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21.
The frequency distribution of the highest temperatures (in °F) in 50 states during a season is shown. Find the percentile rank for a temperature of 117.5°F by constructing a percentile graph.
 Temperatures Number of states 99.5 - 104.5 2 104.5 - 109.5 8 109.5 - 114.5 18 114.5 - 119.5 13 119.5 - 124.5 7 124.5 - 129.5 1 129.5 - 134.5 1 a. 65th percentile b. 55th percentile c. 60th percentile d. 40th percentile

Solution:

Compute the cumulative frequencies for each class (Temperature).

Compute cumulative percents for each class by using cumulative percent = cumulative frequencyTotal frequency × 100%

For first class, cumulative % = 250 Ãƒâ€” 100% = 4% Tabulate the results Draw a graph with the class boundaries on x - axis and the corresponding cumulative percents on y - axis as shown in the graph.

To find the percentile rank for a temperature of 117.5°F, find 117.5 on the x - axis of the percentile graph, and draw a vertical line to the graph. Then move horizontally to the value on the y - axis.

So, the approximate percentile rank for a temperature of 117.5°F is 60th percentile.

22.
Check the following data set for outliers.
25, 43, 16, 4, 11, 39, 28, 8 a. no outlier b. 43 c. 86 d. 4

Solution:

Arrange the data values in increasing order.
4, 8, 11, 16, 25, 28, 39, 43

Q2 = 16+25 / 2= 20.5
[Q2 is the median, which is the mean of the middle values of the data set in the order.]

Q1 = 8+11 / 2= 9.5
[Q1 is the median of the data values less than Q2.]

Q3 = 28+39 / 2= 33.5
[Q3 is the median of the data values greater than Q2.]

Interquartile Range (IQR) = Q3 - Q1= 33.5 - 9.5 = 24

To find the outlier, compute the cut-off points for outliers.

Lower fence = Q1 - 1.5(IQR) = 9.5 - 1.5(24) = - 26.5

Upper fence = Q3 + 1.5(IQR) = 33.5 + 1.5(24) = 69.5

If a data value is less than the lower fence or greater than the upper fence, then it is considered an outlier.

There are no values that are less than - 26.5 or greater than 69.5.

So, there is no outlier.

23.
Find Q1, Q2 and Q3 for the data set.
9, 3, 2, 5, 4, 7, 6, 11, 13. a. Q1 = 4.5, Q2 = 6 and Q3 = 10 b. Q1 = 3.5, Q2 = 6 and Q3 = 11 c. Q1 = 3.5, Q2 = 7 and Q3 = 10 d. Q1 = 3.5, Q2 = 6 and Q3 = 10

Solution:

Arrange the data in increasing order.
2, 3, 4, 5, 6, 7, 9, 11, 13

Median = 6.
[Median is the middle value of the data set in the order.]

So, Q2 = 6
[Q2 is the median.]

The median of the data values less than 6 (median) is Q1.

The data values that are less than 6 (median) are 2, 3, 4, 5.

Q1 = 3+4 / 2 = 3.5

The median of the data values greater than 6 (median) is Q3.

The data values that are greater than 6 (median) are 7, 9, 11, 13.

Q3 = 9+11 / 2 = 10.

So, Q1 = 3.5, Q2 = 6 and Q3 = 10.

24.
Amy recorded the number of different flowers used in a bouquet. Which measure of central tendency is most appropriate to describe the data ? a. mode b. mean c. outlier d. median

25.
Which measure of central tendency is most appropriate for a data that describes the shoe size of the students of your class? a. outlier b. mode c. mean d. median

26.
Use the stem-and-leaf plot to find the number of data items in the interval 20 - 69.  a. 18 b. 14 c. 20 d. 17

27.
The line plot shows the number of chocolates that each student of a class has. How many more chocolates does Smith have than Catherine?  a. 5 b. 3 c. 4

28.
Which line plot displays the data set 45, 49, 53, 54, 57, 49, 58, 52 accurately?  a. Plot 1 b. Plot 2 c. Plot 3 d. Plot 4

29.
Which measure of central tendency is most appropriate for a data that describes the favorite color of 5 girls: blue, red, red, green, orange? a. outlier b. median c. mode d. mean a. median b. mean c. mode d. all of these