91Ó°ÊÓ

$\mathrm{T}$: the percentage of people who use public transportation at least once a week

The number of trials, $n=200$

The Probability of success, $p=34\%=0.34$

The following are four binomial setup conditions:

• Success/failure (binary)
• Trials conducted independently
• Trials are limited to a certain number.
• The likelihood of success (same for each trial)

The criteria has been satisfied because success results in utilising public transportation at least once and failure results in not using public transportation at all.

Independent trials: Because the random sample of 200 persons represents less than$10\%$of the city's total population. As a result, the $10\%$condition is safe to conclude that the trials are independent.

Fixed number of trials: We chose 200 inhabitants, thus the number of trials will be 200 as well. As a result, the criterion has been met.

Probability of success: Because inhabitants have a 34 percent likelihood of using public transportation at least once, the probability of success is also 34 percent. As a result, the criterion has been met.

The provided scenario describes a binomial situation because all four conditions are met.

Thus,

With $n=200$ and $p=0.34$,

T has a Binomial distribution that is close to it.

T: the percentage of people who use public transportation at least once a week.

The number of trials, $n=200$

The probability of success, $p=34\%=0.34$

If the large numbers condition is met, the normal distribution can be used to approximate the binomial distribution.

Thus,

If

$npâ‰\yen 10$

As well as

$nqâ‰\yen 10$

Now,

Calculate:

$np=200(0.34)=68â‰\yen 10$

Also,

$\begin{array}{rcl}nq& =& n(1-p)\\ & =& 200(1-0.34)\\ & =& 200(0.66)\\ & =& 132â‰\yen 10\end{array}$

Therefore, the requirements are met.

As a result, the normal distribution can be used to approximate the binomial distribution.

T: the percentage of people who use public transportation at least once a week.

The number of trials, $n=200$

The Probability of success, $p=34\%=0.34$

From the above result,

We've got

The normal distribution is a good fit for the binomial distribution.

Determine the z - score.

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}=\frac{x-np}{\sqrt{npq}}=\frac{x-np}{\sqrt{np(1-p)}}$

Where,

Mean,

$\mathrm{μ}=np$

Standard deviation,

$\mathrm{σ}=\sqrt{npq}=\sqrt{np(1-p)}$

Thus,

$z=\frac{x-np}{\sqrt{np(1-p)}}=\frac{60-200(0.34)}{\sqrt{200(0.34)(1-0.34)}}≈-1.19$

To find the equivalent probability, use the normal probability table in the appendix.

$P(x≤60)=P(z<-1.19)=0.1170=11.70\%$

Thus,

There are $11.70\%$ chances that at most 60 residents in the sample use public transportation at least once per week and the probabilityis $0.1170$.