Give n = 100, \(\bar x\) =162 cm, s = 8 cm is known in this problem
since σ is unknown , so we consider \(\hat{\sigma}\) = s and \(\varphi\) = 160 cm
\(S.E = \frac { \hat{\sigma} }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 8 }{ \sqrt { 100 } } =0.8\)
Therefore the standard error for the average height of large group of students of 160 cm is 0.8.

Givens \(\sigma^2\) = 4 which implies \(\sigma\) = 2, n = 1 hour = 60 min, \(\bar { X } \) = 20/min
Standard Error \(=\frac { \sigma }{ \sqrt { n } } =\frac { 2 }{ \sqrt { 60 } } =0.2582\)

There many ways to select 15 random samples from the given Tippet’s random number table. Since the population size is 83(two-digit number). Here the door numbers are assigned from 1 to 83. Assume that at random we first choose 2^nd column. So the first sample is 66 and other 14 samples are 74, 52, 39, 15, 34, 11, 14, 13, 27, 61, 79, 72, 35, and 60. If the numbers are above 83, choose the next number ranging from 1 to 83.

2952	6641	3992	9792	7969	5911	3170	5624
4167	9524	1545	1396	7203	5356	1300	2693
2670	7483	3408	2762	3563	1089	6913	7991
0560	5246	1112	6107	6008	8125	4233	8776
2754	9143	1405	9025	7002	6111	8816	6446

There many ways to select 5 random samples from the given Kendall-Babington Smith - Random number table. Assume that at random we select 3^rd column 1^st value. This location gives the digit to be 75. So the first sample will be 75, and then the next choices can be in the same 3^rd column which follows as 55, 16, 67, and 26. Therefore 75, 55,16,67, and 26 will be used as random samples. The various shaded numbers can be taken as 5 random sample numbers. Apart from this, one can select any 5 random sample numbers as they like.

The first thing is to number the population (8585 children). The numbering has already been provided by the frequency table. There are 2 children with height of 105 cm, therefore we assign number 1 and 2 to the children those in the group 105 cm, number 3 to 6 is assigned to those in the group 107 cm and similarly all other children are assigned the numbers. In the last group 145 cms there are two children with assigned number 8584 and 8585.

Height (cm)	Number of children	Cumulative Frequency
105	2	2
107	4	6
109	14	20
111	41	61
113	83	144
115	169	313
117	394	707
119	669	1376
121	990	2366
123	1223	3589
125	1329	4918
127	1230	6148
129	1063	7211
131	646	7857
133	392	8249
135	202	8451
137	79	8530
139	32	8562
141	16	8578
143	5	8583
145	2	8585
Total	8585

Now we take 10 samples from the tables, since the population size is in 4 digits we can use the given random number table. Select the10 random numbers from 1 to 8585 in the table, Here, we consider column wise selection of random numbers, starting from first column.

Tippet’s random number table
2952	6641	3992	9792	7969	5911	3170	5624
4167	9524	1545	1396	7203	5356	1300	2693
2670	7483	3408	2762	3563	1089	6913	7991
0560	5246	1112	6107	6008	8125	4233	8776
2754	9143	1405	9025	7002	6111	8816	6446

The children with assigned number 2952 is selected and then see the cumulative frequency table where 2952 is present, now select the corresponding row height which is 123 cm, similarly all the selected random numbers are considered for the selection of the child with their corresponding height. The following table shows all the selected 10 children with their heights.

Child with assigned Number	2952	4167	2670	0560	2754
Corresponding Height (cms)	123	125	123	117	123
Child with assigned Number	6641	7483	5246	3996	1545
Corresponding Height (cms)	129	131	127	125	121

Sample size n = 50
Sample S.D s = 6.3
Population S.D \(\sigma\) = 6
The standard error for sample S.D is given by 
\(S.E=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } =\frac { 6 }{ \sqrt { 2(50) } } =\frac { 6 }{ \sqrt { 100 } } =0.6\)
Thus standard error for sample S.D = 0.6.

If the occurrence of 3 or 4 on the die is called a success, then
Sample size = 9000; Number of Success = 3240
Sample proportion =\(P=\frac { 3240 }{ 9000 } =0.36\)
Population proportion (P) = Prob(getting 3 or 4 when a die is thrown)
\(=\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } =0.3333\)
Thus P = 0.3333 and Q = 1–P = 1– 0.3333 = 0.6667
The S.E for sample proportion is given by
\(S.E=\sqrt { \frac { PQ }{ N } } =\sqrt { \frac { (0.3333)(0.6667) }{ 9000 } } =0.00496\)
Hence the standard error for sample proportion is S.E = 0.00496.

Given \(\sigma\) = 10, S.E. \(\bar { X } \) = 3
We know that S.E = \(\frac { \sigma }{ \sqrt { n } } \)
Therefore, \(3=\frac { 10 }{ \sqrt { n } } \Rightarrow \sqrt { n } =\frac { 10 }{ 3 } \)
Taking Squaring on both sides we get
\(n=\left(\frac{10}{3}\right)^{2}=\frac{100}{9}=11.11 \cong 11\), 
The required sample size is 11.

Stratum	Homogenous group	Percentage from population	Number of people in each strata	Random Samples
Category 1	Lower income class	39	\(\frac{39}{100}\) \(\times\) 500 = 195	195 \(\times\)\(\frac{68}{500}\) = 26.5~26
Category 2	Middle income class	38	\(\frac{38}{100}\) \(\times\) 500 = 190	190 \(\times\)\(\frac{68}{500} \) = 2.6 ~26
Category 3	Upper income class	23	\(\frac{23}{100}\) \(\times\) 500 = 115	115 \(\times\) \(\frac{68}{500}\) = 15.6~16
Total		100	500	68

Merits :
(a) A random stratified sample is superior to a simple random sample because it ensures representation of all groups and thus it is more representative of the population which is being sampled.
(b) A stratified random sample can be kept small in size without losing its accuracy.
(c) It is easy to administer, if the population under study is sub-divided.
(d) It reduces the time and expenses in dividing the strata into geographical divisions, since the government itself had divided the geographical areas.
Demerits :
(a) To divide the population into homogeneous strata (if not divided), it requires more money, time and statistical experience which is a difficult one.
(b) If proper stratification is not done, the sample will have an effect of bias.
(c) There is always a possibility of faulty classification of strata and hence increases variability.

Here, we have to select 10 random numbers ranging from 1550 to 8000 but the given random number table has only 2 digit numbers. To solve this, two - 2 digit numbers can be combined together to make a four- figure number. Let us select the 5^th and 6^th column and combine them to form a random number, then select the random number with given range. This gives 5 random numbers, similarly, 8^th and 9^th is selected and combined to form a random numbers, then select the random number with given range. This gives 5 random numbers, totally 10 four- figure numbers have been selected. The following table shows the 10 random numbers which are combined and selected.

Therefore the selected 10 random numbers are

5901	4310	5032	5194	1899
7259	7435	4362	1758	5308