Question

1) The following stem and leaf plot displays the number of home runs allowed per nine innings for all pitchers in 1999 who pitched at least 100 innings. Stem (1=0.1 HR/9IP) Leaf (1=0.01 HR/9IP) Stem (1-0.1 HR/91P) Leaf (1=0.01 HR/91P) 37 13 0023667888 45 14 001234777899 50278 15 0111334569 6 0355667799 16 023378 7003345999 174 8 001336888999 18 357 9 02333445567788899 19 2234 1000001334455567799 20 11 | 11223344555569 21 120122233444466789 220 10 a) How many data values are in this data set? b) List the data values with a stem of 8. 80.80.91, 03103, ho 99.9%.36,099 c) Construct a relative frequency histogram for the data using a class width of 0.20. d) Construct a boxplot for the data. Be sure to scale appropriately. e) Does it appear that this data is roughly symmetric, skewed to the left, or skewed to the right? Justify your reasoning. 2) Download Data Set 14 (14 - Oscar Winner Age) from the Triola website. Does it appear that

Answer 1

1.there are are 148 data values

2. Data values with a stem of 8
0.8
0.8
0.81
0.83
0.83
0.86
0.88
0.88
0.88
0.89
0.89
0.89

Box plot Number 37 0.37 45 0.45 50278 0.50 0.52 0.57 0.58 60355667799 0.60 0.63 0.65 0.65 0.66 0.66 0.67 0.67 0.69 0.69 7003345999 0.70 0.70 0.73 0.73 0.74 0.750 0.79 0.79) 0.79 8 001336888999 0.80 0.80 0.81 0.83 0.83 0.86 0.88 0.88 0.88 0.89 902333445567788899 0.90 0.92 0.93 0.93 0.93 0.94 0.94 0.95 0.95 0.96 1000001334455567799 1.00 1.00 1.00 1.00 1.01 1.03 1.03 1.04 1.04 1.05 11 11223344555569 1.11 1.11 1.12 1.12 1.13 1.13 1.14 1.14 1.15 1.15 1201222334444667891.20 1.21 1.22 1.22 1.22 1.23 1.23 1.24 1.24 1.24 13 0023667888 1.30 1.30 1.32 1.33 1.36 1.36 1.37 1.38 1.38 1.38 14 001234777899 1.40 1.40 1.41 1.42 1.43 1.44 1.47 1.47 1.47 1.48 15 '0111334569 1.50 1.51 1.51 1.51 1.53 1.53 1.54 1.55 1.56 1.59 16 023378 1.60 1.62 1.63 1.63 1.67 1.68 174 1.74 18 357 1.83 1.85 1.87 19 2337 1.92 1.93 1.93 1.97 0.89 0.89 0.97 0.97 1.05 1.05 1.15 1.15 1.24 1.26 0.98 0.99 0.99 1.07 1.09 1.09 0.98 1.06 1.16 1.26 0.98 1.07 1.19 1.27 1.28 1.29 1.49 1.49 20) 21. 220 | 2.20

Note : The stem = 0.8 and the values are 0.00
Hence we have 0.80, 0.80,0.81,0.83...

3.Relative frequency histogram

Step 1 : Create the Class intervals as follows and find how many observation fall in each set

Step 2. The Frequency distribution is given below along with formula for relative and cumulative frequency.

Cumulative freqency 0.1 07 o O 0+0.01 20 Interval Frequency 0-0.2 0.2-0.4 1 0.4-0. 6 6 0.6-0.8 0.8-1 31 1-1.2 28 1.2-1.4 1.4-1.6 20 1.6-1.8 7 1.8-2 2-2.2 Mid point (0+0.2)/2 (0.2+0.4)/2 0.3 (0.4+0.6)/2 0.5 (0.6+0.8)/2 0.7 (0.8+1)/2 0.9 (1+1.2)/2 1.1 (1.2+1.4)/2 1.3 (1.4+1.6)/2 1.5 (1.6+1.8)/2 1.7 (1.8+2)/2 (2+2.2)/2 12.1 Relative Frequency 0/148 1/148 0.01 6/148 0.04 20/148 0.14 31/148 0.21 28/148 0.19 27/148 0.18 20/148 0.14 7/148 0.05 7/148 0.05 11/148 0.01 0.01 0.05 0.19 0.4 0.59 0.77 0.91 0.96 1.01 1.02 0.01+0.04 0.05+0.14 0.19+0.21 0.4+0.19 0.59+0.18 0.77+0.14 0.91+0.05 0.96+0.05 1.01+0.01 27 7 Total 148 1.02

0.25 0.2 0.15 Relative Frequency 0.05 2-2.2 0-0.2 0.2-0.40.4-0.60.6-0.8 0.8-1 1-1.2 1.2-1.41.4-1.61.6-1.8 1.8-2 Number of home runs

4 Boxplot

Box plot 2.5 2 1.5 0.5

5.
The data is normally distributed.It is very evident from the histogram and a the boxplot. Also the mean and median of the data is almost equal.