Question

Sometimes, data must be modified before a stem-and-leaf plot may be constructed. For example, the data in the accompanying table represent the five-year rate of return of 20 mutual funds and are reported to the hundredth decimal place. So, if we used the integer portion of the data as the stem and the decimals as the leaves, then the stems would be 8, 9, 10, ..., 19; but the leaves would be two digits (such as 94. 53, and so on). This is not acceptable since each leaf must be a single digit. To resolve this problem, we round the data to the nearest tenth. Complete parts (a) through (c) below. Click the icon to view the table of five-year rates of return. (a) Round the data to the nearest tenth. 11.03 12.93 12.26 9.32 10.14 11.37 9.85 10.15 9.04 11.02 11.72 11.57 12.56 9.58 10.28 10.44 12.37 10.68 12.45 9.44 (b) Draw a stem-and-leaf plot of the modified data. Legend: 90 represents 9.0 (c) Describe the shape of the distribution. Choose the correct answer below. A. The distribution is skewed left because the left tail is longer than the right tail. B. The distribution is bell-shaped because the middle stems in the stem-and-leaf plot have the highest frequencies, and the frequencies tail off to each side of these stems. تم عمل مطلمحطاعمممملمننماطمنمملمممممطقطمنع مصممعلمم Oc . The distribution OD. The distribution i Five-year rate of return of mutual funds - Х 11.03 10.14 9.04 12.56 10.44 12.93 11.37 11.02 9.58 12.37 12.26 9.85 11.72 10.28 10.68 9.32 10.15 11.57 12.45 9.44 Print Done

Answer 1

Problem 1: (a)

11.0
10.1
9.0
12.6
10.4
12.9
11.4
11.0
9.6
12.4
12.3
9.8
11.7
10.3
10.7
9.3
10.2
11.6
12.5
9.4
(b)

Stem	Leaf
9	0,3,4,6,8
10	1,2,3,4,7
11	0,0,4,6,7
12	3,4,5,6
12	9

(c) Option: C. The distribution is skewed right because the right tail is longer than the left tail.

Problem 2: (a) Population mean pulse=(74+90+88+77+60+64+78+75+88)/9=77.1

(b) Mean pulse of sample 1=(77+88+88)/3=84.3

Mean pulse of sample 2=(78+75+90)/3=81
(c) The mean pulse rate of sample 1 overestimates the population mean.

The mean pulse rate of sample 2 overestimates the population mean.

Problem 3: Marissa's course average=(1*0.15+0.93*0.10+0.82*0.5+0.85*0.25)*100=86.55%.