Chapter 7.3, Problem 68E

(a)

To determine

To find: the event is more likely without any calculation.

Answer to Problem 68E

Randomly selecting one car entering the interchange during rush hour and finding 2 or more people in that car

Explanation of Solution

The standard deviation of the sampling distribution of the sample mean is equal to the population standard deviation divide by the square root of sample size.

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

This then means that the standard deviation decreases as the sample size increases and therefore the data values are closer to the expected value when using a larger sample size.

This then means that you are less likely to find a mean of 2 or more people in the cars when the sample size is larger and therefore event with the smaller sample size (1 car) is more likely to occur.

(b)

To determine

To Calculate: the probability of the first event in part (a).

Answer to Problem 68E

Population distribution is unknown and the sample is small

Explanation of Solution

  $n=1$

Centre limit theorem: if the sample size is equal or greater than 30 then the sampling distribution of the sample mean $\overline{x}$ is about normal.

Since the sample size of 1 is less than 30, it cannot be used the central limit theorem. In this case, the sampling distribution of the sample mean has the same shape as the population distribution

Although, do not know the population distribution and therefore it cannot be find the probability when have only 1 car in the sample.

(c)

To determine

To Calculate: the probability of the second event in part (a).

Answer to Problem 68E

0.04%

Explanation of Solution

Given:

  $\begin{array}{c}\mu =1.6\\ \sigma =0.75\\ n=40\\ \overline{x}=2\end{array}$

Formula used:

Calculation:

Since the sample size is equal or greater than 30 then the central limit theorem says that the sampling distribution of the sample mean is about normal.

The Z-score is

  $z=\frac{x-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{2-1.6}{0.75/\sqrt{40}}=3.37$

The associating probability using the normal probability table

  $P\left(Z<3.37\right)$ is given in the row starting with $3.3$ and in the column starting with $.07$ of the standard normal probability table in the appendix.

  $\begin{array}{c}P\left(\overline{X}\ge 2\right)=P\left(Z>3.37\right)\\ =1-P\left(Z<3.37\right)\\ =1-0.9996\\ =0.0004\\ =0.04\%\end{array}$