Question

Critical Thinking. For Exercises 5–20, watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, (b) median, (c) mode, (d) midrange, and then answer the given question

Diamond Ring Listed below in dollars are the amounts it costs for marriage proposal packages at the different Major League Baseball stadiums. Five of the teams don’t allow proposals. Are there any outliers?

39 50 50 50 55 55 75 85 100 115 175 175 200 209 250 250 350 400 450 500 500 500 500 1500 2500

Short answer

(a) The mean is $365.3.

(b) The median is $200.0.

(c) The mode is $500.

(d) The midrange is $1269.5.

The outliers are $1500 and $2500.

Step-by-step solution

Given information

The cost for marriage proposal packages is listed in terms of dollars at various League Baseball stadiums.

39, 50, 50, 50, 55, 55, 75, 85, 100, 115, 175, 175, 200, 209, 250, 250, 350, 400, 450, 500, 500, 500, 500, 1500, 2500

Compute mean

(a)

The mean is computed using observations x and the counts of these observations n as:

$\overline{x}=\frac{\sum x}{n}$

The values are substituted in the formula as follows:

$$\begin{gathered} \overline{x}=\frac{39+50+...+2500}{25} \\ =\frac{9133}{25} \\ \approx 365.32 \end{gathered}$$

Thus, the mean value is approximately $365.3.

Compute median

(b)

The median is computed using the count of observations; n.

• When n is even, the median is the average of the middlemost terms.
• When n is odd, the median is the middlemost term.

The number of observations is25.

Arrange the observations in ascending order.

39
50
50
50
55
55
75
85
100
115
175
175
200
209
250
250
350
400
450
500
500
500
500
1500
2500

The middlemost observation is200.

The median is given as:

$M=200$

Thus, the median is $200.0.

Compute mode

(c)

The mode is the value that is the most frequent in the set of observations.

The frequency distribution for the set of observations is:

Observations	Frequency
39	1
50	3
55	2
75	1
85	1
100	1
115	1
175	2
200	1
209	1
250	2
350	1
400	1
450	1
500	4
1500	1
2500	1

Thus, the mode of the data is $500.

Compute midrange

(d)

The midrange is the average of the two extreme values.

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

Substitute the values in the formula.

$$\begin{gathered} \text{Midrange}=\frac{39+2500}{2} \\ =\frac{2539}{2} \\ =1269.5 \end{gathered}$$

Thus, the midrange is $1269.5.

Determine the outliers in the study

Outliers are the set of values that are more extreme than the rest of the observations. In the data, the values that are extremely large compared to other values are 1500 and 2500.

Thus, the outliers are $1500 and $2500.