91Ó°ÊÓ

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,

There are $30\%$of California residents who have adequate earthquake supplies and we have randomly surveyed $11$California residents.

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 California residents who do have adequate earthquake supplies.

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 $11$then X is given by,

$X=0,1,2,.........,11.$

The random variable is distributed by the data provided X will keep the track of online offerings.

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

Therefore according to the given information, $n=11,p=0.30$

Hence, the distribution of X is$X~B(11,0.30)$

We will use the binomial calculator to find the probability that at least eight have adequate earthquake supplies.

Therefore,

$$\begin{gathered} P(xâ‰\yen 8){=}_{11}{C}_8{(0.3)}^8{(0.7)}^3{+}_{11}{C}_9{(0.3)}^9{(0.7)}^2{+}_{11}{C}_{10}{(0.3)}^{10}{(0.7)}^1{+}_{11}{C}_{11}{(0.3)}^{11} \\ P(xâ‰\yen 8)=0.0043 \end{gathered}$$

It is more likely that none of the residents surveyed will have adequate earthquake supplies.

Therefore,

$P(X=11)=0ornone$

The expected binomial distribution is calculated as:

$\mathrm{μ}=np$, where $\mathrm{μ}$is the number of residents that have adequate earthquake supplies , $n=11$is number of trials and $p=0.30$is probability of success.

Therefore, the number of residents that have adequate earthquake supplies is:

$$\begin{gathered} \mathrm{μ}=np \\ \mathrm{μ}=11×0.30 \\ \mathrm{μ}=3.3 \end{gathered}$$