91影视

The responses for the number of cigarettes smoked in a day are recorded from randomly selected respondents.

9, 10, 10, 20, 40, 50, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

(a)

The mean of observations is calculated as follows:

$\stackrel{炉}{x}=\frac{鈭�x}{n}$

Here, xis the observations, and nis the count of the observations.

Substitute the values in the formula.

$$\begin{gathered} \stackrel{炉}{x}=\frac{9+10+10+...+0}{50} \\ =\frac{139}{50} \\ 鈮�2.78 \end{gathered}$$

Thus, the mean value is approximately 2.8 cigarettes.

(b)

Median is a specific value that is obtained as follows:

• n is even; mean of two middle observations.
• n is odd; the middle observation.

The number of observations is50.

Arrange the observations in ascending order.

The middlemost observations are 0 and 0.

The median is given as:

$$\begin{gathered} M=\frac{0+0}{2} \\ =0 \end{gathered}$$

Thus, the median is 0.0 cigarettes.

(c)

The value which repeats the most is mode.

The frequency distribution is:

As 0 repeats the most, the mode is 0.

Thus, the mode is 0 cigarettes.

(d)

The midrange is determined by the following formula as the average of extreme values:

$\text{Midrange}=\frac{\text{Minimum}\text{value}+\text{Maximum}\text{value}}{2}$

Substitute the values in the formula.

$$\begin{gathered} \text{Midrange}=\frac{0+50}{2} \\ =\frac{50}{2} \\ =25 \end{gathered}$$

Thus, the midrange is 25.0 cigarettes.

The respondents were randomly selected and self-reported the number of cigarettes smoked. It is highly likely that they would report incorrect numbers as they were not observed for the same. Also, it is unlikely that an individual smokes 40 or 50 cigarettes in one day.

Therefore, the data can be inaccurate.

Thus, it can be concluded that the data may not be an accurate representation of the population of all adults in California. Consequently, the results are not a correct representation of the population.