Chapter 6, Problem R6.8RE

(a)

To determine

Type of probability distribution for T .

Answer to Problem R6.8RE

T has approximately a Binomial distribution.

Explanation of Solution

Given information:

T : the number who use public transport at least once per week

Number of trials, $n=200$

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

Four conditions of binomial setting are as follows:

• Binary (success/failure)
• Independent trials
• Fixed number of trials
• Probability of success (same for each trial)

Binary: Since the success results in using public transport at least once and failure results in not using public transport at least once, the condition has been satisfied.

Independent trials: Since the random sample of 200 residents is less than 10% of all residents of the city. Thus, it is safe to assume that the trials are independent by the 10% condition.

Fixed number of trials: Since we selected 200 residents and the number of trials is also 200. Thus, the condition has been satisfied.

Probability of success: Since there are 34% chances for the residents using public transport at least once and the probability of success is also 34%. Thus, the condition has been satisfied.

Since all 4 conditions are satisfied,

The given scenario describes a binomial setting.

Thus,

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

T has approximately a Binomial distribution.

(b)

To determine

Approximation of T by a Normal distribution.

Answer to Problem R6.8RE

Large counts condition is satisfied.

Explanation of Solution

Given information:

T : the number who use public transport at least once per week

Number of trials, $n=200$

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

It is suitable to approximate the binomial distribution by the normal distribution, if the large counts condition is satisfied.

Thus,

If

  $np\ge 10$

As well as

  $nq\ge 10$

Now,

Calculate:

  $np=200\left(0.34\right)=68\ge 10$

Also,

  $nq=n\left(1-p\right)=200\left(1-0.34\right)=200\left(0.66\right)=132\ge 10$

Thus,

The requirements are satisfied.

Thereby,

We can approximate the binomial distribution by the normal distribution.

(c)

To determine

Probability for at most 60 residents in the sample use public transportation at least once per week.

Answer to Problem R6.8RE

Probability that at most 60 residents in the sample use public transportation at least once per week is 0.1170.

Explanation of Solution

Given information:

T : the number who use public transport at least once per week

Number of trials, $n=200$

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

From Part (b) result,

We have

The binomial distribution can be approximated by the normal distribution.

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{x-np}{\sqrt{npq}}=\frac{x-np}{\sqrt{np\left(1-p\right)}}$

Where,

Mean,

  $\mu =np$

Standard deviation,

  $\sigma =\sqrt{npq}=\sqrt{np\left(1-p\right)}$

Thus,

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

Use normal probability table in the appendix, to find the corresponding probability.

  $P\left(x\le 60\right)=P\left(z<-1.19\right)=0.1170=11.70\%$

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.