Sample size n = 1600
Sample mean \(\bar { X } \) = 99
Population mean μ = 100
Population standard deviation σ = 15
Null Hypotheses H_o : μ = 100
Alternative Hypotheses H₁ : μ ≠ 100
Level of significance α = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 99-100 }{ \frac { 15 }{ \sqrt { 1600 } } } =\frac { -1 }{ \frac { 15 }{ 40 } } =-\frac { 40 }{ 15 } =-2.667\)
∴ |Z| = 2.667
Critical value at 5% level of significance is \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here |Z|>\({ Z }_{ \frac { \alpha }{ 2 } }\) as 2.667>1.96
Inference: since |Z|>\({ Z }_{ \frac { \alpha }{ 2 } }\) at 5% level of sinificance, the null hypothesis is rejected. 
Hence, we can conclude that the sample has not been taken from the population with mean μ = 100.

Sample size n=100
Sample mean \(\bar { X } \) = 67.45
Sample variance s² = 9
∴ Sample standard deviation s=√9=3
Let μ the population mean.
Standard error \(S.E=\frac { s }{ \sqrt { n } } =\frac { 3 }{ \sqrt { 100 } } =\frac { 3 }{ 10 } =0.3\)
a) The 95% confidence limits for μ are given by
\(\bar { X } -{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { X } +{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\)
As the level of significance is α = 0.05,
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
∴ (1) becomes,
67.45-(1.96)(0.3) ≤ μ ≤ 67.45+(1.96)(0.3)
⇒ 67.45-0.588 ≤ μ ≤ 67.45+0.588
⇒ 66.862 ≤ μ ≤ 68.038
Thus, the 95% confidence intervals for estimating μ is given by (66.86, 68.04).
b) The 99% confidence limits for estimating μ are given by
\(\bar { X } -{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\le \mu \le \bar { X } +{ Z }_{ \frac { \alpha }{ 2 } }(S.E)\)
As the level of significance is α = 0.01,
\({ Z }_{ \frac { \alpha }{ 2 } }=2.58\)
∴ 67.45-2.58(0.3) ≤ μ ≤ 67.45+2.58(0.3)
⇒ 67.45-0.774 ≤ μ ≤ 67.45+0.774
⇒ 66.676 ≤ μ ≤ 68.224
Thus, the 99% confidence intervals for estimating μ is given by (66.67, 68.22).

Let x_i, (i = 1,2,3....n) is a ran dom sample of size n from a normal population with mean μ and variance σ², then the sample mean is distributed
normally with mean μ and variance \(\frac { { \sigma }^{ 2 } }{ n } \)
Thus for large samples, the standard normal variate corresponding to \(\bar { X } \) is:
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
Under the null hypothesis, that the sample has been drawn from a population with mean μ and variance σ², the test statistic is \(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)

The types of sampling are :
(i) Non- Random sampling or Non-probability sampling.
(ii) Random Sampling or Probability sampling.
Different types of probability sampling are :
(i) Simple random sampling
(ii) Stratified random sampling
(iii) Systematic sampling Simple random sampling has
(a) Lottery method and
(b) Table of random number method

Sample size n = 50,
Sample mean \(\bar { X } \) = 1850
Population mean μ = 1800
Population standard deviation σ = 100
Null Hypotheses H₀:
μ = 1800 (i.e., To claim that the breaking strength of the cables have increased)
Alternative Hypotheses H₁:
μ≠1800(To claim that the breaking strength of the cables have not increased)
The level of significance ∝ = 1% =.001
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 1850-1800 }{ \frac { 100 }{ \sqrt { 50 } } } =\frac { 50 }{ \frac { 100 }{ 7.07 } } =\frac { 50 }{ 14.144 } =3.535\)
\(\Rightarrow \therefore Z=3.535\)
The Significant value \({ Z }_{ \frac { \alpha }{ 2 } }=2.58\)
Here \(Z<{ Z }_{ \frac { \alpha }{ 2 } }i.e.,3.535<2.58\)
Inference: Since \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 1% level of significance, the null hypothesis Ho is rejected.
Hence, we conclude that μ≠ 1800 and we cannot support the claim that the breaking strength of the cables have increased.

