Consider the EVEN-ODD language of strings, defined over ∑= {a,b}, having even number of a’s and odd number of b,s.
a) Build an FA for the given language
b) Build a Transition Graph (TG) that accepts the same language but has fewer states than FA.
c) Find the Regular Expression (RE) corresponding to TG accepting EVEN-ODD language (Show all possible steps)
Sol: The Regular Expression (RE) corresponding to TG accepting EVEN-ODD language.
An FA that accepts the EVEN-ODD language of strings.
Question 1: (Marks: 5+10=15)
a) Compute the mean deviation from the following data.
| Groups | 30-34 | 35-39 | 40-44 | 45-49 | 50-54 |
| Frequency | 10 | 15 | 5 | 11 | 9 |
Sol:
Necessary calculation for “mean deviation”.
| Groups | Frequency ( |
|
|
i.e. |
|
|
| 30-34 | 10 | 32 | 320 | -9.4 | 9.4 | 94 |
| 35-39 | 15 | 37 | 555 | -4.4 | 4.4 | 66 |
| 40-44 | 5 | 42 | 210 | 0.6 | 0.6 | 3 |
| 45-49 | 11 | 47 | 517 | 5.6 | 5.6 | 61.6 |
| 50-54 | 9 | 52 | 468 | 10.6 | 10.6 | 95.4 |
| Total |
| |
| | |
|
Mean (Arithmetic mean) is
So mean deviation is
=
Ans.
b) The mean of 5 observations is 15 and variance is 9. If two more observations having values -3 and 10 are combined with these 5 observations, what will be the new mean and variance of 7 observations?
Sol:
Now sum of seven observations is i.e.
So the mean of seven observations is
Now variance of seven observations is so 
Now 
Ans.
Question 2: (Marks: 5+10=15)
a) Mean and the standard deviation of a set of values is 100 and 5. Compute mean ±2S.D and mean ±3S.D. Also interpret the results in the light of (i) empirical rule (ii) Chebychev’s inequality.
Sol: given 
So calculations of intervals are
(i) Empirical rule
For interval (90,110)
Approximately 95% of the measurements will fall within standard deviations of the mean.
For interval (85,115)
Approximately 100% of measurements will fall within standard deviations of the mean.
(ii) Chebychev’s inequality
For interval (90,110)
At least 
of the observations will fall within standard deviations of the mean,
For interval (85,115)
At least 
of the observations will fall within standard deviations of the mean.
b) Find the first four moments about X= 174.5 as well as about mean for the following frequency distribution.
| Weights (pounds) | Frequency |
| 110-119 | 1 |
| 120-129 | 4 |
| 130-139 | 17 |
| 140-149 | 28 |
| 150-159 | 25 |
| 160-169 | 18 |
| 170-179 | 13 |
| 180-189 | 6 |
| 190-199 | 5 |
| 200-209 | 2 |
| 210-219 | 1 |
| Total | … |
-End-
Sol:
| Weight |
|
|
|
|
|
|
|
| 110-119 | 114.5 | 1 | -60 | -60 | 3600 | -216000 | 12960000 |
| 120-129 | 124.5 | 4 | -50 | -200 | 10000 | -500000 | 25000000 |
| 130-139 | 134.5 | 17 | -40 | -680 | 27200 | -1088000 | 43520000 |
| 140-149 | 144.5 | 28 | -30 | -840 | 25200 | -756000 | 22680000 |
| 150-159 | 154.5 | 25 | -20 | -500 | 10000 | -200000 | 4000000 |
| 160-169 | 164.5 | 18 | -10 | -180 | 1800 | -18000 | 180000 |
| 170-179 | 174.5 | 13 | 0 | 0 | 0 | 0 | 0 |
| 180-189 | 184.5 | 6 | 10 | 60 | 600 | 6000 | 60000 |
| 190-199 | 194.5 | 5 | 20 | 100 | 2000 | 40000 | 800000 |
| 200-209 | 204.5 | 2 | 30 | 60 | 1800 | 54000 | 1620000 |
| 210-219 | 214.5 | 1 | 40 | 40 | 1600 | 64000 | 2560000 |
| Total | |
| | -2200 | 83800 | -2614000 | 113380000 |
| total/n | | | |
| 698.33 | -21783.33 | 944833.33 |
