Question

lesson 6.2 - continuous random variables
quicknotes
discrete
only takes in
fixed values
continuous: uncountable values

• density curve

probability: area under curve
normal: ( z=\frac{\text{value}-\text{mean}}{\text{sd}} )
check your understanding
among those who play minecraft, the amount of time they spend playing per day is approximately normally distributed with mean ( mu = 150 ) minutes and standard deviation ( sigma = 42.7 ) minutes. suppose we choose a minecraft player at random and let ( m = ) the amount of time they spend playing minecraft (in minutes).
a. is ( m ) a discrete or continuous random variable? explain.
b. find ( p(mleq90) ). interpret this value.
c. find ( p(90leq mleq120) ). interpret this value.

Explanation:

Step1: Determine the type of random variable

A discrete random variable takes on a countable number of distinct values. A continuous random variable can take on any value in an interval. Time is a continuous quantity (e.g., it can be 150.5 minutes, 150.55 minutes etc.). So, \(M\) is a continuous random variable.

Step2: Standardize for \(P(M\leq90)\)

The formula for standardization is \(z=\frac{x - \mu}{\sigma}\). Here, \(\mu = 150\), \(\sigma=42.7\), and \(x = 90\).

$$z=\frac{90 - 150}{42.7}=\frac{- 60}{42.7}\approx - 1.405$$

Using a standard normal table (or a calculator with a normal - cdf function, for \(Z\sim N(0,1)\)), \(P(Z\leq - 1.405)\).
Using a calculator: \(P(M\leq90)=\text{normalcdf}(-\infty,90,150,42.7)\approx0.080\)
Interpretation: Approximately \(8\%\) of Minecraft players spend 90 minutes or less playing per day.

Step3: Standardize for \(P(90\leq M\leq120)\)

For \(x = 90\), \(z_1=\frac{90 - 150}{42.7}\approx - 1.405\)
For \(x = 120\), \(z_2=\frac{120 - 150}{42.7}=\frac{-30}{42.7}\approx - 0.703\)
\(P(90\leq M\leq120)=P(-1.405\leq Z\leq - 0.703)\)
Using a calculator: \(P(90\leq M\leq120)=\text{normalcdf}(90,120,150,42.7)\approx0.207\)
Interpretation: Approximately \(20.7\%\) of Minecraft players spend between 90 and 120 minutes playing per day.

Answer:

a. \(M\) is a continuous random variable because time is a continuous quantity (it can take on any value within an interval).
b. \(P(M\leq90)\approx0.080\). Approximately \(8\%\) of Minecraft players spend 90 minutes or less playing per day.
c. \(P(90\leq M\leq120)\approx0.207\). Approximately \(20.7\%\) of Minecraft players spend between 90 and 120 minutes playing per day.