Sample size n = 100
Sample mean \(\bar { X } \)= 72
Population mean μ = 76
Population standard deviation σ = 8
Null Hypotheses H₀:
μ = 76(i.e., There is no Significant difference between the state scores and the national scores)
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 72-76 }{ \frac { 8 }{ \sqrt { 100 } } } =\frac { -4 }{ \frac { 8 }{ 10 } } =\frac { -4 }{ 8 } =-5\)
\(\Rightarrow |Z|=5\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(Z>{ Z }_{ \frac { \alpha }{ 2 } }i.e.,\ 5>1.96\)
Inference : Since \(Z>{ Z }_{ \frac { \alpha }{ 2 } }\)  at 5% level of significance, the null hypothesis H₀ is rejected. 
Hence, we conclude that there is significant difference between the state scores and the national scores.

Given Sample size n = 400
Sample mean \(\\ \bar { X } =67.47\)inches
Population mean μ = 67.39 & σ = 1.30 inches)
Null Hypotheses H_o:
μ = 67.39 inches (i.e., the sample has been drawn from the population with μ = 67.39 & σ = 1.30 inches)
Alternative Hypotheses H₁:
μ ≠ 67.39 inches (two tail test)
(i.e., the sample has not been drawn from the population with μ = 67.39 & σ = 1.30 inches) The level of significance a = 5% = 0.05
Applying the test statistic,
\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 67.47-67.39 }{ \frac { 1.30 }{ \sqrt { 400 } } } =\frac { 0.08 }{ 0.065 } =1.2308\)
\(\therefore |Z|=1.2308\)
The significant value \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here \(Z={ Z }_{ \frac { \alpha }{ 2 } }i.e.,1.2308<1.96\)
Inference: Since \({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance the null hypothesis H_o is accepted.
Hence, we conclude that the sample has been drawn from the population with mean height 67.39 inches and standard deviation 1.30 inches.

Sample size n = 400
Probability of vegetarian = \(\frac{230}{400}=0.575\)
∴ P = 0.575
Probability of non-vegetarian q = 1 - P = 1 - 0.575
⇒ q = 0.425
Standard error for sample proportion =\(\\ \\ \sqrt { \frac { pq }{ n } } \)
\(\sqrt { \frac { (0.575)(0.425) }{ 400 } } \)
\(=\sqrt { 0.000610 } \)
= 0.0246
S.E = 0.025(app)

Sample size n = 60
Population standard deviation σ = 3
The standard error for sample standard deviation
\(=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } \)
\(=\sqrt { \frac { { 3 }^{ 2 } }{ 2(60) } } =\sqrt { \frac { 9 }{ 120 } } =\sqrt { 0.075 } \)
S.E.of sample standard deviation = 0.2739

In systematic sampling, randomly select the first sample from the first k units. Then every k^th member, starting with the first selected sample, is included in the sample.
Procedure for selection of samples by systematic sampling method.
(I) If we want to select a sample of 10 students from a class of 100 students, then \(k=\frac{N}{n}=\frac{100}{10}=10\)
Thus, sampling interval = 10 denotes that for every 10 samples one sample has to be selected.
(II) If the selected first random sample is 5, then the rest of the samples are automatically selected as 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.[∵ k = 10]

When the population is heterogeneous with respect to the variable, then Stratified Random Sampling method is used.
Following steps are involved:
a) The population is divided into different classes so that each stratum will consist of more or less homogeneous elements. The strata are so designed that they do not overlap each other.
b) After the population is stratified, a sample is drawn at random from each stratum using Lottery Method or Table of Random Number Method.
Example:
From the following data, select 68 random samples from the population of heterogeneous group with size of 500 through stratified random sampling, considering the following categories as strata.
Category 1 : Lower income class - 39%
Category 2 : Middle income class - 38%
Category 3 : Upper income class- 23%
solution:

