Question

Working out Choose a person aged 19 to 25 years at random and ask, “In the past seven days, how many times did you go to an exercise or fitness center or work out?” Call the response Y for short. Based on a large sample survey, here is the probability distribution of Y

Part (a). A histogram of the probability distribution is shown. Describe its shape.

Part (b). Calculate and interpret the expected value of Y.

Short answer

Part (a)

Rightward skewed

The most common number of days working out is unimodal 0

The number of days spent exercising ranges from 0 to 7.

Part (b)$\mu =1.03$

Step-by-step solution

Part (a) Step 1. Given information

Here is the probability distribution of Y.

Part (a) Step 2. Describe its form.

Because the highest bar in the histogram is to the left, and there is a tail of smaller bars to its right, the distribution is skewed to the right.

Because there is only one peak in the histogram, the distribution is unimodal.

Because the highest bar in the histogram is centred at 0, the most common number of days working out is 0.

The number of days spent exercising ranges from 0 to 7.

As a result:

Rightward skewed

The most common number of days working out is unimodal 0

The number of days spent exercising ranges from 0 to 7.

Part (b) Step 1. Interpret the expected value of Y. 

Days $y_i$	0	1	2	3	4	5	6	7
Probability $p_i$	0.68	0.05	0.07	0.08	0.05	0.04	0.01	0.02

The expected value (or mean) is calculated by multiplying each possibility y by its probability $p_i$

$$\begin{gathered} \mu =\sum _{}{y}_i{p}_i \\ \mu ==0x0.68+1x0.05+2x0.07+3x0.08+4x0.05+ \\ 5x0.04+6x0.01+7x0.02 \\ \mu =1.03 \end{gathered}$$

The average number of days spent working out is 1.03 days.