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Q. 96

Expert-verifiedFound in: Page 294

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen.

a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X. X ~ _____(_____,_____)

d. How many are expected to attend their graduation?

e. Find the probability that 17 or 18 attend.

f. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically

a. The random variable X is the number of students that attended their graduation.

b. The values that X may take on are $0,1,2,.....,22.$

c. The distribution $X~B(22,0.85)$

d. $18.7$ are expected to attend the graduation.

e. The probability of $17or18$ attend is $0.3249$

f. Yes, it will be surprising if all $22$ attended graduation as it is unusual.

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 $p=\frac{85}{100}\phantom{\rule{0ex}{0ex}}p=0.85$

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

The expected binomial distribution is calculated as:

$\mu =np$, where,

$\mu $ is average number of students attending graduation,

$n$ is number of trials

$p$ is probability of success.

Therefore,

$\mu =np\phantom{\rule{0ex}{0ex}}\mu =22\times 0.85\phantom{\rule{0ex}{0ex}}\mu =18.7$

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

$=P\left(17\right)+P\left(18\right)\phantom{\rule{0ex}{0ex}}{=}_{22}{C}_{17}{(0.85)}^{17}{(0.15)}^{5}{+}_{22}{C}_{18}{(0.85)}^{18}{(0.15)}^{5}\phantom{\rule{0ex}{0ex}}=0.3249$

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.

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