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

In simple random sampling the samples are selected in such a way that each and every unit in the population has an equal and independent chance of being selected as a sample. It can be done with or without replacement of the samples selected. If the sampling is With replacement, there is a possibility of selecting the same sample any number of times. So, in simple random sampling without replacement is followed. Thus in simple random sampling from a population of N units, the probability is 1/ N the probability of drawing any unit in the second draw from among the available (N - 1) units is 1/(N - 1) and so on.
Simple random sampling without replacement is followed. The following two methods are generally used.
(A) Lottery method:
This is the most popular and simplest method when the population is finite. In this method, all the items of the population are numbered on separate slips of paper of same size, shape and colour. They are folded and placed in a container and shuffled thoroughly. Then the required numbers of slips are selected.
(B) Table of Random number:
The random number table has been so constructed that each of the digits 0,1,2, ... ,9 will appear approximately with the same frequency and independently of each other.
The various random number tables available are
a. L.H.C. Tippet random number series
b. Fisher and Yates random number series
c. Kendall and Smith random number series
d. Rand Corporation random number series.
Example: Tippett's table of random numbers is 20 items out of 6000.
Here we consider row wise selection of random numbers.

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

Sample size n = 50
Sample mean \(\bar { x } =9.3\) minutes
Sample S.D s = 1.6 minutes
Population mean  μ = 8.9 minutes
Null hypothesis H₀:  μ = 8.9
Alternative hypothesis H₁:  μ = 8.9 (two tail)
Level of significance  μ = 0.05 
Test statistic  \(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(\\ Z=\frac { 9.3-8.9 }{ \frac { 1.6 }{ \sqrt { 50 } } } =\frac { 0.4 }{ 0.2263 } =1.7676\)
Calculated value Z = 1.7676
Critical value at 5% level of significance is \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Inference: Since the calculated value is less than table value i.e., \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) at 5% level of significance, the null hypothesis is accepted.
Therefore we conclude that an ambulance service claims on the average 8.9 minutes to reach its destination in emergency calls.

Sample size n = 50 workers
Total wages  \(\sum { x } =2550\)
Sample mean \(\bar { x } =\frac { \text{total wages} }{ n } -\frac { \sum { x } }{ n } =\frac { 2550 }{ 50 } \) = 51units
Population mean \(\mu\) = 52 
Population variance \(\sigma ^{ 2 }=25\)
Population SD  \( \sigma =5\)
Under the null hypothesis H₀:\(\mu\) = 52
Against the alternative hypothesis H₁:\(\mu\) \(\neq\) 52 (Two tail)
Level of significance  \(\mu\) = 0.01
Test statistic \(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(Z=\frac { 51-52 }{ \frac { 5 }{ \sqrt { 50 } } } =\frac { -1 }{ 0.7071 } =-1.4142\)
Since alternative hypothesis is of two tailed test we can take |Z| = 1.4142
Critical value at 1% level of significance is \(\\ { Z }_{ \frac { \alpha }{ 2 } }=2.58\)
Inference: Since the calculated value is less than table value i.e., Z<\({ Z }_{ \frac { \alpha }{ 2 } }\) at 1% level  of significance, the null hypothesis H₀ is accepted.
Therefore, we conclude that there is no significant difference between the sample mean and population mean \(\mu\) = 52 and SD \(\sigma\) = 5.Therefore μ = 49 is rejected.

Sample size n = 400 stores
Sample mean \(\bar x\) = 153.7 bars
Sample SD s = 17.2 bars
Population mean m = 146.3 bars
Since population SD is unknown we can consider the sample SD s = \(\sigma\)
Null Hypothesis :
The advertising campaign is not successful i.e, H₀: \(\mu\) = 146.3 
(There is no significant difference between the mean weekly sales of soap bars in department stores before and after advertising campaign)
Alternative Hypothesis H₁:
\(\mu\) >143.3 (Right tail test). The advertising campaign was successful
Level of significance \(\sigma\) = 0.05
Test statistic :
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1)\)
\(Z=\frac { 153.7-146.3 }{ \frac { 17.2 }{ \sqrt { 400 } } } \)
\(=\frac { 7.4 }{ 0.86 } =8.605\)
\(\therefore\) Z = 8.605
Comparing the calculated value Z=8.605 and the significant value or table value \({ Z }_{ \alpha }=1.645.\) We get 8.605 > 1.645 
Inference:
Since, the calculated value is much greater than table value i.e., Z > \({ Z }_{ \alpha }\), it is highly significant at 5% level of significance.
Hence we reject the null hypothesis H₀ and conclude that the advertising campaign was definitely successful in promoting sales.

