variable. duan has basketball practice two days a week. ninety percent of the time he attends one practice…
variable. duan has basketball practice two days a week. ninety percent of the time he attends one practice. four percent of the time he attends both practices. six percent of the time he does not attend either practice. complete the probability distribution table. probability distribution table\n|x|p(x)|\n|2|\n|1|\n|0|\nhow many practices should we expect duan to attend any given week?\nwhat is the standard deviation?\nquestion help: message instructor post to forum
Answer
Explanation:
Step1: Fill in probability - distribution table
The probability of attending 2 practices $P(2)=0.9$, the probability of attending 1 practice $P(1) = 0.06$, and the probability of attending 0 practices $P(0)=0.04$.
Step2: Calculate the mean ($\mu$)
The formula for the mean of a discrete - random variable is $\mu=\sum_{i}x_{i}P(x_{i})$. So, $\mu=(2\times0.9)+(1\times0.06)+(0\times0.04)=1.8 + 0.06+0=1.86$.
Step3: Calculate the variance ($\sigma^{2}$)
The formula for the variance is $\sigma^{2}=\sum_{i}(x_{i}-\mu)^{2}P(x_{i})$. $(2 - 1.86)^{2}\times0.9+(1 - 1.86)^{2}\times0.06+(0 - 1.86)^{2}\times0.04=(0.14)^{2}\times0.9+( - 0.86)^{2}\times0.06+( - 1.86)^{2}\times0.04=0.0196\times0.9 + 0.7396\times0.06+3.4596\times0.04=0.01764+0.044376 + 0.138384=0.2004$.
Step4: Calculate the standard deviation ($\sigma$)
The standard deviation is the square - root of the variance. So, $\sigma=\sqrt{0.2004}\approx0.448$.
Answer:
The probability - distribution table:
| $x$ | $P(x)$ |
|---|---|
| 2 | 0.9 |
| 1 | 0.06 |
| 0 | 0.04 |
| The expected number of practices attended per week is $1.86$. | |
| The standard deviation is approximately $0.448$. |