin{array}{c}7\\ 5\end{array}\right){\left(0.09\right)}^5{\left(0.91\right)}^2=0.000$
6	$\left(\begin{array}{c}7\\ 6\end{array}\right){\left(0.09\right)}^6{\left(0.91\right)}^1=0.000$
7	$\left(\begin{array}{c}7\\ 7\end{array}\right){\left(0.09\right)}^7{\left(0.91\right)}^0=0.000$

03

Part (b) Step 1. Calculate and interpret X's mean.

$$\begin{gathered} \mathrm{μ}=np \\ =7\left(0.09\right) \\ =0.63 \end{gathered}$$

We can interpret from it that 0.63 children are the users of the video games.

04

Part (c) Step 1. Explanation.

First, calculate $P\left(Xâ‰\yen 3\right)$.

role="math" localid="1652944741579" $$\begin{gathered} P\left(Xâ‰\yen 3\right)=1-P\left(X<3\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)\right] \\ =1-\left[0.517+0.358+0.106\right] \\ =1-0.981 \\ =0.019 \end{gathered}$$

It has just happened that the probability of three or more youngsters being pathological video game players is 0.019. Furthermore, it is a rare occurrence. Instead of assuming that a once-in-a-lifetime event occurred, it is now possible to conclude that the probability of success has grown in 2011.

05

Part (d) Step 1. Explanation.

First, calculate $P\left(Xâ‰\yen 2\right)$.

$$\begin{gathered} P\left(Xâ‰\yen 2\right)=1-P\left(X<2\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)\right] \\ =1-\left[0.517+0.358\right] \\ =1-0.875 \\ =0.125 \end{gathered}$$

It has just happened that the probability of two or more youngsters being pathological video game players is 0.125. Furthermore, it occurs frequently. As a result, it is impossible to consider 2 out of 7 as proof of an increase in the rate in 2011.

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