Exercise 89 Sampling Words 41 42 43 44 47 48 89 915 1023 146

Exercise 8-9: Sampling Words 4-1, 4-2, 4-3, 4-4, 4-7, 4-8, 8-9, 9-15, 10-23, 14-6, 20-26 Reconsider the distribution of lengths of the 268 words in the Gettysburg Address. The frequency distribution is given here: 8 ua Word Length Frequency ller jra 49 54 59 34 27 15 2.36 170 62 4g /3 10 294 10 an live Without performing any calculations, how do you expect the mean and median to compare? Explain. mian w an a. w below b. Determine the mean and median word length. Explain how you performed these calculations. Also, comment on whether or not the calculations verified your predictio # length, to be low. The onewf4s 1s lou ot

Solution

We are given that Word length and their corresponding frequencies.

Mean and median length is,

mean = x*f / f = 1151 / 268 = 4.2948

median =

We find the cumulative frequency just more than N/2 that is 268 / 2 = 134 and the value of x that corresponding to 169 is 4.

Median = 4

Mean > median

The mean word length is greator than median wotd length

Here we calculate statistic that is function of observations.

here statistic is mean.

In the fourth part suppose a student mistakenly calculate mean = 24.36 letters.

mean = 24.36 is not possible because our calculated mean = 4.2948

Which is large as compare to calculated mean.

x	f	xf	cumulative
1	7	7	7
2	49	98	56
3	54	162	110
4	59	236	169
5	34	170	203
6	27	162	230
7	15	105	245
8	6	48	251
9	10	90	261
10	4	40	265
11	3	33	268
	268	1151