(i) Given:
Sample size n = 900, Sample mean \(\bar x\) = 3.4 cm, sample SD s = 2.61 cm.
Population mean \(\mu\) = 3.25 cm, Population SD \(\sigma\) = 2.61 cm
Null Hypothesis H₀ : \(\mu\) = 3.25 cm(the sample has been drawn from the population mean \(\mu\) =3.25 cm and SD \(\sigma\) = 2.61 cm)
Alternative Hypothesis H₁ : \(\mu \neq\)3.25 (two tail) i.e., the sample has not been drawn from the population mean μ = 3.25 cm and SD σ = 2.61 cm.
The level of significance \(\alpha\) = 5% = 0.05
Teststatistic:
\(Z=\frac { 3.4-3.25 }{ \frac { 2.61 }{ \sqrt { 900 } } } =\frac { 0.15 }{ 0.087 } =1.724\)
\(\therefore\) Z = 1.724
Thus the calculated and the significant value or table value \({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Comparing the calculated and table values, \(Z<{ Z }_{ \frac { \alpha }{ 2 } }\) i.e., 1.724 < 1.96
Inference: Since the calculated value is less than table value i.e., Z < \({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance, the null hypothesis is accepted.
Hence we conclude that the data doesn’t provide us any evidence against the null hypothesis.
Therefore, the sample has been drawn from the population mean \(\mu\) = 3.25 cm and SD, \(\sigma\) = 2.61 cm.
(ii) Confidence limits
95% confidential limits for the population mean \(\mu\) are :
\(\bar { x } -{ Z }_{ \frac { \alpha }{ 2 } }SE\le \mu \le \bar { x } +{ Z }_{ \frac { \alpha }{ 2 } }SE\)
\(3.4-(1.96\times 0.087)\le \mu \le 3.4+(1.96\times 0.087)\)
\(3.229\le \mu \le 3.571\)
98% confidential limits for the population mean μ are :
\(\bar { x } -{ Z }_{ \frac { \alpha }{ 2 } }SE\le \mu \le \bar { x } +{ Z }_{ \frac { \alpha }{ 2 } }SE\)
\(3.4-(2.33\times 0.087)\le \mu \le 3.4+(2.33\times 0.087)\)
\(3.197\le \mu \le 3.603\)
Therefore,95% confidential limits is (3.229, 3.571) and 98% confidential limits is (3.197, 3.603).

Sample size n =100, Sample mean \(\bar x\) = 390 pages, Population mean \(\mu\) = 400 pages
Population SD \(\sigma\) = 20 pages
The sample is a large sample and so we apply Z -test
Null Hypothesis:
There is no significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₀ : \(\mu\) = 400
Alternative Hypothesis:
There is significant difference between the sample mean and the population mean of writing life of pen he manufactures, i.e., H₁:\(\mu\neq\) 400 (two tailed test)
The level of significance \(\alpha\) = 1% = 0.01
Applying the test statistic
\(Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1) ; \)
\(Z=\frac{390-400}{\frac{20}{\sqrt{100}}}=\frac{-10}{2}=-5, \therefore|Z|=5\)
Thus the calculated value |Z| = 5 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=2.58\)
Comparing the calculated and table values, we found Z > \({ Z }_{ \frac { \sigma }{ 2 } }\) i.e., 5 > 2.58
Inference: Since the calculated value is greater than table value i.e., \(Z>{ Z }_{ \frac { \sigma }{ 2 } }\) at 1% level of significance, the null hypothesis is rejected and Therefore we concluded that \(\mu \neq400\) and the manufacturer’s claim is rejected at 1% level of significance.

