91影视

The binomial distribution determines the probability of looking at a specific quantity of a hit results in a specific quantity of trials.

We are given,

$85\%$of graduating students attend their graduation and a group of $22$graduating students is randomly chosen.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case the random variable X is the number of students that attended their graduation.

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand$22$, then X is given by$X=0,1,2,.....,22.$

The random variable is distributed by the data provided X is the number of students that attended their graduation.

According to the given information, $85\%$students attend their graduation.

The probability distribution of binomial distribution has two parameters $n=numberoftrials$and $p=probabilityofsuccess.$The binomial distribution is of the form: $X~B(n,p)$

Therefore, according to given information $n=22$and $$\begin{gathered} p=\frac{85}{100} \\ p=0.85 \end{gathered}$$

The distribution of X is $X~B(22,0.85)$

The expected binomial distribution is calculated as:

$渭=np$, where,

$渭$is average number of students attending graduation,

$n$ is number of trials

$p$is probability of success.

Therefore,

$$\begin{gathered} 渭=np \\ 渭=22脳0.85 \\ 渭=18.7 \end{gathered}$$

The probability that $17or18$ attend is calculated as follow:

$$\begin{gathered} =P\left(17\right)+P\left(18\right) \\ {=}_{22}{C}_{17}{(0.85)}^{17}{(0.15)}^5{+}_{22}{C}_{18}{(0.85)}^{18}{(0.15)}^5 \\ =0.3249 \end{gathered}$$

Using $TI-83$calculator we have found that if all $22$students attends graduation which comes out to be:

$P(X=22)=0.0280$, less than $3\%$.

Therefore, it is unusual.