A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that this count has mean 1.6 and standard deviation 0.75 in the population of all cars that enter at this interchange during rush hour.

a. Without doing any calculations, explain which event is more likely:

• randomly selecting 1 car entering this interchange during rush hour and finding 2 or more people in the car
• randomly selecting 35 cars entering this interchange during rush hour and finding an average of 2 or more people in the cars

b. Explain why you cannot use a Normal distribution to calculate the probability of the first event in part (a).

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

The mean is 1.6 and standard deviation is 0.75

The population standard deviation divided by the square root of sample size equals the standard deviation of the sampling distribution of the sample mean.

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

As a result, the standard deviation lowers as the sample size grows, and the data values become closer to the predicted value as the sample size grows.

This means that when the sample size is bigger, you are less likely to find a mean of two or more individuals in the automobiles, and hence an occurrence with a smaller sample size (1 car) is more likely to occur.

The mean is 1.6 and standard deviation is 0.75

$\mathrm{n}=1$

The centre limit theorem states that if the sample size is more than 30, the sampling distribution of the sample mean $\overline{x}$is approximately normal.

The central limit theorem cannot be applied since the sample size of 1 is less than 30. The sample mean sampling distribution has the same shape as the population distribution in this example.

Although the population distribution is unknown, the form of the sampling distribution of the sample mean is also unknown, which means that the probability cannot be calculated.

The mean is 1.6 and standard deviation is 0.75

Consider that,

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

The following concept was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

So, the z-score is

localid="1657533591800" role="math" $$\begin{gathered} \text{z=}\frac{2-1.6}{\frac{0.75}{\sqrt{40}}} \\ =3.37 \end{gathered}$$

The z-two score's found expressions must then be equal: $P(Z<3.37)$is standard distribution of the sample mean is roughly normal, as shown in the row beginning with 3.3 and the column beginning with.07.