Sample size n = 50 Sample mean \(\bar x\) = 10 km sample standard deviation s = 3.5 km 
Population mean \(\mu\) = 9.5km
Since population SD is unknown we consider \(\sigma\) = s
The sample is a large sample and so we apply Z-test
Null Hypothesis :
There is no significant difference between the sample average and the company’s claim, i.e., \(H_0 : \mu=9.5\)
Alternative Hypothesis :
There is significant difference between the sample average and the company’s claim, i.e., H₁ : \(\mu\)\(\neq \) 9.5 (two tailed test)
The level of significance \(\alpha\) = 5% = 0.05
Applying the test statistic
\(Z=\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \sim N(0,1);\ \)
\(Z=\frac { 10-9.5 }{ \frac { 3.5 }{ \sqrt { 50 } } } \sim N(0,1)=\frac { 0.5 }{ 0.495 } =1.01\)
Thus the calculated value 1.01 and the significant value or table value \({ Z }_{ \frac { \sigma }{ 2 } }=1.96\)
Comparing the calculated and table value ,Here Z<\({ Z }_{ \frac { \sigma }{ 2 } }\)i.e., 1.01<1.96.
Inference : Since the calculated value is less than table value i.e.,  Z < \({ Z }_{ \frac { \sigma }{ 2 } }\) at 5% level of sinificance, the null hypothesis H₀ is accepted. Hence we conclude that the company’s claim that the new car petrol consumption is 9.5 km per litre is acceptable.

Given: n = 169, \(\bar x\) =1350 hours, \(\sigma\) =100 hours, since the level of significance is (100-90)% =10% thus \(\alpha\) is 0.1, hence the significant value at 10% is \({ Z }_{ \frac { \alpha }{ 2 } }\)=1.645
\(S.E.=\frac { \sigma }{ \sqrt { n } } =\frac { 100 }{ \sqrt { 169 } } =7.69\)
Hence 90% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \sigma }{ 2 } }<\mu <\bar { x } +{ Z }_{ \frac { \sigma }{ 2 } }SE\)
\(\begin{gathered} 1350-(1.645 \times 7.69) \leq \mu \leq 1350+(1.645 \times 7.69) \end{gathered}\)
\(1337.35 \leq \mu \leq 1362.65\)
Hence the mean life time of light bulbs is expected to lie between the interval (1337.35, 1362.65).

Given, sample size = 100, \(\bar x\) = 7.4, since σ is unknown but s = 1.2 is known.
In this problem, we consider \(\breve { \sigma } =s\quad { Z }_{ \frac { \alpha }{ 2 } }=1.96\)
\(S.E=\frac { \breve { \sigma } }{ \sqrt { n } } =\frac { s }{ \sqrt { n } } =\frac { 1.2 }{ \sqrt { 100 } } =0.12\)
Hence 95% confidence limits for the population mean are
\(\bar { x } -{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } <\mu <\bar { x } +{ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } } \)
\(7.4-(1.96\times 0.12)\le \mu \le 7.4+(1.96\times 0.12)\)
\(7.4-0.2352\le \mu \le 7.4+0.2352\)
\(7.165\le \mu \le 7.635\)
This implies that the probability that the true value of the population mean breaking strength of the cotton threads will fall in this interval (7.165,7.635) at 95%.

Here φ is the mean length of the components in the population.
The formula for the confidence interval is
\(\bar{x}-Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+Z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\)
\({ Here } \ \sigma=1.6, Z_{\alpha / 2}=1.96, \bar{x}=90 \text { and } \mathrm{n}=64\)
Then \(S.E=\frac { \sigma }{ \sqrt { n } } =\frac { 1.6 }{ \sqrt { 64 } } =0.2\)
Therefore, 90 - (1.96 x 0.2)\(\le φ \le\) 90 + (1.96 x 0.2)
\(\text { i.e. } \ (89.61 \leq \varphi \leq 90.39)\)
This implies that the probability that the true value of the population mean length of the components will fall in this interval (89.61,90.39) at 95% . Hence we concluded that 95% confidence interval ensures acceptance of the component by the consumer.

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