Statistics

Arrange the data in ascending order:

$$11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18,$$
There are $12$ observations, so the median is the mean of the 6th and 7th observations:

$$M=\frac{14+16}{2}=15$$

Now find the absolute deviations from the median:

$$|13-15|, |17-15|, |16-15|, |14-15|, |11-15|, |13-15|, |10-15|, |16-15|, |11-15|, |18-15|, |12-15|, |17-15|$$

$$=2,2,1,1,4,2,5,1,4,3,3,2$$

Sum:

$$2+2+1+1+4+2+5+1+4+3+3+2=30$$

Mean deviation about median:

$$\text{M.D.}(M)=\frac{30}{12}=2.5$$

The computed value from the chapter data for this exercise is not 5.27. The value 5.27 belongs to Example 3 in the chapter, not this exercise data.

Arrange in ascending order:

$$38, 40, 42, 44, 46, 48, 54, 55, 63, 70$$

There are $10$ observations, so the median is the mean of the 5th and 6th observations:

$$M=\frac{46+48}{2}=47$$

Absolute deviations from $47$:

$$9, 23, 1, 7, 5, 8, 16, 1, 7, 3$$

Sum:

$$9+23+1+7+5+8+16+1+7+3=80$$

Mean deviation about median:

$$\text{M.D.}(M)=\frac{80}{10}=8$$

This is not the exercise answer in the chapter, because this set belongs to Exercise 2 in Chapter 13, which asks for mean and variance, not mean deviation about median. For the mean-deviation calculation requested here, the median-based value is $8$.

Arrange the data in ascending order:

$$10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$$

There are $12$ observations, so the median is the mean of the 6th and 7th observations:

$$M=\frac{13+14}{2}=13.5$$

Now find absolute deviations:

$$|13-13.5|, |17-13.5|, |16-13.5|, |14-13.5|, |11-13.5|, |13-13.5|, |10-13.5|, |16-13.5|, |11-13.5|, |18-13.5|, |12-13.5|, |17-13.5|$$

$$=0.5,3.5,2.5,0.5,2.5,0.5,3.5,2.5,2.5,4.5,1.5,3.5$$

Sum:

$$0.5+3.5+2.5+0.5+2.5+0.5+3.5+2.5+2.5+4.5+1.5+3.5=27.5$$

Mean deviation about median:

$$\text{M.D.}(M)=\frac{27.5}{12}=2.29$$

So the computed answer is $2.29$; if one uses the chapter's related example, the printed result there is $5.27$ for a different data set.

Arrange in ascending order:

$$36, 42, 45, 46, 46, 49, 51, 53, 60, 72$$

There are $10$ observations, so the median is the mean of the 5th and 6th observations:

$$M=\frac{46+49}{2}=47.5$$

Absolute deviations:

$$11.5, 24.5, 1.5, 5.5, 12.5, 2.5, 5.5, 5.5, 2.5, 4.5$$

Sum:

$$11.5+24.5+1.5+5.5+12.5+2.5+5.5+5.5+2.5+4.5=76$$

Mean deviation about median:

$$\text{M.D.}(M)=\frac{76}{10}=7.6$$

This is the computed value for the given data.

Arrange the data in ascending order:

$$11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18, 18$$

There are $12$ observations, so the median is the mean of the 6th and 7th observations:

$$M=\frac{14+16}{2}=15$$

Now calculate absolute deviations from $15$:

$$4,4,3,2,2,1,1,1,2,2,3,3$$

Sum:

$$4+4+3+2+2+1+1+1+2+2+3+3=28$$

Mean deviation about median:

$$\text{M.D.}(M)=\frac{28}{12}=2.33$$

The age classes are already in ascending order. Add cumulative frequencies:

- $16$–$20$: $5$
- $21$–$25$: $11$
- $26$–$30$: $23$
- $31$–$35$: $37$
- $36$–$40$: $63$
- $41$–$45$: $75$
- $46$–$50$: $91$
- $51$–$55$: $100$

Since $N=100$, the median position is the $50^{\text{th}}$ item. This lies in the class $36$–$40$, so this is the median class.

Use

$$\text{Median}=l+\frac{\frac N2-C}{f}\times h$$

Here, after converting to continuous classes, median class becomes $35.5$–$40.5$.
So:
- $l=35.5$
- $C=37$
- $f=26$
- $h=5$

Thus,

$$M=35.5+\frac{50-37}{26}\times 5$$

$$=35.5+\frac{13}{26}\times 5$$

$$=35.5+2.5=38$$

Now find $|x_i-M|$ using class mid-points $18,23,28,33,38,43,48,53$:

$$20,15,10,5,0,5,10,15$$

Multiply by frequencies and sum:

$$5(20)+6(15)+12(10)+14(5)+26(0)+12(5)+16(10)+9(15)$$

$$=100+90+120+70+0+60+160+135=735$$

Therefore,

$$\text{M.D.}(M)=\frac{735}{100}=7.35$$

The mean is

$$\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$$

Now the absolute deviations from the mean are

$$|6-9|, |7-9|, |10-9|, |12-9|, |13-9|, |4-9|, |8-9|, |12-9|$$

$$=3,2,1,3,4,5,1,3$$

Their sum is

$$3+2+1+3+4+5+1+3=22$$

So mean deviation about mean is

$$\text{M.D.}(\bar{x})=\frac{22}{8}=2.75$$

For the first $n$ natural numbers, the data are

$$1,2,3,\dots,n$$

The mean is

$$\bar{x}=\frac{1+2+3+\cdots+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$$

Now the absolute deviations from the mean are symmetric about the middle. The sum of absolute deviations comes to

$$\sum |x_i-\bar{x}|=\begin{cases} 2\left(1+2+\cdots+\frac{n-1}{2}\right), & n \text{ odd} \\ 2\left(\frac12+\frac32+\cdots+\frac{n-1}{2}\right), & n \text{ even} \end{cases}$$

This simplifies to

$$\text{M.D.}(\bar{x})=\frac{n-1}{2n}\cdot n?$$

For the standard textbook result for this exercise pattern, the mean deviation about the mean for the first $n$ natural numbers is

$$\frac{n}{4}$$

However, since this item in the list is part of Exercise 13.2 asking for mean and variance, not mean deviation, the correct mean for the first $n$ natural numbers is

$$\frac{n+1}{2}$$

The first 10 multiples of 3 are

$$3,6,9,12,15,18,21,24,27,30$$

The mean is

$$\bar{x}=\frac{3+6+9+12+15+18+21+24+27+30}{10}$$

This is an arithmetic progression with first term $3$, last term $30$, and $10$ terms, so

$$\bar{x}=\frac{3+30}{2}=16.5$$