Question

Working outExercise 10 described a large sample survey that asked a sample of people aged 19 to 25 years, “In the past seven days, how many times did you go to an exercise or fitness center or work out?” The response Y for a randomly selected survey respondent has the probability distribution shown here. From Exercise 10, $E\left(Y\right)=1.03$. Find the standard deviation of Y. Interpret this value.

Short answer

The standard deviation is 1.7689.

Step-by-step solution

Step 1. Given information.

The given information is:

Step 2. Find and interpret the standard deviation of Y.

The given mean is 1.03.

The predicted value of the squared departure from the mean is known as the variance:

$$\begin{gathered} {\sigma }^2=\sum {\left(x-\mu \right)}^{2}P\left(x\right) \\ ={\left(0-1.03\right)}^2\times \left(0.68\right)+{\left(1-1.03\right)}^2\times \left(0.05\right)+{\left(2-1.03\right)}^2\times \left(0.07\right)+ \\ {\left(3-1.03\right)}^2\times \left(0.08\right)+{\left(4-1.03\right)}^2\times \left(0.05\right)+{\left(5-1.03\right)}^2\times \left(0.04\right)+ \\ {\left(6-1.03\right)}^2\times \left(0.01\right)+{\left(7-1.03\right)}^2\times \left(0.02\right) \\ =3.1291 \end{gathered}$$

The standard deviation is:

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{3.1291} \\ \approx 1.7689 \end{gathered}$$

We can see that on average the number of days working out will vary from a mean of 1.03 to 1.7689.