Question

Watching TV (6.1, 7.3) Choose a young person (aged $19$to $25$) at random and ask, “In the past seven days, how many days did you watch television?” Call the response $X$for short. Here is the probability distribution for $X$.

(a) What is the probability that $X=7$? Justify your answer.

(b) Calculate the mean of the random variable X. Interpret this value in context.

(c) Suppose that you asked $100$randomly sele

cted young people (aged $19$to $25$) to respond to the question and found that the mean $x$of their responses was $4.96$. Would this result surprise you? Justify your answer.

Short answer

a). The probability when $\mathrm{X}$is equal to $7$is $0.57$.

b). The mean of the random variable $X$ is $5.44$.

c). The sample mean of $4.96$is within standard deviation of the mean $2.1369$.

Step-by-step solution

Part (a) Step 1: Given Information 

Probability of sample is:

$\begin{array}{|ll|}\hline \text{Days X}& \text{Probabilities}\\ 0& 0.04\\ 1& 0.03\\ 2& 0.06\\ 3& 0.08\\ 4& 0.09\\ 5& 0.08\\ 6& 0.05\\ \hline\end{array}$

Part (a) Step 2: Explanation 

Add all probabilities are:

$P(X\le 6)=0.04+0.03+0.06+0.08+0.09+0.08+0.05$

$=0.43$

The sum of all probabilities should be equal to $1$.

Now, find the missing probabilities.

$P(X=7)=1-P(X\le 6)$

$=1-0.43$

$=0.57$

Part (b) Step 1: Given Information

Probability of sample is:

$\begin{array}{|ll|}\hline \text{Days X}& \text{Probabilities}\\ 0& 0.04\\ 1& 0.03\\ 2& 0.06\\ 3& 0.08\\ 4& 0.09\\ 5& 0.08\\ 6& 0.05\\ \hline\end{array}$

Part (b) Step 2: Explanation 

The sample mean $\mu =\sum xP\left(x\right)$

Calculation:

Find the mean, use the formula $\mu =\sum xP\left(x\right)$.

$\mu =\sum xP\left(x\right)$

$=0\times 0.04+1\times 0.03+2\times 0.06+3\times 0.08+4\times 0.09+5\times 0.08+6\times 0.05+7\times 0.57$

$=5.44$

Hence, the mean of the random variable $X$is $5.44$.

Part (c) Step 1: Given Information 

Probability of sample is

$\begin{array}{|ll|}\hline \text{Days X}& \text{Probabilities}\\ 0& 0.04\\ 1& 0.03\\ 2& 0.06\\ 3& 0.08\\ 4& 0.09\\ 5& 0.08\\ 6& 0.05\\ \hline\end{array}$

Part (c) Step 2: Explanation 

Find the standard deviation, use the formula $\sigma =\sqrt{\sum }(x-\mu {)}^{2}P(x)$.

$\sigma =\sqrt{\sum (x-\mu {)}^{2}P(x)}$

$=\sqrt{\left[\begin{array}{c}(0-5.44{)}^2\times 0.04+(1-5.44{)}^2\times 0.03+(2-5.44{)}^2\times 0.06\\ +(3-5.44{)}^2\times 0.08+(4-5.44{)}^2\times 0.09+(5-5.44{)}^2\times 0.08\\ +(6-5.44{)}^2\times 0.05+(7-5.44{)}^2\times 0.57\\ =\end{array}\right]}$

localid="1649851646550" $=2.1369$