Question

variable. compute the mean and standard deviation of a discrete random mato has basketball practice two days a week. eighty - two percent of the time he attends both practices. seventeen percent of the time he attends one practice. one percent of the time he does not attend either practice. complete the probability distribution table. probability distribution table

x	p(x)

|2|
|1|
|0|
how many practices should we expect mato to attend any given week?
what is the standard deviation?
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Explanation:

Step1: Fill probability - distribution table

Given that 82% of the time he attends both practices, so $P(2)=0.82$. Seventeen percent of the time he attends one practice, so $P(1) = 0.17$. One percent of the time he does not attend either practice, so $P(0)=0.01$.

Step2: Calculate the mean $\mu$

The formula for the mean of a discrete - random variable is $\mu=\sum_{x}x\cdot P(x)$. So, $\mu=(2\times0.82)+(1\times0.17)+(0\times0.01)=1.64 + 0.17+0=1.81$.

Step3: Calculate the variance $\sigma^{2}$

The formula for the variance is $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
$(2 - 1.81)^{2}\times0.82+(1 - 1.81)^{2}\times0.17+(0 - 1.81)^{2}\times0.01=(0.19)^{2}\times0.82+( - 0.81)^{2}\times0.17+( - 1.81)^{2}\times0.01=(0.0361\times0.82)+(0.6561\times0.17)+(3.2761\times0.01)=0.029602+0.111537+0.032761 = 0.1739$.

Step4: Calculate the standard deviation $\sigma$

The standard deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{0.1739}\approx0.417$.

Answer:

The probability - distribution table:

$x$	$P(x)$
1	0.17
0	0.01

The expected number of practices: 1.81
The standard deviation: 0.417